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Page No 93:
Question 1:
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
⇒ 3y = (2 − 2x)
⇒ 3y = 2(1 − x)
⇒ y = ...(i)
Putting x = 1, we get y = 0
Putting x = − 2, we get y = 2
Putting x = 4, we get y = −2
Thus, we have the following table for the equation 2x + 3y = 2.
x | 1 | −2 | 4 |
y | 0 | 2 | −2 |
Now, plot the points A(1, 0), B(− 2 , 2) and C(4, − 2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.
Graph of x − 2y = 8
x − 2y = 8
⇒ 2y = ( x − 8)
⇒ ...(ii)
Putting x = 2, we get y = −3
Putting x = 4, we get y = −2
Putting x = 0, we get y = −4
Thus, we have the following table for the equation x - 2y = 8.
x | 2 | 4 | 0 |
y | − 3 | − 2 | − 4 |
Thus, line PC is the graph of x − 2y = 8.
The two graph lines intersect at C(4, −2).
∴ x = 4 and y = −2 are the solutions of the given system of equations.
Page No 93:
Question 2:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
⇒ 3y = (2 − 2x)
⇒ 3y = 2(1 − x)
⇒ y = ...(i)
Putting x = 1, we get y = 0
Putting x = − 2, we get y = 2
Putting x = 4, we get y = −2
Thus, we have the following table for the equation 2x + 3y = 2.
x | 1 | −2 | 4 |
y | 0 | 2 | −2 |
Now, plot the points A(1, 0), B(− 2 , 2) and C(4, − 2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.
Graph of x − 2y = 8
x − 2y = 8
⇒ 2y = ( x − 8)
⇒ ...(ii)
Putting x = 2, we get y = −3
Putting x = 4, we get y = −2
Putting x = 0, we get y = −4
Thus, we have the following table for the equation x - 2y = 8.
x | 2 | 4 | 0 |
y | − 3 | − 2 | − 4 |
Thus, line PC is the graph of x − 2y = 8.
The two graph lines intersect at C(4, −2).
∴ x = 4 and y = −2 are the solutions of the given system of equations.
Answer:
Page No 93:
Question 3:
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
⇒ 3y = (8 − 2x)
∴ ...........(i)
Putting x = 1, we get y = 2.
Putting x = −5, we get y = 6.
Putting x = 7, we get y = −2.
Thus, we have the following table for the equation 2x + 3y = 8.
x | 1 | −5 | 7 |
y | 2 | 6 | −2 |
Now, plot the points A(1, 2), B(5, −6) and C(7, −2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.
Graph of x − 2y + 3 = 0
x − 2y + 3 = 0
⇒ 2y = (x + 3)
∴ ..........(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = −3, we get y = 0.
Thus, we have the following table for the equation x − 2y + 3 = 0.
x | 1 | 3 | −3 |
y | 2 | 3 | 0 |
Thus, PQ is the graph of x − 2y + 3 = 0.
The two graph lines intersect at A(1, 2).
∴ x = 1 and y = 2
Page No 93:
Question 4:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
⇒ 3y = (8 − 2x)
∴ ...........(i)
Putting x = 1, we get y = 2.
Putting x = −5, we get y = 6.
Putting x = 7, we get y = −2.
Thus, we have the following table for the equation 2x + 3y = 8.
x | 1 | −5 | 7 |
y | 2 | 6 | −2 |
Now, plot the points A(1, 2), B(5, −6) and C(7, −2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.
Graph of x − 2y + 3 = 0
x − 2y + 3 = 0
⇒ 2y = (x + 3)
∴ ..........(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = −3, we get y = 0.
Thus, we have the following table for the equation x − 2y + 3 = 0.
x | 1 | 3 | −3 |
y | 2 | 3 | 0 |
Thus, PQ is the graph of x − 2y + 3 = 0.
The two graph lines intersect at A(1, 2).
∴ x = 1 and y = 2
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x − 5y + 4 = 0
2x − 5y + 4 = 0
⇒ 5y = (2x + 4)
∴ ...........(i)
Putting x = −2, we get y = 0.
Putting x = 3, we get y = 2.
Putting x = 8, we get y = 4.
Thus, we have the following table for the equation 2x − 5y + 4 = 0.
x | −2 | 3 | 8 |
y | 0 | 2 | 4 |
Now, plot the points A(−2 , 0), B(3, 2) and C(8, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x − 5y + 4 = 0.
Graph of 2x + y − 8 = 0
2x + y − 8 = 0
⇒ y = (8 − 2x) ..........(ii)
Putting x = 1, we get y = 6.
Putting x = 3, we get y = 2.
Putting x = 2, we get y = 4.
Thus, we have the following table for the equation 2x + y − 8 = 0.
x | 1 | 3 | 2 |
y | 6 | 2 | 4 |
Thus, PB is the graph of 2x + y − 8 = 0.
The two graph lines intersect at B(3, 2).
∴ x = 3 and y = 2
Page No 93:
Question 5:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x − 5y + 4 = 0
2x − 5y + 4 = 0
⇒ 5y = (2x + 4)
∴ ...........(i)
Putting x = −2, we get y = 0.
Putting x = 3, we get y = 2.
Putting x = 8, we get y = 4.
Thus, we have the following table for the equation 2x − 5y + 4 = 0.
x | −2 | 3 | 8 |
y | 0 | 2 | 4 |
Now, plot the points A(−2 , 0), B(3, 2) and C(8, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x − 5y + 4 = 0.
Graph of 2x + y − 8 = 0
2x + y − 8 = 0
⇒ y = (8 − 2x) ..........(ii)
Putting x = 1, we get y = 6.
Putting x = 3, we get y = 2.
Putting x = 2, we get y = 4.
Thus, we have the following table for the equation 2x + y − 8 = 0.
x | 1 | 3 | 2 |
y | 6 | 2 | 4 |
Thus, PB is the graph of 2x + y − 8 = 0.
The two graph lines intersect at B(3, 2).
∴ x = 3 and y = 2
Answer:
The given equations are:
From (i), write y in terms of x
Now, substitute different values of x in (iii) to get different values of y
For x = 0,
For x = 2,
For x = 4,
Thus, the table for the first equation (3x + 2y = 12) is
x | 0 | 2 | 4 |
y | 6 | 3 | 0 |
Now, plot the points A(0, 6), B(2, 3) and C(4, 0) on a graph paper and join
A, B and C to get the graph of 3x + 2y = 12.
From (ii), write y in terms of x
Now, substitute different values of x in (iv) to get different values of y
For x = 0,
For x = 2,
For x = 4,
Thus, the table for the first equation (5x − 2y = 4) is
x | 0 | 2 | 4 |
y | −2 | 3 | 8 |
Now, plot the points D(0, −2), E(2, 3) and F(4, 8) on the same graph paper and join
D, E and F to get the graph of 5x − 2y = 4.
From the graph it is clear that, the given lines intersect at (2, 3).
Hence, the solution of the given system of equations is (2, 3).
Page No 93:
Question 6:
The given equations are:
From (i), write y in terms of x
Now, substitute different values of x in (iii) to get different values of y
For x = 0,
For x = 2,
For x = 4,
Thus, the table for the first equation (3x + 2y = 12) is
x | 0 | 2 | 4 |
y | 6 | 3 | 0 |
Now, plot the points A(0, 6), B(2, 3) and C(4, 0) on a graph paper and join
A, B and C to get the graph of 3x + 2y = 12.
From (ii), write y in terms of x
Now, substitute different values of x in (iv) to get different values of y
For x = 0,
For x = 2,
For x = 4,
Thus, the table for the first equation (5x − 2y = 4) is
x | 0 | 2 | 4 |
y | −2 | 3 | 8 |
Now, plot the points D(0, −2), E(2, 3) and F(4, 8) on the same graph paper and join
D, E and F to get the graph of 5x − 2y = 4.
From the graph it is clear that, the given lines intersect at (2, 3).
Hence, the solution of the given system of equations is (2, 3).
Answer:
n a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x - axis and y - axis respectively.
Graph of 3x + y + 1 = 0
3x + y + 1 = 0
⇒ y = (−3x − 1) ...........(i)
Putting x = 0, we get y = −1.
Putting x = −1, we get y = 2.
Putting x = 1, we get y = −4.
Thus, we have the following table for the equation 3x + y + 1 = 0.
x | 0 | −1 | 1 |
y | −1 | 2 | −4 |
Now, plot the points A(0, −1), B(−1, 2) and C(1, −4) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + y + 1 = 0.
Graph of 2x − 3y + 8 = 0
2x − 3y + 8 = 0
⇒ 3y = (2x + 8)
∴
Putting x = −1, we get y = 2.
Putting x = 2, we get y = 4.
Putting x = −4, we get y = 0.
Thus, we have the following table for the equation 2x − 3y + 8 = 0.
x | −1 | 2 | −4 |
y | 2 | 4 | 0 |
Thus, PQ is the graph of 2x − 3y + 8 = 0.
The two graph lines intersect at B(−1, 2).
∴ x = −1 and y = 2
Page No 93:
Question 7:
n a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x - axis and y - axis respectively.
Graph of 3x + y + 1 = 0
3x + y + 1 = 0
⇒ y = (−3x − 1) ...........(i)
Putting x = 0, we get y = −1.
Putting x = −1, we get y = 2.
Putting x = 1, we get y = −4.
Thus, we have the following table for the equation 3x + y + 1 = 0.
x | 0 | −1 | 1 |
y | −1 | 2 | −4 |
Now, plot the points A(0, −1), B(−1, 2) and C(1, −4) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + y + 1 = 0.
Graph of 2x − 3y + 8 = 0
2x − 3y + 8 = 0
⇒ 3y = (2x + 8)
∴
Putting x = −1, we get y = 2.
Putting x = 2, we get y = 4.
Putting x = −4, we get y = 0.
Thus, we have the following table for the equation 2x − 3y + 8 = 0.
x | −1 | 2 | −4 |
y | 2 | 4 | 0 |
Thus, PQ is the graph of 2x − 3y + 8 = 0.
The two graph lines intersect at B(−1, 2).
∴ x = −1 and y = 2
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation ( 2x + 3y + 5 = 0 ) is
x | −1 | 2 | 5 |
y | −1 | −3 | −5 |
Now, plot the points A(−1,−1), B(2,−3) and C(5,−5) on a graph paper and join
them to get the graph of 2x + 3y + 5 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( 3x − 2y − 12 = 0 ) is
x | 0 | 2 | 4 |
y | −6 | −3 | 0 |
Now, plot the points D(0,−6), E(2,−3) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 3x − 2y − 12 = 0.
From the graph it is clear that, the given lines intersect at (2,−3).
Hence, the solution of the given system of equation is (2,−3).
Page No 93:
Question 8:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation ( 2x + 3y + 5 = 0 ) is
x | −1 | 2 | 5 |
y | −1 | −3 | −5 |
Now, plot the points A(−1,−1), B(2,−3) and C(5,−5) on a graph paper and join
them to get the graph of 2x + 3y + 5 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( 3x − 2y − 12 = 0 ) is
x | 0 | 2 | 4 |
y | −6 | −3 | 0 |
Now, plot the points D(0,−6), E(2,−3) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 3x − 2y − 12 = 0.
From the graph it is clear that, the given lines intersect at (2,−3).
Hence, the solution of the given system of equation is (2,−3).
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation ( 2x − 3y + 13 = 0 ) is
x | −5 | 1 | 4 |
y | 1 | 5 | 7 |
Now, plot the points A(−5,1), B(1,5) and C(4,7) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 13 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( 3x − 2y + 12 = 0 ) is
x | −4 | −2 | 0 |
y | 0 | 3 | 6 |
Now, plot the points D(−4,0), E(−2,3) and F(0,6) on the same graph paper and join
D, E and F to get the graph of 3x − 2y + 12 = 0.
From the graph it is clear that, the given lines intersect at (−2,3).
Hence, the solution of the given system of equation is (−2,3).
Page No 93:
Question 9:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation ( 2x − 3y + 13 = 0 ) is
x | −5 | 1 | 4 |
y | 1 | 5 | 7 |
Now, plot the points A(−5,1), B(1,5) and C(4,7) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 13 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( 3x − 2y + 12 = 0 ) is
x | −4 | −2 | 0 |
y | 0 | 3 | 6 |
Now, plot the points D(−4,0), E(−2,3) and F(0,6) on the same graph paper and join
D, E and F to get the graph of 3x − 2y + 12 = 0.
From the graph it is clear that, the given lines intersect at (−2,3).
Hence, the solution of the given system of equation is (−2,3).
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 4
2x + 3y = 4
⇒ 3y = (4 − 2x)
∴ ...(i)
Putting x = −1, we get y = 2
Putting x = 2, we get y = 0
Putting x = 5, we get y = −2
Thus, we have the following table for the equation 2x + 3y = 4.
x | −1 | 2 | 5 |
y | 2 | 0 | −2 |
Now, plot the points A(−1 , 2), B(2 , 0) and C(5, −2) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x + 3y = 4.
Graph of 3x − y = −5
3x − y = −5
⇒ y = (3x + 5) ...(ii)
Putting x = −1, we get y = 2
Putting x = 0, we get y = 5
Putting x = −2, we get y = −1
Thus, we have the following table for the equation 3x − y = − 5 = 0.
x | −1 | 0 | −2 |
y | 2 | 5 | −1 |
Thus, PQ is the graph of 3x − y = −5.
The two graph lines intersect at A(−1 , 2).
∴ x = −1 and y = 2 are the solutions of the given system of equations.
Page No 93:
Question 10:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 4
2x + 3y = 4
⇒ 3y = (4 − 2x)
∴ ...(i)
Putting x = −1, we get y = 2
Putting x = 2, we get y = 0
Putting x = 5, we get y = −2
Thus, we have the following table for the equation 2x + 3y = 4.
x | −1 | 2 | 5 |
y | 2 | 0 | −2 |
Now, plot the points A(−1 , 2), B(2 , 0) and C(5, −2) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x + 3y = 4.
Graph of 3x − y = −5
3x − y = −5
⇒ y = (3x + 5) ...(ii)
Putting x = −1, we get y = 2
Putting x = 0, we get y = 5
Putting x = −2, we get y = −1
Thus, we have the following table for the equation 3x − y = − 5 = 0.
x | −1 | 0 | −2 |
y | 2 | 5 | −1 |
Thus, PQ is the graph of 3x − y = −5.
The two graph lines intersect at A(−1 , 2).
∴ x = −1 and y = 2 are the solutions of the given system of equations.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of x + 2y + 2 = 0
x + 2y + 2 = 0
⇒ 2y = (−2 − x)
∴ ...............(i)
Putting x = −2, we get y = 0.
Putting x = 0, we get y = −1.
Putting x = 2, we get y = −2.
Thus, we have the following table for the equation x + 2y + 2 = 0.
x | −2 | 0 | 2 |
y | 0 | −1 | −2 |
Now, plot the points A(−2, 0) , B(0 , −1) and C(2, −2) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of x + 2y + 2 = 0.
Graph of 3x + 2y − 2 = 0
3x + 2y − 2 = 0
⇒ 2y = (2 − 3x)
∴ ...............(ii)
Putting x = 0, we get y = 1.
Putting x = 2, we get y = −2.
Putting x = 4, we get y = −5.
Thus, we have the following table for the equation 3x + 2y − 2 = 0.
x | 0 | 2 | 4 |
y | 1 | −2 | −5 |
Thus, PQ is the graph of 3x + 2y − 2 = 0.
The two graph lines intersect at A(2, −2).
∴ x = 2 and y = −2
Page No 93:
Question 11:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of x + 2y + 2 = 0
x + 2y + 2 = 0
⇒ 2y = (−2 − x)
∴ ...............(i)
Putting x = −2, we get y = 0.
Putting x = 0, we get y = −1.
Putting x = 2, we get y = −2.
Thus, we have the following table for the equation x + 2y + 2 = 0.
x | −2 | 0 | 2 |
y | 0 | −1 | −2 |
Now, plot the points A(−2, 0) , B(0 , −1) and C(2, −2) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of x + 2y + 2 = 0.
Graph of 3x + 2y − 2 = 0
3x + 2y − 2 = 0
⇒ 2y = (2 − 3x)
∴ ...............(ii)
Putting x = 0, we get y = 1.
Putting x = 2, we get y = −2.
Putting x = 4, we get y = −5.
Thus, we have the following table for the equation 3x + 2y − 2 = 0.
x | 0 | 2 | 4 |
y | 1 | −2 | −5 |
Thus, PQ is the graph of 3x + 2y − 2 = 0.
The two graph lines intersect at A(2, −2).
∴ x = 2 and y = −2
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − y + 3 = 0) is
x | −3 | −1 | 1 |
y | 0 | 2 | 4 |
Now, plot the points A(−3,0), B(−1,2) and C(1,4) on a graph paper and join
A, B and C to get the graph of x − y + 3 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( 2x + 3y − 4 = 0 ) is
x | −4 | −1 | 2 |
y | 4 | 2 | 0 |
Now, plot the points D(−4,4), E(−1,2) and F(2,0) on the same graph paper and join
D, E and F to get the graph of 2x − 3y − 4 = 0.
From the graph it is clear that, the given lines intersect at (−1,2).
So, the solution of the given system of equation is (−1,2).
The vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2) and (2,0).
Now, draw a perpendicular from the intersection point E on the x-axis. So,
Hence, the vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2)
and (2,0) and its area is 5 sq. units.
Page No 93:
Question 12:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − y + 3 = 0) is
x | −3 | −1 | 1 |
y | 0 | 2 | 4 |
Now, plot the points A(−3,0), B(−1,2) and C(1,4) on a graph paper and join
A, B and C to get the graph of x − y + 3 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( 2x + 3y − 4 = 0 ) is
x | −4 | −1 | 2 |
y | 4 | 2 | 0 |
Now, plot the points D(−4,4), E(−1,2) and F(2,0) on the same graph paper and join
D, E and F to get the graph of 2x − 3y − 4 = 0.
From the graph it is clear that, the given lines intersect at (−1,2).
So, the solution of the given system of equation is (−1,2).
The vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2) and (2,0).
Now, draw a perpendicular from the intersection point E on the x-axis. So,
Hence, the vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2)
and (2,0) and its area is 5 sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 3y + 4 = 0) is
x | −2 | 1 | 4 |
y | 0 | 2 | 4 |
Now, plot the points A(−2,0), B(1,2) and C(4,4) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 4 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( x + 2y − 5 = 0 ) is
x | −3 | 1 | 5 |
y | 4 | 2 | 0 |
Now, plot the points D(−3,4), B(1,2) and F(5,0) on the same graph paper and join
D, E and F to get the graph of x + 2y − 5 = 0.
From the graph it is clear that, the given lines intersect at (1,2).
So, the solution of the given system of equations is (1,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (1,2) and (5,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,
Hence, the vertices of the triangle formed by the given lines and the x-axis are (−2,0), (1,2) and (5,0)
and the area of the triangle is 7 sq. units.
Page No 93:
Question 13:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 3y + 4 = 0) is
x | −2 | 1 | 4 |
y | 0 | 2 | 4 |
Now, plot the points A(−2,0), B(1,2) and C(4,4) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 4 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation ( x + 2y − 5 = 0 ) is
x | −3 | 1 | 5 |
y | 4 | 2 | 0 |
Now, plot the points D(−3,4), B(1,2) and F(5,0) on the same graph paper and join
D, E and F to get the graph of x + 2y − 5 = 0.
From the graph it is clear that, the given lines intersect at (1,2).
So, the solution of the given system of equations is (1,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (1,2) and (5,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,
Hence, the vertices of the triangle formed by the given lines and the x-axis are (−2,0), (1,2) and (5,0)
and the area of the triangle is 7 sq. units.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 4x − 3y + 4 = 0
4x − 3y + 4 = 0
⇒ 3y = (4x + 4)
∴ ............(i)
Putting x = −1, we get y = 0.
Putting x = 2, we get y = 4.
Putting x = 5, we get y = 8.
Thus, we have the following table for the equation 4x − 3y + 4 = 0
x | −1 | 2 | 5 |
y | 0 | 4 | 8 |
Now, plot the points A(−1, 0), B( 2, 4) and C(5, 8) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 4x − 3y + 4 = 0.
Graph of 4x + 3y − 20 = 0
4x + 3y − 20 = 0
⇒ 3y = (−4x + 20)
∴ ............(ii)
Putting x = 2, we get y = 4.
Putting x = −1, we get y = 8.
Putting x = 5, we get y = 0.
Thus, we have the following table for the equation 4x + 3y − 20 = 0.
x | 2 | −1 | 5 |
y | 4 | 8 | 0 |
Then, line PQ is the graph of the equation 4x + 3y − 20 = 0.
The two graph lines intersect at B(2, 4).
∴ The solution of the given system of equations is x = 2 and y = 4.
Clearly, the vertices of ΔABQ formed by these two lines and x-axis are Q(5, 0), B(2, 4) and A(−1, 0).
Now, consider ΔABQ.
Here, height = 4 units and base (AQ) = 6 units
∴ Area of ΔABQ = sq. units
= sq. units.
Page No 93:
Question 14:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 4x − 3y + 4 = 0
4x − 3y + 4 = 0
⇒ 3y = (4x + 4)
∴ ............(i)
Putting x = −1, we get y = 0.
Putting x = 2, we get y = 4.
Putting x = 5, we get y = 8.
Thus, we have the following table for the equation 4x − 3y + 4 = 0
x | −1 | 2 | 5 |
y | 0 | 4 | 8 |
Now, plot the points A(−1, 0), B( 2, 4) and C(5, 8) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 4x − 3y + 4 = 0.
Graph of 4x + 3y − 20 = 0
4x + 3y − 20 = 0
⇒ 3y = (−4x + 20)
∴ ............(ii)
Putting x = 2, we get y = 4.
Putting x = −1, we get y = 8.
Putting x = 5, we get y = 0.
Thus, we have the following table for the equation 4x + 3y − 20 = 0.
x | 2 | −1 | 5 |
y | 4 | 8 | 0 |
Then, line PQ is the graph of the equation 4x + 3y − 20 = 0.
The two graph lines intersect at B(2, 4).
∴ The solution of the given system of equations is x = 2 and y = 4.
Clearly, the vertices of ΔABQ formed by these two lines and x-axis are Q(5, 0), B(2, 4) and A(−1, 0).
Now, consider ΔABQ.
Here, height = 4 units and base (AQ) = 6 units
∴ Area of ΔABQ = sq. units
= sq. units.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of x − y + 1 = 0
x − y + 1 = 0
⇒ y = (x + 1) .........(i)
Putting x = −1, we get y = 0.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation x − y + 1 = 0.
x | −1 | 1 | 2 |
y | 0 | 2 | 3 |
Now, plot the points A(-1, 0), B( 1, 2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of x − y + 1 = 0.
Graph of 3x + 2y − 12 = 0
3x + 2y − 12 = 0
⇒ 2y = (−3x + 12)
∴ ............(ii)
Putting x = 0, we get y = 6.
Putting x = 2, we get y = 3.
Putting x = 4, we get y = 0.
Thus, we have the following table for the equation 3x + 2y − 12 = 0.
x | 0 | 2 | 4 |
y | 6 | 3 | 0 |
Then, PQ is the graph of the equation 3x + 2y − 12 = 0.
The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔACQ formed by these two lines and the x-axis are Q(4, 0), C(2, 3) and A(−1, 0).
Now, consider ΔACQ.
Here, height = 3 units and base (AQ) = 5 units
Area of ΔACQ = sq. units
= sq. units.
Page No 93:
Question 15:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of x − y + 1 = 0
x − y + 1 = 0
⇒ y = (x + 1) .........(i)
Putting x = −1, we get y = 0.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation x − y + 1 = 0.
x | −1 | 1 | 2 |
y | 0 | 2 | 3 |
Now, plot the points A(-1, 0), B( 1, 2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of x − y + 1 = 0.
Graph of 3x + 2y − 12 = 0
3x + 2y − 12 = 0
⇒ 2y = (−3x + 12)
∴ ............(ii)
Putting x = 0, we get y = 6.
Putting x = 2, we get y = 3.
Putting x = 4, we get y = 0.
Thus, we have the following table for the equation 3x + 2y − 12 = 0.
x | 0 | 2 | 4 |
y | 6 | 3 | 0 |
Then, PQ is the graph of the equation 3x + 2y − 12 = 0.
The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔACQ formed by these two lines and the x-axis are Q(4, 0), C(2, 3) and A(−1, 0).
Now, consider ΔACQ.
Here, height = 3 units and base (AQ) = 5 units
Area of ΔACQ = sq. units
= sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − 2y + 2 = 0) is
x | −2 | 2 | 4 |
y | 0 | 2 | 3 |
Now, plot the points A(−2,0), B(2,2) and C(4,3) on a graph paper and join
A, B and C to get the graph of x − 2y + 2 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + y − 6 = 0 ) is
x | 1 | 3 | 4 |
y | 4 | 0 | −2 |
Now, plot the points D(1,4), E(3,0) and F(4,−2) on the same graph paper and join
D, E and F to get the graph of 2x + y − 6 = 0.
From the graph it is clear that, the given lines intersect at (2,2).
So, the solution of the given system of equations is (2,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,
Hence, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0) and the area of the triangle is 5 sq. units.
Page No 93:
Question 16:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − 2y + 2 = 0) is
x | −2 | 2 | 4 |
y | 0 | 2 | 3 |
Now, plot the points A(−2,0), B(2,2) and C(4,3) on a graph paper and join
A, B and C to get the graph of x − 2y + 2 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + y − 6 = 0 ) is
x | 1 | 3 | 4 |
y | 4 | 0 | −2 |
Now, plot the points D(1,4), E(3,0) and F(4,−2) on the same graph paper and join
D, E and F to get the graph of 2x + y − 6 = 0.
From the graph it is clear that, the given lines intersect at (2,2).
So, the solution of the given system of equations is (2,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,
Hence, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0) and the area of the triangle is 5 sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 3y + 6 = 0) is
x | −3 | 0 | 3 |
y | 0 | 2 | 4 |
Now, plot the points A(−3,0), B(0,2) and C(3,4) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 6 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + 3y − 18 = 0 ) is
x | 0 | 3 | 9 |
y | 6 | 4 | 0 |
Now, plot the points D(0,6), E(3,4) and F(9,0) on the same graph paper and join
D, E and F to get the graph of 2x + 3y − 18 = 0.
From the graph it is clear that, the given lines intersect at (3,4).
So, the solution of the given system of equations is (3,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4).
Now, draw a perpendicular from the intersection point E (or C) on the y-axis. So,
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4) and its area is 6 sq. units.
Page No 93:
Question 17:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 3y + 6 = 0) is
x | −3 | 0 | 3 |
y | 0 | 2 | 4 |
Now, plot the points A(−3,0), B(0,2) and C(3,4) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 6 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + 3y − 18 = 0 ) is
x | 0 | 3 | 9 |
y | 6 | 4 | 0 |
Now, plot the points D(0,6), E(3,4) and F(9,0) on the same graph paper and join
D, E and F to get the graph of 2x + 3y − 18 = 0.
From the graph it is clear that, the given lines intersect at (3,4).
So, the solution of the given system of equations is (3,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4).
Now, draw a perpendicular from the intersection point E (or C) on the y-axis. So,
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4) and its area is 6 sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (4x − y − 4 = 0) is
x | 0 | 1 | 2 |
y | −4 | 0 | 4 |
Now, plot the points A(0,−4), B(1,0) and C(2,4) on a graph paper and join
A, B and C to get the graph of 4x − y − 4 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x + 2y − 14 = 0 ) is
x | 0 | 4 | |
y | 7 | 1 | 0 |
Now, plot the points D(0,7), E(4,1) and on the same graph paper and join
D, E and F to get the graph of 3x + 2y − 14 = 0.
From the graph it is clear that, the given lines intersect at (,4).
So, the solution of the given system of equations is (2,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4).
Now, draw a perpendicular from the intersection point C on the y-axis. So,
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4) and its area is 11 sq. units.
Page No 93:
Question 18:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (4x − y − 4 = 0) is
x | 0 | 1 | 2 |
y | −4 | 0 | 4 |
Now, plot the points A(0,−4), B(1,0) and C(2,4) on a graph paper and join
A, B and C to get the graph of 4x − y − 4 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x + 2y − 14 = 0 ) is
x | 0 | 4 | |
y | 7 | 1 | 0 |
Now, plot the points D(0,7), E(4,1) and on the same graph paper and join
D, E and F to get the graph of 3x + 2y − 14 = 0.
From the graph it is clear that, the given lines intersect at (,4).
So, the solution of the given system of equations is (2,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4).
Now, draw a perpendicular from the intersection point C on the y-axis. So,
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4) and its area is 11 sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − y − 5 = 0) is
x | 0 | 2 | 5 |
y | −5 | −3 | 0 |
Now, plot the points A(0,−5), B(2,−3) and C(5,0) on a graph paper and join
A, B and C to get the graph of x − y − 5 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x + 5y − 15 = 0 ) is
x | −5 | 0 | 5 |
y | 6 | 3 | 0 |
Now, plot the points D(−5,6), E(0,3) and F(5,0) on the same graph paper and join
D, E and F to get the graph of 3x + 5y − 15 = 0.
From the graph it is clear that, the given lines intersect at (5,0).
So, the solution of the given system of equations is (5,0).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0).
Now,
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0) and its area is 20 sq. units.
Page No 93:
Question 19:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − y − 5 = 0) is
x | 0 | 2 | 5 |
y | −5 | −3 | 0 |
Now, plot the points A(0,−5), B(2,−3) and C(5,0) on a graph paper and join
A, B and C to get the graph of x − y − 5 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x + 5y − 15 = 0 ) is
x | −5 | 0 | 5 |
y | 6 | 3 | 0 |
Now, plot the points D(−5,6), E(0,3) and F(5,0) on the same graph paper and join
D, E and F to get the graph of 3x + 5y − 15 = 0.
From the graph it is clear that, the given lines intersect at (5,0).
So, the solution of the given system of equations is (5,0).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0).
Now,
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0) and its area is 20 sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 5y + 4 = 0) is
x | −2 | 0 | 3 |
y | 0 | 2 |
Now, plot the points A(−2,0), and C(3,2) on a graph paper and join
A, B and C to get the graph of 2x − 5y + 4 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + y − 8 = 0 ) is
x | 0 | 2 | 4 |
y | 8 | 4 | 0 |
Now, plot the points D(0,8), E(2,4) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 2x + y − 8 = 0.
From the graph it is clear that, the given lines intersect at (3,2).
So, the solution of the given system of equations is (3,2).
The vertices of the triangle formed by the system of equations and y-axis are (0,8), and (3,2).
Draw a perpendicular from point C to the y-axis. So,
Hence, the veritices of the triangle are (0,8), and (3,2) and its area is sq. units.
Page No 93:
Question 20:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 5y + 4 = 0) is
x | −2 | 0 | 3 |
y | 0 | 2 |
Now, plot the points A(−2,0), and C(3,2) on a graph paper and join
A, B and C to get the graph of 2x − 5y + 4 = 0.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + y − 8 = 0 ) is
x | 0 | 2 | 4 |
y | 8 | 4 | 0 |
Now, plot the points D(0,8), E(2,4) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 2x + y − 8 = 0.
From the graph it is clear that, the given lines intersect at (3,2).
So, the solution of the given system of equations is (3,2).
The vertices of the triangle formed by the system of equations and y-axis are (0,8), and (3,2).
Draw a perpendicular from point C to the y-axis. So,
Hence, the veritices of the triangle are (0,8), and (3,2) and its area is sq. units.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 5x − y = 7
5x − y = 7
⇒ y = (5x − 7) .........(i)
Putting x = 0, we get y = −7.
Putting x = 1, we get y = −2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation 5x − y = 7.
x | 0 | 1 | 2 |
y | −7 | −2 | 3 |
Now, plot the points A(0, −7) , B( 1, −2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 5x − y = 7.
Graph of x − y + 1 = 0
x − y + 1 = 0
⇒ y = (x + 1) ...........(ii)
Putting x = 0, we get y = 1.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation x − y + 1 = 0.
x | 0 | 1 | 2 |
y | 1 | 2 | 3 |
Then, PC is the graph of the equation x − y + 1 = 0.
The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔAPC formed by these two lines and the y-axis are P(0, 1), C(2, 3) and A(0, −7).
Now, consider ΔAPC.
Here, height = 2 units and base (AP) = 8 units
∴ Area of ΔAPC = sq. units
= sq. units.
Page No 93:
Question 21:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 5x − y = 7
5x − y = 7
⇒ y = (5x − 7) .........(i)
Putting x = 0, we get y = −7.
Putting x = 1, we get y = −2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation 5x − y = 7.
x | 0 | 1 | 2 |
y | −7 | −2 | 3 |
Now, plot the points A(0, −7) , B( 1, −2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 5x − y = 7.
Graph of x − y + 1 = 0
x − y + 1 = 0
⇒ y = (x + 1) ...........(ii)
Putting x = 0, we get y = 1.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation x − y + 1 = 0.
x | 0 | 1 | 2 |
y | 1 | 2 | 3 |
Then, PC is the graph of the equation x − y + 1 = 0.
The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔAPC formed by these two lines and the y-axis are P(0, 1), C(2, 3) and A(0, −7).
Now, consider ΔAPC.
Here, height = 2 units and base (AP) = 8 units
∴ Area of ΔAPC = sq. units
= sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 3y = 12) is
x | 0 | 3 | 6 |
y | −4 | −2 | 0 |
Now, plot the points A(0,−4), B(3,−2) and C(6,0) on a graph paper and join
A, B and C to get the graph of 2x − 3y = 12.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (x + 3y = 6 ) is
x | 0 | 3 | 6 |
y | 2 | 1 | 0 |
Now, plot the points D(0,2), E(3,1) and F(6,0) on the same graph paper and join
D, E and F to get the graph of x + 3y = 6.
From the graph it is clear that, the given lines intersect at (6,0).
So, the solution of the given system of equations is (6,0).
The vertices of the triangle formed by the system of equations and y-axis are (0,2), (6,0) and (0,−4).
Now,
Hence, the veritices of the triangle are (0,2), (6,0) and (0,−4) and its area is 18 sq. units.
Page No 94:
Question 22:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x − 3y = 12) is
x | 0 | 3 | 6 |
y | −4 | −2 | 0 |
Now, plot the points A(0,−4), B(3,−2) and C(6,0) on a graph paper and join
A, B and C to get the graph of 2x − 3y = 12.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (x + 3y = 6 ) is
x | 0 | 3 | 6 |
y | 2 | 1 | 0 |
Now, plot the points D(0,2), E(3,1) and F(6,0) on the same graph paper and join
D, E and F to get the graph of x + 3y = 6.
From the graph it is clear that, the given lines intersect at (6,0).
So, the solution of the given system of equations is (6,0).
The vertices of the triangle formed by the system of equations and y-axis are (0,2), (6,0) and (0,−4).
Now,
Hence, the veritices of the triangle are (0,2), (6,0) and (0,−4) and its area is 18 sq. units.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x + 3y = 6) is
x | −3 | 3 | 6 |
y | 4 | 0 | −2 |
Now, plot the points A(−3,4), B(3,0) and C(6,−2) on a graph paper and join
A, B and C to get the graph of 2x + 3y = 6.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (4x + 6y = 12 ) is
x | −6 | 0 | 9 |
y | 6 | 2 | −4 |
Now, plot the points D(−6,6), E(0,2) and F(9,−4) on the same graph paper and join
D, E and F to get the graph of 4x + 6y = 12.
From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.
Page No 94:
Question 23:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x + 3y = 6) is
x | −3 | 3 | 6 |
y | 4 | 0 | −2 |
Now, plot the points A(−3,4), B(3,0) and C(6,−2) on a graph paper and join
A, B and C to get the graph of 2x + 3y = 6.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (4x + 6y = 12 ) is
x | −6 | 0 | 9 |
y | 6 | 2 | −4 |
Now, plot the points D(−6,6), E(0,2) and F(9,−4) on the same graph paper and join
D, E and F to get the graph of 4x + 6y = 12.
From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 3x − y = 5
3x − y = 5
⇒ y = (3x − 5) .........(i)
Putting x = 1, we get y = −2.
Putting x = 0, we get y = −5.
Putting x = 2, we get y = 1.
Thus, we have the following table for the equation 3x − y = 5.
x | 1 | 0 | 2 |
y | −2 | −5 | 1 |
Now, plot the points A(1, −2), B(0, −5) and C(2, 1) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x − y = 5.
Graph of 6x − 2y = 10
6x − 2y = 10
⇒ 2y = (6x − 10)
∴ ...........(ii)
Putting x = 0, we get y = −5.
Putting x = 1, we get y = −2.
Putting x = 2, we get y = 1.
Thus, we have the following table for the equation 6x − 2y = 10.
x | 0 | 1 | 2 |
y | −5 | −2 | 1 |
It is clear from the graph that these two lines coincide.
Hence, the given system of equations has infinitely many solutions.
Page No 94:
Question 24:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 3x − y = 5
3x − y = 5
⇒ y = (3x − 5) .........(i)
Putting x = 1, we get y = −2.
Putting x = 0, we get y = −5.
Putting x = 2, we get y = 1.
Thus, we have the following table for the equation 3x − y = 5.
x | 1 | 0 | 2 |
y | −2 | −5 | 1 |
Now, plot the points A(1, −2), B(0, −5) and C(2, 1) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x − y = 5.
Graph of 6x − 2y = 10
6x − 2y = 10
⇒ 2y = (6x − 10)
∴ ...........(ii)
Putting x = 0, we get y = −5.
Putting x = 1, we get y = −2.
Putting x = 2, we get y = 1.
Thus, we have the following table for the equation 6x − 2y = 10.
x | 0 | 1 | 2 |
y | −5 | −2 | 1 |
It is clear from the graph that these two lines coincide.
Hence, the given system of equations has infinitely many solutions.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + y = 6
2x + y = 6 ⇒ y = (6 − 2x) ...(i)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 2x + y = 6.
x | 3 | 1 | 2 |
y | 0 | 4 | 2 |
Now, plot the points A(3, 0), B(1, 4) and C(2, 2) on the graph paper.
Join AC and CB to get the graph line AB. Extend it on both ways.
Thus, AB is the graph of 2x + y = 6.
Graph of 6x + 3y = 18
6x + 3y = 18 ⇒ 3y = (18 − 6x)
∴ ...........(ii)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 6x + 3y = 18.
x | 3 | 1 | 2 |
y | 0 | 4 | 2 |
It is clear from the graph that these two lines coincide.
Hence, the given system of equations has an infinite number of solutions.
Page No 94:
Question 25:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + y = 6
2x + y = 6 ⇒ y = (6 − 2x) ...(i)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 2x + y = 6.
x | 3 | 1 | 2 |
y | 0 | 4 | 2 |
Now, plot the points A(3, 0), B(1, 4) and C(2, 2) on the graph paper.
Join AC and CB to get the graph line AB. Extend it on both ways.
Thus, AB is the graph of 2x + y = 6.
Graph of 6x + 3y = 18
6x + 3y = 18 ⇒ 3y = (18 − 6x)
∴ ...........(ii)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 6x + 3y = 18.
x | 3 | 1 | 2 |
y | 0 | 4 | 2 |
It is clear from the graph that these two lines coincide.
Hence, the given system of equations has an infinite number of solutions.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − 2y = 5) is
x | −5 | 1 | 3 |
y | −5 | −2 | −1 |
Now, plot the points A(−5,−5), B(1,−2) and C(3,−1) on a graph paper and join
A, B and C to get the graph of x − 2y = 5.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x − 6y = 15 ) is
x | −3 | −1 | 5 |
y | −4 | −3 | 0 |
Now, plot the points D(−3,−4), E(−1,−3) and F(5,0) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 15.
From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.
Page No 94:
Question 26:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − 2y = 5) is
x | −5 | 1 | 3 |
y | −5 | −2 | −1 |
Now, plot the points A(−5,−5), B(1,−2) and C(3,−1) on a graph paper and join
A, B and C to get the graph of x − 2y = 5.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x − 6y = 15 ) is
x | −3 | −1 | 5 |
y | −4 | −3 | 0 |
Now, plot the points D(−3,−4), E(−1,−3) and F(5,0) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 15.
From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − 2y = 6) is
x | −2 | 0 | 2 |
y | −4 | −3 | −2 |
Now, plot the points A(−2,−4), B(0,−3) and C(2,−2) on a graph paper and join
A, B and C to get the graph of x − 2y = 6.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x − 6y = 0 ) is
x | −4 | 0 | 4 |
y | −2 | 0 | 2 |
Now, plot the points D(−4,−2), O(0,0) and E(4,2) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 0.
From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.
Page No 94:
Question 27:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (x − 2y = 6) is
x | −2 | 0 | 2 |
y | −4 | −3 | −2 |
Now, plot the points A(−2,−4), B(0,−3) and C(2,−2) on a graph paper and join
A, B and C to get the graph of x − 2y = 6.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (3x − 6y = 0 ) is
x | −4 | 0 | 4 |
y | −2 | 0 | 2 |
Now, plot the points D(−4,−2), O(0,0) and E(4,2) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 0.
From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 4
2x + 3y = 4
⇒ 3y = (−2x + 4)
∴ .........(i)
Putting x = 2, we get y = 0.
Putting x = −1, we get y = 2.
Putting x = −4, we get y = 4.
Thus, we have the following table for the equation 2x + 3y = 4.
x | 2 | −1 | −4 |
y | 0 | 2 | 4 |
Now, plot the points A(2, 0), B(−1, 2) and C(−4, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x + 3y = 4.
Graph of 4x + 6y = 12
4x + 6y = 12
⇒ 6y = (−4x + 12)
∴ ...........(ii)
Putting x = 3, we get y = 0.
Putting x = 0, we get y = 2.
Putting x = 6, we get y = −2.
Thus, we have the following table for the equation 4x + 6y = 12.
x | 3 | 0 | 6 |
y | 0 | 2 | −2 |
Now, on the same graph, plot the points P(3, 0), Q(0, 2) and R (6, −2).
Join PQ and PR to get the graph line QR. Extend it on both ways.
Then, QR is the graph of the equation 4x + 6y = 12.
It is clear from the graph that these two lines are parallel and do not intersect when produced.
Hence, the given system of equations is inconsistent.
Page No 94:
Question 28:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 4
2x + 3y = 4
⇒ 3y = (−2x + 4)
∴ .........(i)
Putting x = 2, we get y = 0.
Putting x = −1, we get y = 2.
Putting x = −4, we get y = 4.
Thus, we have the following table for the equation 2x + 3y = 4.
x | 2 | −1 | −4 |
y | 0 | 2 | 4 |
Now, plot the points A(2, 0), B(−1, 2) and C(−4, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x + 3y = 4.
Graph of 4x + 6y = 12
4x + 6y = 12
⇒ 6y = (−4x + 12)
∴ ...........(ii)
Putting x = 3, we get y = 0.
Putting x = 0, we get y = 2.
Putting x = 6, we get y = −2.
Thus, we have the following table for the equation 4x + 6y = 12.
x | 3 | 0 | 6 |
y | 0 | 2 | −2 |
Now, on the same graph, plot the points P(3, 0), Q(0, 2) and R (6, −2).
Join PQ and PR to get the graph line QR. Extend it on both ways.
Then, QR is the graph of the equation 4x + 6y = 12.
It is clear from the graph that these two lines are parallel and do not intersect when produced.
Hence, the given system of equations is inconsistent.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x + y = 6) is
x | 0 | 2 | 4 |
y | 6 | 2 | −2 |
Now, plot the points A(0,6), B(2,2) and C(4,−2) on a graph paper and join
A, B and C to get the graph of 2x + y = 6.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (6x + 3y = 20 ) is
x | 0 | 5 | |
y | 0 |
Now, plot the points on the same graph paper and join
D, E and F to get the graph of 6x + 3y = 20.
From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.
Page No 94:
Question 29:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x + y = 6) is
x | 0 | 2 | 4 |
y | 6 | 2 | −2 |
Now, plot the points A(0,6), B(2,2) and C(4,−2) on a graph paper and join
A, B and C to get the graph of 2x + y = 6.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (6x + 3y = 20 ) is
x | 0 | 5 | |
y | 0 |
Now, plot the points on the same graph paper and join
D, E and F to get the graph of 6x + 3y = 20.
From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.
Answer:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x + y = 2) is
x | 0 | 1 | 2 |
y | 2 | 0 | −2 |
Now, plot the points A(0,2), B(1,0) and C(2,−2) on a graph paper and join
A, B and C to get the graph of 2x + y = 2.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + y = 6 ) is
x | 0 | 1 | 3 |
y | 6 | 4 | 0 |
Now, plot the points D(0,6), E(1,4) and F(3,0)on the same graph paper and join
D, E and F to get the graph of 2x + y = 6.
From the graph it is clear that, the given lines do not intersect at all when produced. So, these lines are
parallel to each other and therefore, the quadrilateral DABF is a trapezium. The vertices of the
required trapezium are D(0,6), A(0,2), B(1,0) and F(3,0).
Now,
Hence, the area of the rquired trapezium is 8 sq. units.
Page No 109:
Question 1:
From the first equation, write y in terms of x
Substitute different values of x in (i) to get different values of y
Thus, the table for the first equation (2x + y = 2) is
x | 0 | 1 | 2 |
y | 2 | 0 | −2 |
Now, plot the points A(0,2), B(1,0) and C(2,−2) on a graph paper and join
A, B and C to get the graph of 2x + y = 2.
From the second equation, write y in terms of x
Now, substitute different values of x in (ii) to get different values of y
So, the table for the second equation (2x + y = 6 ) is
x | 0 | 1 | 3 |
y | 6 | 4 | 0 |
Now, plot the points D(0,6), E(1,4) and F(3,0)on the same graph paper and join
D, E and F to get the graph of 2x + y = 6.
From the graph it is clear that, the given lines do not intersect at all when produced. So, these lines are
parallel to each other and therefore, the quadrilateral DABF is a trapezium. The vertices of the
required trapezium are D(0,6), A(0,2), B(1,0) and F(3,0).
Now,
Hence, the area of the rquired trapezium is 8 sq. units.
Answer:
The given system of equation is:
x + y = 3 .........(i)
4x − 3y = 26 ........(ii)
On multiplying (i) by 3, we get:
3x + 3y = 9 .......(iii)
On adding (ii) and (iii), we get:
7x = 35
⇒ x = 5
On substituting the value of x = 5 in (i), we get:
5 + y = 3
⇒ y = (3 − 5) = −2
Hence, the solution is x = 5 and y = −2
Page No 109:
Question 2:
The given system of equation is:
x + y = 3 .........(i)
4x − 3y = 26 ........(ii)
On multiplying (i) by 3, we get:
3x + 3y = 9 .......(iii)
On adding (ii) and (iii), we get:
7x = 35
⇒ x = 5
On substituting the value of x = 5 in (i), we get:
5 + y = 3
⇒ y = (3 − 5) = −2
Hence, the solution is x = 5 and y = −2
Answer:
The given system of equations is
From (i), write y in terms of x to get
Substituting y = x − 3 in (ii), we get
Now, substituting x = 9 in (i), we have
Hence, x = 9 and y = 6.
Page No 110:
Question 3:
The given system of equations is
From (i), write y in terms of x to get
Substituting y = x − 3 in (ii), we get
Now, substituting x = 9 in (i), we have
Hence, x = 9 and y = 6.
Answer:
The given system of equation is:
2x + 3y = 0 .........(i)
3x + 4y = 5 ........(ii)
On multiplying (i) by 4 and (ii) by 3, we get:
8x + 12y = 0 .......(iii)
9x + 12y = 15 ......(iv)
On subtracting (iii) from (iv) we get:
x = 15
On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
⇒ 3y = −30
⇒ y = −10
Hence, the solution is x = 15 and y = −10.
Page No 110:
Question 4:
The given system of equation is:
2x + 3y = 0 .........(i)
3x + 4y = 5 ........(ii)
On multiplying (i) by 4 and (ii) by 3, we get:
8x + 12y = 0 .......(iii)
9x + 12y = 15 ......(iv)
On subtracting (iii) from (iv) we get:
x = 15
On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
⇒ 3y = −30
⇒ y = −10
Hence, the solution is x = 15 and y = −10.
Answer:
The given system of equation is:
2x − 3y = 13 .........(i)
7x − 2y = 20 ........(ii)
On multiplying (i) by 2 and (ii) by 3, we get:
4x − 6y = 26 .......(iii)
21x − 6y = 60 ......(iv)
On subtracting (iii) from (iv) we get:
17x = (60 − 26) = 34
⇒ x = 2
On substituting the value of x = 2 in (i), we get:
4 − 3y = 13
⇒ 3y = (4 − 13) = −9
⇒ y = −3
Hence, the solution is x = 2 and y = −3
Page No 110:
Question 5:
The given system of equation is:
2x − 3y = 13 .........(i)
7x − 2y = 20 ........(ii)
On multiplying (i) by 2 and (ii) by 3, we get:
4x − 6y = 26 .......(iii)
21x − 6y = 60 ......(iv)
On subtracting (iii) from (iv) we get:
17x = (60 − 26) = 34
⇒ x = 2
On substituting the value of x = 2 in (i), we get:
4 − 3y = 13
⇒ 3y = (4 − 13) = −9
⇒ y = −3
Hence, the solution is x = 2 and y = −3
Answer:
The given system of equation is:
3x − 5y − 19 = 0 .........(i)
−7x + 3y + 1 = 0 ........(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 57 or .........(iii)
−35x + 15y = −5 ........(iv)
On adding (iii) from (iv) we get:
−26x = (57 − 5)) = 52
⇒ x = −2
On substituting the value of x = −2 in (i), we get:
−6 − 5y − 19 = 0
⇒ 5y = (− 6 − 19) = −25
⇒ y = −5
Hence, the solution is x = −2 and y = −5.
Page No 110:
Question 6:
The given system of equation is:
3x − 5y − 19 = 0 .........(i)
−7x + 3y + 1 = 0 ........(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 57 or .........(iii)
−35x + 15y = −5 ........(iv)
On adding (iii) from (iv) we get:
−26x = (57 − 5)) = 52
⇒ x = −2
On substituting the value of x = −2 in (i), we get:
−6 − 5y − 19 = 0
⇒ 5y = (− 6 − 19) = −25
⇒ y = −5
Hence, the solution is x = −2 and y = −5.
Answer:
Page No 110:
Question 7:
Answer:
The given system of equations can be written as
Multiplying (i) by 7 and (ii) by 2, we get
Now, substituting x = 14 in (1), we have
Hence, x = 14 and y = 9.
Page No 110:
Question 8:
The given system of equations can be written as
Multiplying (i) by 7 and (ii) by 2, we get
Now, substituting x = 14 in (1), we have
Hence, x = 14 and y = 9.
Answer:
The given equations are:
⇒ 4x + 3y = 132 ........(i)
and
⇒ 5x − 2y = −42..........(ii)
On multiplying (i) by 2 and (ii) by 3, we get:
8x + 6y = 264 ........(iii)
15x − 6y = −126 ...........(iv)
On adding (iii) and (iv), we get:
23x = 138
⇒ x = 6
On substituting x = 6 in (i), we get:
24 + 3y = 132
⇒ 3y = (132 − 24) = 108
⇒ y = 36
Hence, the solution is x = 6 and y = 36.
Page No 110:
Question 9:
The given equations are:
⇒ 4x + 3y = 132 ........(i)
and
⇒ 5x − 2y = −42..........(ii)
On multiplying (i) by 2 and (ii) by 3, we get:
8x + 6y = 264 ........(iii)
15x − 6y = −126 ...........(iv)
On adding (iii) and (iv), we get:
23x = 138
⇒ x = 6
On substituting x = 6 in (i), we get:
24 + 3y = 132
⇒ 3y = (132 − 24) = 108
⇒ y = 36
Hence, the solution is x = 6 and y = 36.
Answer:
The given system of equation is:
4x − 3y = 8 .........(i)
........(ii)
On multiplying (ii) by 3, we get:
18x − 3y = 29.........(iii)
On subtracting (iii) from (i) we get:
−14x = −21
⇒ x =
On substituting the value of x = in (i), we get:
Hence, the solution is x = and y = .
Page No 110:
Question 10:
The given system of equation is:
4x − 3y = 8 .........(i)
........(ii)
On multiplying (ii) by 3, we get:
18x − 3y = 29.........(iii)
On subtracting (iii) from (i) we get:
−14x = −21
⇒ x =
On substituting the value of x = in (i), we get:
Hence, the solution is x = and y = .
Answer:
The given equations are:
........(i)
5x = 2y + 7 ............(ii)
On multiplying (i) by 2 and (ii) by , we get:
.......(iii)
.......(iv)
On subtracting (iii) from (iv), we get:
On substituting x = 3 in (i), we get:
Hence, the solution is x = 3 and y = 4.
Page No 110:
Question 11:
The given equations are:
........(i)
5x = 2y + 7 ............(ii)
On multiplying (i) by 2 and (ii) by , we get:
.......(iii)
.......(iv)
On subtracting (iii) from (iv), we get:
On substituting x = 3 in (i), we get:
Hence, the solution is x = 3 and y = 4.
Answer:
The given equations are:
........(i)
..........(ii)
On multiplying (i) by 2 and (ii) by 5, we get:
........(iii)
...........(iv)
On adding (iii) and (iv), we get:
On substituting x = in (i), we get:
Hence, the solution is x = and y = .
Page No 110:
Question 12:
The given equations are:
........(i)
..........(ii)
On multiplying (i) by 2 and (ii) by 5, we get:
........(iii)
...........(iv)
On adding (iii) and (iv), we get:
On substituting x = in (i), we get:
Hence, the solution is x = and y = .
Answer:
The given equations are:
⇒ 4x + 3y = 7 .......(i)
and 2x + 3y + 1 = 0
⇒ 2x + 3y = −1 .............(ii)
On subtracting (ii) from (i), we get:
2x = 8
⇒ x = 4
On substituting x = 4 in (i), we get:
16 + 3y = 7
⇒ 3y = (7 − 16) = −9
⇒ y = −3
Hence, the solution is x = 4 and y = −3.
Page No 110:
Question 13:
The given equations are:
⇒ 4x + 3y = 7 .......(i)
and 2x + 3y + 1 = 0
⇒ 2x + 3y = −1 .............(ii)
On subtracting (ii) from (i), we get:
2x = 8
⇒ x = 4
On substituting x = 4 in (i), we get:
16 + 3y = 7
⇒ 3y = (7 − 16) = −9
⇒ y = −3
Hence, the solution is x = 4 and y = −3.
Answer:
The given system of equations is
Multiplying (i) by 0.2 and (ii) by 0.3 and adding them, we get
Now, substituting x = 2 in (i), we have
Hence, x = 2 and y = 3.
Page No 110:
Question 14:
The given system of equations is
Multiplying (i) by 0.2 and (ii) by 0.3 and adding them, we get
Now, substituting x = 2 in (i), we have
Hence, x = 2 and y = 3.
Answer:
The given system of equations is
Multiplying (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get
Now, substituting y = 0.7 in (i), we have
Hence, x = 0.5 and y = 0.7.
Page No 110:
Question 15:
The given system of equations is
Multiplying (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get
Now, substituting y = 0.7 in (i), we have
Hence, x = 0.5 and y = 0.7.
Answer:
The given equations are:
7(y + 3) − 2(x + 2) = 14
⇒ 7y + 21 − 2x − 4 = 14
⇒ −2x + 7y = −3 ..........(i)
and 4(y − 2) + 3(x − 3) = 2
⇒ 4y − 8 + 3x − 9 = 2
⇒ 3x + 4y = 19 .........(ii)
On multiplying (i) by 4 and (ii) by 7, we get:
−8x + 28y = −12 ........(iii)
21x + 28y = 133 ...........(iv)
On subtracting (iii) from (iv), we get:
29x = 145
⇒ x = 5
On substituting x = 5 in (i), we get:
−10 + 7y = −3
⇒ 7y = (−3 + 10) = 7
⇒ y = 1
Hence, the solution is x = 5 and y = 1.
Page No 110:
Question 16:
The given equations are:
7(y + 3) − 2(x + 2) = 14
⇒ 7y + 21 − 2x − 4 = 14
⇒ −2x + 7y = −3 ..........(i)
and 4(y − 2) + 3(x − 3) = 2
⇒ 4y − 8 + 3x − 9 = 2
⇒ 3x + 4y = 19 .........(ii)
On multiplying (i) by 4 and (ii) by 7, we get:
−8x + 28y = −12 ........(iii)
21x + 28y = 133 ...........(iv)
On subtracting (iii) from (iv), we get:
29x = 145
⇒ x = 5
On substituting x = 5 in (i), we get:
−10 + 7y = −3
⇒ 7y = (−3 + 10) = 7
⇒ y = 1
Hence, the solution is x = 5 and y = 1.
Answer:
The given equations are:
6x + 5y = 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 6x + 5y = 2(x + 6y − 1)
⇒ 6x + 5y = 2x + 12y − 2
⇒ 6x − 2x + 5y − 12y = − 2
⇒ 4x − 7y = −2 ..........(i)
and 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 7x + 3y + 1 = 2x + 12y − 2
⇒ 7x − 2x + 3y − 12y = −2 − 1
⇒ 5x − 9y = −3 .........(ii)
On multiplying (i) by 9 and (ii) by 7, we get:
36x − 63y = −18 ..........(iii)
35x − 63y = −21.........(iv)
On subtracting (iv) from (iii), we get:
x = (−18 + 21) = 3
On substituting x = 3 in (i), we get:
12 − 7y = −2
⇒ 7y = (2 + 12) = 14
⇒ y = 2
Hence, the solution is x = 3 and y = 2.
Page No 110:
Question 17:
The given equations are:
6x + 5y = 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 6x + 5y = 2(x + 6y − 1)
⇒ 6x + 5y = 2x + 12y − 2
⇒ 6x − 2x + 5y − 12y = − 2
⇒ 4x − 7y = −2 ..........(i)
and 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 7x + 3y + 1 = 2x + 12y − 2
⇒ 7x − 2x + 3y − 12y = −2 − 1
⇒ 5x − 9y = −3 .........(ii)
On multiplying (i) by 9 and (ii) by 7, we get:
36x − 63y = −18 ..........(iii)
35x − 63y = −21.........(iv)
On subtracting (iv) from (iii), we get:
x = (−18 + 21) = 3
On substituting x = 3 in (i), we get:
12 − 7y = −2
⇒ 7y = (2 + 12) = 14
⇒ y = 2
Hence, the solution is x = 3 and y = 2.
Answer:
The given equations are:
i.e.,
By cross multiplication, we get:
11x + 11y − 88 = 6x + 2y − 24
⇒ 11x − 6x + 11y − 2y = −24 + 88
⇒ 5x + 9y = 64 .........(i)
and
⇒ 11x + 22y − 154 = 9x + 3y − 36
⇒ 11x − 9x + 22y − 3y = −36 + 154
⇒ 2x + 19y = 118 .........(ii)
On multiplying (i) by 19 and (ii) by 9, we get:
95x + 171y = 1216..........(iii)
18x + 171y = 1062.........(iv)
On subtracting (iv) from (iii), we get:
77x = 154
⇒ x = 2
On substituting x = 2 in (i), we get:
10 + 9y = 64
⇒ 9y = (64 − 10) = 54
⇒ y = 6
Hence, the solution is x = 2 and y = 6.
Page No 110:
Question 18:
The given equations are:
i.e.,
By cross multiplication, we get:
11x + 11y − 88 = 6x + 2y − 24
⇒ 11x − 6x + 11y − 2y = −24 + 88
⇒ 5x + 9y = 64 .........(i)
and
⇒ 11x + 22y − 154 = 9x + 3y − 36
⇒ 11x − 9x + 22y − 3y = −36 + 154
⇒ 2x + 19y = 118 .........(ii)
On multiplying (i) by 19 and (ii) by 9, we get:
95x + 171y = 1216..........(iii)
18x + 171y = 1062.........(iv)
On subtracting (iv) from (iii), we get:
77x = 154
⇒ x = 2
On substituting x = 2 in (i), we get:
10 + 9y = 64
⇒ 9y = (64 − 10) = 54
⇒ y = 6
Hence, the solution is x = 2 and y = 6.
Answer:
The given equations are:
............(i)
.............(ii)
Putting , we get:
5u + 6y = 13 .............(iii)
3u + 4y = 7 ...........(iv)
On multiplying (iii) by 4 and (iv) by 6, we get:
20u + 24y = 52 ...........(v)
18u + 24y = 42 ............(vi)
On subtracting (vi) from (v), we get:
2u = 10 ⇒ u = 5
On substituting in (i), we get:
⇒ 25 + 6y = 13
⇒ 6y = (13 − 25) = −12
⇒ y = −2
Hence, the required solution is and y = −2.
Page No 110:
Question 19:
The given equations are:
............(i)
.............(ii)
Putting , we get:
5u + 6y = 13 .............(iii)
3u + 4y = 7 ...........(iv)
On multiplying (iii) by 4 and (iv) by 6, we get:
20u + 24y = 52 ...........(v)
18u + 24y = 42 ............(vi)
On subtracting (vi) from (v), we get:
2u = 10 ⇒ u = 5
On substituting in (i), we get:
⇒ 25 + 6y = 13
⇒ 6y = (13 − 25) = −12
⇒ y = −2
Hence, the required solution is and y = −2.
Answer:
The given equations are:
............(i)
.............(ii)
Putting , we get:
x + 6v = 6 .............(iii)
3x − 8v = 5 ...........(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
4x + 24v = 24 ...........(v)
9x − 24v = 15 ............(vi)
On adding (v) and (vi), we get:
13x = 39 ⇒ x = 3
On substituting x = 3 in (i), we get:
Hence, the required solution is x = 3 and y = 2.
Page No 110:
Question 20:
The given equations are:
............(i)
.............(ii)
Putting , we get:
x + 6v = 6 .............(iii)
3x − 8v = 5 ...........(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
4x + 24v = 24 ...........(v)
9x − 24v = 15 ............(vi)
On adding (v) and (vi), we get:
13x = 39 ⇒ x = 3
On substituting x = 3 in (i), we get:
Hence, the required solution is x = 3 and y = 2.
Answer:
The given equations are:
............(i)
.............(ii)
Putting , we get:
2x − 3v = 9 .............(iii)
3x + 7v = 2 ...........(iv)
On multiplying (iii) by 7 and (iv) by 3, we get:
14x − 21v = 63 .............(v)
9x + 21v = 6............(vi)
On adding (v) and (vi), we get:
23x = 69 ⇒ x = 3
On substituting x = 3 in (i), we get:
Hence, the required solution is x = 3 and y = −1.
Page No 110:
Question 21:
The given equations are:
............(i)
.............(ii)
Putting , we get:
2x − 3v = 9 .............(iii)
3x + 7v = 2 ...........(iv)
On multiplying (iii) by 7 and (iv) by 3, we get:
14x − 21v = 63 .............(v)
9x + 21v = 6............(vi)
On adding (v) and (vi), we get:
23x = 69 ⇒ x = 3
On substituting x = 3 in (i), we get:
Hence, the required solution is x = 3 and y = −1.
Answer:
Page No 110:
Question 22:
Answer:
The given equations are:
............(i)
.............(ii)
Putting and , we get:
9u − 4v = 8 .............(iii)
13u + 7v = 101 ...........(iv)
On multiplying (iii) by 7 and (iv) by 4, we get:
63u − 28v = 56 .............(v)
52u + 28v = 404............(vi)
On adding (v) and (vi), we get:
115u = 460 ⇒ u = 4
On substituting in (i), we get:
Hence, the required solution is and .
Page No 110:
Question 23:
The given equations are:
............(i)
.............(ii)
Putting and , we get:
9u − 4v = 8 .............(iii)
13u + 7v = 101 ...........(iv)
On multiplying (iii) by 7 and (iv) by 4, we get:
63u − 28v = 56 .............(v)
52u + 28v = 404............(vi)
On adding (v) and (vi), we get:
115u = 460 ⇒ u = 4
On substituting in (i), we get:
Hence, the required solution is and .
Answer:
The given equations are:
............(i)
.............(ii)
Putting and , we get:
5u − 3v = 1 .............(iii)
⇒
...............(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
20u − 12v = 4 .............(v)
27u + 12v = 90 ...........(vi)
On adding (iv) and (v), we get:
47u = 94 ⇒ u = 2
On substituting in (i), we get:
Hence, the required solution is and .
Page No 110:
Question 24:
The given equations are:
............(i)
.............(ii)
Putting and , we get:
5u − 3v = 1 .............(iii)
⇒
...............(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
20u − 12v = 4 .............(v)
27u + 12v = 90 ...........(vi)
On adding (iv) and (v), we get:
47u = 94 ⇒ u = 2
On substituting in (i), we get:
Hence, the required solution is and .
Answer:
Multiplying equation (i) and (ii) by 6, we get
Multiplying (i) by 3 and (ii) by 2 and subtracting (ii) from (i), we get
Now, substituting in (i), we have
Hence, and .
Page No 110:
Question 25:
Multiplying equation (i) and (ii) by 6, we get
Multiplying (i) by 3 and (ii) by 2 and subtracting (ii) from (i), we get
Now, substituting in (i), we have
Hence, and .
Answer:
The given equations are:
4x + 6y = 3xy .......(i)
8x + 9y = 5xy .........(ii)
From equation (i), we have:
.............(iii)
For equation (ii), we have:
.............(iv)
On substituting , we get:
4v + 6u = 3 ...........(v)
8v + 9u = 5 ...........(vi)
On multiplying (v) by 9 and (vi) by 6, we get:
36v + 54u = 27 ..............(vii)
48v + 54u = 30 ...............(viii)
On subtracting (vii) from (viii), we get:
12v = 3
On substituting y = 4 in (iii), we get:
Hence, the required solution is x = 3 and y = 4.
Page No 110:
Question 26:
The given equations are:
4x + 6y = 3xy .......(i)
8x + 9y = 5xy .........(ii)
From equation (i), we have:
.............(iii)
For equation (ii), we have:
.............(iv)
On substituting , we get:
4v + 6u = 3 ...........(v)
8v + 9u = 5 ...........(vi)
On multiplying (v) by 9 and (vi) by 6, we get:
36v + 54u = 27 ..............(vii)
48v + 54u = 30 ...............(viii)
On subtracting (vii) from (viii), we get:
12v = 3
On substituting y = 4 in (iii), we get:
Hence, the required solution is x = 3 and y = 4.
Answer:
The given equations are:
x + y = 5xy .......(i)
3x + 2y = 13xy .........(ii)
From equation (i), we have:
.............(iii)
From equation (ii), we have:
.............(iv)
On substituting , we get:
v + u = 5 ...........(v)
3v + 2u = 13 ...........(vi)
On multiplying (v) by 2, we get:
2v + 2u = 10 ..............(vii)
On subtracting (vii) from (vi), we get:
v = 3
On substituting in (iii), we get:
Hence, the required solution is and or x = 0 and y = 0.
Page No 110:
Question 27:
The given equations are:
x + y = 5xy .......(i)
3x + 2y = 13xy .........(ii)
From equation (i), we have:
.............(iii)
From equation (ii), we have:
.............(iv)
On substituting , we get:
v + u = 5 ...........(v)
3v + 2u = 13 ...........(vi)
On multiplying (v) by 2, we get:
2v + 2u = 10 ..............(vii)
On subtracting (vii) from (vi), we get:
v = 3
On substituting in (iii), we get:
Hence, the required solution is and or x = 0 and y = 0.
Answer:
The given equations are
Substituting in (i) and (ii), we get
Multiplying (iii) by 3 and subtracting it from (iv), we get
Now, substituting v = 1 in (iii), we get
Adding (v) and (vi), we get
Substituting x = 3 in (vi), we have
Hence, x = 3 and y = 2.
Page No 111:
Question 28:
The given equations are
Substituting in (i) and (ii), we get
Multiplying (iii) by 3 and subtracting it from (iv), we get
Now, substituting v = 1 in (iii), we get
Adding (v) and (vi), we get
Substituting x = 3 in (vi), we have
Hence, x = 3 and y = 2.
Answer:
The given equations are:
...(i)
...(ii)
Putting and , we get:
3u + 2v = 2 ...(iii)
9u − 4v = 1 ...(iv)
On multiplying (iii) by 2, we get:
6u + 4v = 4 ...(v)
On adding (iv) and (v), we get:
15u = 5
...(vi)
On substituting in (iii), we get:
1 + 2v = 2
⇒ 2v = 1
...(vii)
On adding (vi) and (vii), we get:
2x = 5
⇒
On substituting in (vi), we get:
Hence, the required solution is .
Page No 111:
Question 29:
The given equations are:
...(i)
...(ii)
Putting and , we get:
3u + 2v = 2 ...(iii)
9u − 4v = 1 ...(iv)
On multiplying (iii) by 2, we get:
6u + 4v = 4 ...(v)
On adding (iv) and (v), we get:
15u = 5
...(vi)
On substituting in (iii), we get:
1 + 2v = 2
⇒ 2v = 1
...(vii)
On adding (vi) and (vii), we get:
2x = 5
⇒
On substituting in (vi), we get:
Hence, the required solution is .
Answer:
The given equations are:
.............(i)
..............(ii)
Putting and , we get:
.................(iii)
................(iv)
On adding (iii) and (iv), we get:
15u = 3
⇒
⇒
On substituting in (iii), we get:
Hence, the required solution is x = 4 and y = 5.
Page No 111:
Question 30:
The given equations are:
.............(i)
..............(ii)
Putting and , we get:
.................(iii)
................(iv)
On adding (iii) and (iv), we get:
15u = 3
⇒
⇒
On substituting in (iii), we get:
Hence, the required solution is x = 4 and y = 5.
Answer:
The given equations are:
...(i)
...(ii)
Putting and , we get:
44u + 30v = 10 ...(iii)
55u + 40v = 13 ...(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
176u + 120v = 40 ...(v)
165u + 120v = 39 ...(vi)
On subtracting (vi) from (v), we get:
11u = 1
...(vii)
On substituting in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
...(viii)
On adding (vii) and (viii), we get:
2x = 16
⇒ x = 8
On substituting x = 8 in (vii), we get:
8 + y = 11
⇒ y = (11 − 8) = 3
Hence, the required solution is x = 8 and y = 3.
Page No 111:
Question 31:
The given equations are:
...(i)
...(ii)
Putting and , we get:
44u + 30v = 10 ...(iii)
55u + 40v = 13 ...(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
176u + 120v = 40 ...(v)
165u + 120v = 39 ...(vi)
On subtracting (vi) from (v), we get:
11u = 1
...(vii)
On substituting in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
...(viii)
On adding (vii) and (viii), we get:
2x = 16
⇒ x = 8
On substituting x = 8 in (vii), we get:
8 + y = 11
⇒ y = (11 − 8) = 3
Hence, the required solution is x = 8 and y = 3.
Answer:
The given equations are
Substituting in (i) and (ii), we get
Multiplying (iii) by 9 and (iv) by 2 and adding, we get
Now, substituting in (iii), we get
Adding (v) and (vi), we get
Substituting in (v), we have
Hence, .
Page No 111:
Question 32:
The given equations are
Substituting in (i) and (ii), we get
Multiplying (iii) by 9 and (iv) by 2 and adding, we get
Now, substituting in (iii), we get
Adding (v) and (vi), we get
Substituting in (v), we have
Hence, .
Answer:
The given equations are:
71x + 37y = 253 ..............(i)
37x + 71y = 287 ............(ii)
On adding (i) and (ii), we get:
108x + 108y = 540
⇒ 108(x + y) = 540
⇒ (x + y) = 5................(iii)
On subtracting (ii) from (i), we get:
34x − 34y = −34
⇒ 34(x − y) = −34
⇒ (x − y) = −1...........(iv)
On adding (iii) and (iv), we get:
2x = 5 − 1= 4
⇒ x = 2
On subtracting (iv) from (iii), we get:
2y = 5 + 1 = 6
⇒ y = 3
Hence, the required solution is x = 2 and y = 3.
Page No 111:
Question 33:
The given equations are:
71x + 37y = 253 ..............(i)
37x + 71y = 287 ............(ii)
On adding (i) and (ii), we get:
108x + 108y = 540
⇒ 108(x + y) = 540
⇒ (x + y) = 5................(iii)
On subtracting (ii) from (i), we get:
34x − 34y = −34
⇒ 34(x − y) = −34
⇒ (x − y) = −1...........(iv)
On adding (iii) and (iv), we get:
2x = 5 − 1= 4
⇒ x = 2
On subtracting (iv) from (iii), we get:
2y = 5 + 1 = 6
⇒ y = 3
Hence, the required solution is x = 2 and y = 3.
Answer:
The given equations are:
217x + 131y = 913 .............(i)
131x + 217y = 827 .............(ii)
On adding (i) and (ii), we get:
348x + 348y = 1740
⇒ 348(x + y) = 1740
⇒ x + y = 5 ............(iii)
On subtracting (ii) from (i), we get:
86x − 86y = 86
⇒ 86(x − y) = 86
⇒ x − y = 1 ...............(iv)
On adding (iii) and (iv), we get :
2x = 6 ⇒ x = 3
On substituting x = 3 in (iii), we get:
3 + y = 5
⇒ y = 5 − 3 = 2
Hence, the required solution is x = 3 and y = 2.
Page No 111:
Question 34:
The given equations are:
217x + 131y = 913 .............(i)
131x + 217y = 827 .............(ii)
On adding (i) and (ii), we get:
348x + 348y = 1740
⇒ 348(x + y) = 1740
⇒ x + y = 5 ............(iii)
On subtracting (ii) from (i), we get:
86x − 86y = 86
⇒ 86(x − y) = 86
⇒ x − y = 1 ...............(iv)
On adding (iii) and (iv), we get :
2x = 6 ⇒ x = 3
On substituting x = 3 in (iii), we get:
3 + y = 5
⇒ y = 5 − 3 = 2
Hence, the required solution is x = 3 and y = 2.
Answer:
The given equations are
Adding (i) and (ii), we get
Subtracting (i) from (ii), we get
Now, adding equation (iii) and (iv), we get
Substituting x = 3 in (iv), we have
Hence, .
Page No 111:
Question 35:
The given equations are
Adding (i) and (ii), we get
Subtracting (i) from (ii), we get
Now, adding equation (iii) and (iv), we get
Substituting x = 3 in (iv), we have
Hence, .
Answer:
The given equations can be written as
Adding (i) and (ii), we get
Substituting y = 2 in (i), we have
Hence, x = 1 and y = 2..
Page No 111:
Question 36:
The given equations can be written as
Adding (i) and (ii), we get
Substituting y = 2 in (i), we have
Hence, x = 1 and y = 2..
Answer:
The given equations are
Substituting in (i) and (ii), we get
Adding (iii) and (iv), we get
Now, substituting in (iii), we get
Adding (v) and (vi), we get
Substituting x = 1 in (v), we have
Hence, x = 1 and y = 1.
Page No 111:
Question 37:
The given equations are
Substituting in (i) and (ii), we get
Adding (iii) and (iv), we get
Now, substituting in (iii), we get
Adding (v) and (vi), we get
Substituting x = 1 in (v), we have
Hence, x = 1 and y = 1.
Answer:
The given equations are:
...(i)
...(ii)
Putting and , we get:
...(iii)
...(iv)
On multiplying (iii) by 6 and (iv) by 20, we get:
3u + 10v = −9 ...(v)
...(vi)
On multiplying (v) by 6 and (vi) by 5, we get:
18u + 60v = −54 ...(vii)
...(viii)
On adding (vii) and (viii), we get:
⇒ x + 2y = 3 ...(ix)
On substituting in (v), we get:
1 + 10v = −9
⇒ 10v = −10
⇒ v = −1
...(x)
On adding (ix) and (x), we get:
4x = 2
⇒
On substituting in (x), we get:
Hence, the required solution is .
Page No 111:
Question 38:
The given equations are:
...(i)
...(ii)
Putting and , we get:
...(iii)
...(iv)
On multiplying (iii) by 6 and (iv) by 20, we get:
3u + 10v = −9 ...(v)
...(vi)
On multiplying (v) by 6 and (vi) by 5, we get:
18u + 60v = −54 ...(vii)
...(viii)
On adding (vii) and (viii), we get:
⇒ x + 2y = 3 ...(ix)
On substituting in (v), we get:
1 + 10v = −9
⇒ 10v = −10
⇒ v = −1
...(x)
On adding (ix) and (x), we get:
4x = 2
⇒
On substituting in (x), we get:
Hence, the required solution is .
Answer:
The given equations are
Substituting in (i) and (ii), we get
Multiplying (iv) by 3 and subtracting from(iii), we get
Now, substituting in (iv), we get
Adding (v) and (vi), we get
Substituting x = 1 in (v), we have
Hence, x = 1 and y = 1.
Page No 111:
Question 39:
The given equations are
Substituting in (i) and (ii), we get
Multiplying (iv) by 3 and subtracting from(iii), we get
Now, substituting in (iv), we get
Adding (v) and (vi), we get
Substituting x = 1 in (v), we have
Hence, x = 1 and y = 1.
Answer:
The given equations can be written as
Multiplying (i) by 3 and subtracting (ii) from it, we get
Substituting in (i), we have
Hence, .
Page No 111:
Question 40:
The given equations can be written as
Multiplying (i) by 3 and subtracting (ii) from it, we get
Substituting in (i), we have
Hence, .
Answer:
The given equations are
Multiplying (i) by b and adding it with (ii), we get
Substituting x = a in (i), we have
Hence, x = a and y = b.
Page No 111:
Question 41:
The given equations are
Multiplying (i) by b and adding it with (ii), we get
Substituting x = a in (i), we have
Hence, x = a and y = b.
Answer:
The given equations are:
⇒ [Taking LCM]
⇒bx + ay = 2ab .......(i)
Again, ax − by = (a2 − b2) ........(ii)
On multiplying (i) by b and (ii) by a, we get:
b2x + bay = 2ab2.........(iii)
a2x − bay = a(a2 − b2) ........(iv)
On adding (iii) from (iv), we get:
(b2 + a2)x = 2a2b + a(a2 − b2)
⇒ (b2 + a2)x = 2ab2 + a3 − ab2
⇒ (b2 + a2)x = ab2+ a3
⇒ (b2 + a2)x = a(b2 + a2)
⇒
On substituting x = a in (i), we get:
ba + ay = 2ab
⇒ ay = ab
⇒ y = b
Hence, the solution is x = a and y = b.
Page No 111:
Question 42:
The given equations are:
⇒ [Taking LCM]
⇒bx + ay = 2ab .......(i)
Again, ax − by = (a2 − b2) ........(ii)
On multiplying (i) by b and (ii) by a, we get:
b2x + bay = 2ab2.........(iii)
a2x − bay = a(a2 − b2) ........(iv)
On adding (iii) from (iv), we get:
(b2 + a2)x = 2a2b + a(a2 − b2)
⇒ (b2 + a2)x = 2ab2 + a3 − ab2
⇒ (b2 + a2)x = ab2+ a3
⇒ (b2 + a2)x = a(b2 + a2)
⇒
On substituting x = a in (i), we get:
ba + ay = 2ab
⇒ ay = ab
⇒ y = b
Hence, the solution is x = a and y = b.
Answer:
The given equations are
Multiplying (i) by p and (ii) by q and adding them, we get
Substituting x = 1 in (i), we have
Hence, x = 1 and .
Page No 111:
Question 43:
The given equations are
Multiplying (i) by p and (ii) by q and adding them, we get
Substituting x = 1 in (i), we have
Hence, x = 1 and .
Answer:
The given equations are
From (i)
Substituting in (ii), we get
Now, substitute x = a in (ii) to get
Hence, x = a and y = b.
Page No 111:
Question 44:
The given equations are
From (i)
Substituting in (ii), we get
Now, substitute x = a in (ii) to get
Hence, x = a and y = b.
Answer:
The given equations are:
6(ax + by) = 3a + 2b
⇒ 6ax + 6by = 3a + 2b ...............(i)
and 6(bx − ay) = 3b − 2a
⇒ 6bx − 6ay = 3b − 2a ...................(ii)
On multiplying (i) by a and (ii) by b, we get:
6a2x + 6aby = 3a2 + 2ab ................(iii)
6b2x − 6aby = 3b2 − 2ab ....................(iv)
On adding (iii) and (iv), we get:
6(a2 + b2)x = 3(a2 + b2)
On substituting in (i), we get:
⇒ 3a + 6by = 3a + 2b
⇒ 6by = 2b
⇒ y =
Hence, the required solution is and .
Page No 111:
Question 45:
The given equations are:
6(ax + by) = 3a + 2b
⇒ 6ax + 6by = 3a + 2b ...............(i)
and 6(bx − ay) = 3b − 2a
⇒ 6bx − 6ay = 3b − 2a ...................(ii)
On multiplying (i) by a and (ii) by b, we get:
6a2x + 6aby = 3a2 + 2ab ................(iii)
6b2x − 6aby = 3b2 − 2ab ....................(iv)
On adding (iii) and (iv), we get:
6(a2 + b2)x = 3(a2 + b2)
On substituting in (i), we get:
⇒ 3a + 6by = 3a + 2b
⇒ 6by = 2b
⇒ y =
Hence, the required solution is and .
Answer:
The given equations are
From (ii)
Substituting in (i), we get
Now, substitute x = a + b in (ii) to get
Hence, .
Page No 111:
Question 46:
The given equations are
From (ii)
Substituting in (i), we get
Now, substitute x = a + b in (ii) to get
Hence, .
Answer:
The given equations are:
By taking LCM, we get:
b2x − a2y = −a2b − b2a .......(i)
and bx − ay + 2ab = 0
bx − ay = −2ab ........(ii)
On multiplying (ii) by a, we get:
abx − a2y = −2a2b .......(iii)
On subtracting (i) from (iii), we get:
abx − b2x = − 2a2b + a2b + b2a = −a2b + b2a
⇒ x(ab − b2) = −ab(a − b)
⇒ x(a − b)b = −ab(a − b)
∴
On substituting x = −a in (i), we get:
b2(−a) − a2y = −a2b − b2a
⇒ −b2a − a2y = −a2b − b2a
⇒ −a2y = −a2b
⇒ y = b
Hence, the solution is x = −a and y = b.
Page No 111:
Question 47:
The given equations are:
By taking LCM, we get:
b2x − a2y = −a2b − b2a .......(i)
and bx − ay + 2ab = 0
bx − ay = −2ab ........(ii)
On multiplying (ii) by a, we get:
abx − a2y = −2a2b .......(iii)
On subtracting (i) from (iii), we get:
abx − b2x = − 2a2b + a2b + b2a = −a2b + b2a
⇒ x(ab − b2) = −ab(a − b)
⇒ x(a − b)b = −ab(a − b)
∴
On substituting x = −a in (i), we get:
b2(−a) − a2y = −a2b − b2a
⇒ −b2a − a2y = −a2b − b2a
⇒ −a2y = −a2b
⇒ y = b
Hence, the solution is x = −a and y = b.
Answer:
The given equations are:
By taking LCM, we get:
⇒ b2x + a2y = (ab)a2 + b2
⇒ b2x + a2y = a3b + ab3 .......(i)
Also, x + y = 2ab........(ii)
On multiplying (ii) by a2, we get:
a2x + a2y = 2a3b.........(iii)
On subtracting (iii) from (i), we get:
(b2 − a2)x = a3b + ab3 − 2a3b
⇒ (b2 − a2)x = −a3b + ab3
⇒ (b2 − a2)x = ab(b2 − a2)
⇒ (b2 − a2)x = ab(b2 − a2)
⇒
On substituting x = ab in (i), we get:
b2(ab) + a2y = a3b + ab3
⇒ a2y = a3b
⇒
Hence, the solution is x = ab and y = ab.
Page No 111:
Question 48:
The given equations are:
By taking LCM, we get:
⇒ b2x + a2y = (ab)a2 + b2
⇒ b2x + a2y = a3b + ab3 .......(i)
Also, x + y = 2ab........(ii)
On multiplying (ii) by a2, we get:
a2x + a2y = 2a3b.........(iii)
On subtracting (iii) from (i), we get:
(b2 − a2)x = a3b + ab3 − 2a3b
⇒ (b2 − a2)x = −a3b + ab3
⇒ (b2 − a2)x = ab(b2 − a2)
⇒ (b2 − a2)x = ab(b2 − a2)
⇒
On substituting x = ab in (i), we get:
b2(ab) + a2y = a3b + ab3
⇒ a2y = a3b
⇒
Hence, the solution is x = ab and y = ab.
Answer:
The given equations are
From (i)
Substituting in (ii), we get
Now, substitute x = a in (i) to get
Hence, x = a and y = b.
Page No 111:
Question 49:
The given equations are
From (i)
Substituting in (ii), we get
Now, substitute x = a in (i) to get
Hence, x = a and y = b.
Answer:
The given equations are
Multiplying (i) by a2 and (ii) by b2 and subtracting, we get
Now, multiplying (i) by b2 and (ii) by a2 and subtracting, we get
Hence, .
Page No 111:
Question 50:
The given equations are
Multiplying (i) by a2 and (ii) by b2 and subtracting, we get
Now, multiplying (i) by b2 and (ii) by a2 and subtracting, we get
Hence, .
Answer:
The given equations are
Multiplying (i) by b and (ii) by b2 and subtracting, we get
Now, substituting x = a2 in (i), we get
Hence, .
Page No 117:
Question 1:
The given equations are
Multiplying (i) by b and (ii) by b2 and subtracting, we get
Now, substituting x = a2 in (i), we get
Hence, .
Answer:
The given equations are:
x + 2y + 1 = 0 ...(i)
2x − 3y − 12 = 0 ...(ii)
Here, a1 = 1, b1= 2, c1 = 1, a2 = 2, b2 = −3 and c2 = −12
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 3 and y = −2 is the required solution.
Page No 117:
Question 2:
The given equations are:
x + 2y + 1 = 0 ...(i)
2x − 3y − 12 = 0 ...(ii)
Here, a1 = 1, b1= 2, c1 = 1, a2 = 2, b2 = −3 and c2 = −12
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 3 and y = −2 is the required solution.
Answer:
The given equations are:
3x − 2y + 3 = 0 ...(i)
4x + 3y − 47 = 0 ...(ii)
Here, a1 = 3, b1= −2 , c1 = 3, a2 = 4, b2 = 3 and c2 = −47
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 5 and y = 9 is the required solution.
Page No 117:
Question 3:
The given equations are:
3x − 2y + 3 = 0 ...(i)
4x + 3y − 47 = 0 ...(ii)
Here, a1 = 3, b1= −2 , c1 = 3, a2 = 4, b2 = 3 and c2 = −47
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 5 and y = 9 is the required solution.
Answer:
The given equations are:
6x − 5y − 16 = 0 ...(i)
7x − 13y + 10 = 0 ...(ii)
Here, a1 = 6, b1= −5 , c1 = −16, a2 = 7, b2 = −13 and c2 = 10
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 6 and y = 4 is the required solution.
Page No 117:
Question 4:
The given equations are:
6x − 5y − 16 = 0 ...(i)
7x − 13y + 10 = 0 ...(ii)
Here, a1 = 6, b1= −5 , c1 = −16, a2 = 7, b2 = −13 and c2 = 10
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 6 and y = 4 is the required solution.
Answer:
The given equations are:
3x + 2y + 25 = 0 ...(i)
2x + y + 10 = 0 ...(ii)
Here, a1 = 3, b1= 2 , c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 5 and y = −20 is the required solution.
Page No 117:
Question 5:
The given equations are:
3x + 2y + 25 = 0 ...(i)
2x + y + 10 = 0 ...(ii)
Here, a1 = 3, b1= 2 , c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 5 and y = −20 is the required solution.
Answer:
The given equations may be written as:
2x + 5y − 1 = 0 ...(i)
2x + 3y − 3 = 0 ...(ii)
Here, a1 = 2, b1= 5, c1 = −1, a2 = 2, b2 = 3 and c2 = −3
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 3 and y = −1 is the required solution.
Page No 117:
Question 6:
The given equations may be written as:
2x + 5y − 1 = 0 ...(i)
2x + 3y − 3 = 0 ...(ii)
Here, a1 = 2, b1= 5, c1 = −1, a2 = 2, b2 = 3 and c2 = −3
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 3 and y = −1 is the required solution.
Answer:
The given equations may be written as:
2x + y − 35 = 0 ...(i)
3x + 4y − 65 = 0 ...(ii)
Here, a1 = 2, b1= 1, c1 = −35, a2 = 3, b2 = 4 and c2 = −65
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 15 and y = 5 is the required solution.
Page No 117:
Question 7:
The given equations may be written as:
2x + y − 35 = 0 ...(i)
3x + 4y − 65 = 0 ...(ii)
Here, a1 = 2, b1= 1, c1 = −35, a2 = 3, b2 = 4 and c2 = −65
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 15 and y = 5 is the required solution.
Answer:
The given equations may be written as:
7x − 2y − 3 = 0 ...(i)
...(ii)
Here, a1 = 7, b1= −2 , c1 = −3, a2 = 22, b2 = and c2 = −16
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 1 and y = 2 is the required solution.
Page No 117:
Question 8:
The given equations may be written as:
7x − 2y − 3 = 0 ...(i)
...(ii)
Here, a1 = 7, b1= −2 , c1 = −3, a2 = 22, b2 = and c2 = −16
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 1 and y = 2 is the required solution.
Answer:
The given equations may be written as:
...(i)
...(ii)
Here,
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 18 and y = 15 is the required solution.
Page No 117:
Question 9:
The given equations may be written as:
...(i)
...(ii)
Here,
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 18 and y = 15 is the required solution.
Answer:
Taking and , the given equations become:
u + v = 7
2u + 3v = 17
The given equations may be written as:
u + v − 7 = 0 ...(i)
2u + 3v − 17 = 0 ...(ii)
Here, a1= 1, b1 = 1, c1 = −7, a2 = 2, b2 = 3 and c2 = −17
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
⇒
Hence, and is the required solution.
Page No 117:
Question 10:
Taking and , the given equations become:
u + v = 7
2u + 3v = 17
The given equations may be written as:
u + v − 7 = 0 ...(i)
2u + 3v − 17 = 0 ...(ii)
Here, a1= 1, b1 = 1, c1 = −7, a2 = 2, b2 = 3 and c2 = −17
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
⇒
Hence, and is the required solution.
Answer:
Taking and , the given equations become:
5u − 2v + 1 = 0 ...(i)
15u + 7v − 10 = 0 ...(ii)
Here, a1= 5, b1 = −2, c1 = 1, a2 = 15, b2 = −7 and c2 = −10
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
So, (x + y) = 5 ...(iii)
and (x − y) = 1 ...(iv)
Again, the above equations (iii) and (iv) may be written as:
x + y − 5 = 0 ...(v)
x − y − 1 = 0 ...(vi)
Here, a1= 1, b1 = 1, c1 = −5, a2 = 1, b2 = −1 and c2 = −1
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 3 and y = 2 is the required solution.
Page No 117:
Question 11:
Taking and , the given equations become:
5u − 2v + 1 = 0 ...(i)
15u + 7v − 10 = 0 ...(ii)
Here, a1= 5, b1 = −2, c1 = 1, a2 = 15, b2 = −7 and c2 = −10
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
So, (x + y) = 5 ...(iii)
and (x − y) = 1 ...(iv)
Again, the above equations (iii) and (iv) may be written as:
x + y − 5 = 0 ...(v)
x − y − 1 = 0 ...(vi)
Here, a1= 1, b1 = 1, c1 = −5, a2 = 1, b2 = −1 and c2 = −1
By cross multiplication, we have:
∴
⇒
⇒
⇒
Hence, x = 3 and y = 2 is the required solution.
Answer:
The given equations may be written as:
...(i)
...(ii)
Here, a1= , b1 = , c1 = −(a + b), a2 = a, b2 = −b and c2 = −2ab
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
⇒
Hence, x = b and y = −a is the required solution.
Page No 117:
Question 12:
The given equations may be written as:
...(i)
...(ii)
Here, a1= , b1 = , c1 = −(a + b), a2 = a, b2 = −b and c2 = −2ab
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
⇒
Hence, x = b and y = −a is the required solution.
Answer:
The given equations may be written as:
2ax + 3by − (a + 2b) = 0 ...(i)
3ax + 2by − (2a + b) = 0 ...(ii)
Here, a1= 2a, b1 = 3b, c1 = −(a + 2b), a2 = 3a, b2 = 2b and c2 = −(2a + b)
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
Hence, and is the required solution.
Page No 117:
Question 13:
The given equations may be written as:
2ax + 3by − (a + 2b) = 0 ...(i)
3ax + 2by − (2a + b) = 0 ...(ii)
Here, a1= 2a, b1 = 3b, c1 = −(a + 2b), a2 = 3a, b2 = 2b and c2 = −(2a + b)
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
Hence, and is the required solution.
Answer:
Substituting in the given equations, we get
Here, .
So, by cross-multiplication, we have
Hence, x = a and y = b.
Page No 128:
Question 1:
Substituting in the given equations, we get
Here, .
So, by cross-multiplication, we have
Hence, x = a and y = b.
Answer:
The given system of equations is:
3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 3, b1= 5, c1 = −12 and a2 = 5, b2= 3, c2 = −4
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
3x + 5y = 12 ...(i)
5x + 3y = 4 ...(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36 ...(iii)
25x + 15y = 20 ...(iv)
On subtracting (iii) from (iv), we get:
16x = −16
⇒ x = −1
On substituting x = −1 in (i), we get:
3(−1) + 5y = 12
⇒ 5y = (12 + 3) = 15
⇒ y = 3
Hence, x = −1 and y = 3 is the required solution.
Page No 128:
Question 2:
The given system of equations is:
3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 3, b1= 5, c1 = −12 and a2 = 5, b2= 3, c2 = −4
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
3x + 5y = 12 ...(i)
5x + 3y = 4 ...(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36 ...(iii)
25x + 15y = 20 ...(iv)
On subtracting (iii) from (iv), we get:
16x = −16
⇒ x = −1
On substituting x = −1 in (i), we get:
3(−1) + 5y = 12
⇒ 5y = (12 + 3) = 15
⇒ y = 3
Hence, x = −1 and y = 3 is the required solution.
Answer:
The system of equations can be written as
The given equations are of the form
where
Now,
Since, , therefore the system of equations has unique solution.
Using cross multiplication method, we have
Hence, .
Page No 128:
Question 3:
The system of equations can be written as
The given equations are of the form
where
Now,
Since, , therefore the system of equations has unique solution.
Using cross multiplication method, we have
Hence, .
Answer:
The given system of equations are:
⇒
2x + 3y = 18
⇒ 2x + 3y − 18 = 0 ...(i)
and
x − 2y = 2
x − 2y − 2 = 0 ...(ii)
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 2, b1= 3, c1 = −18 and a2 = 1, b2= −2, c2 = −2
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
2x + 3y − 18 = 0 ...(iii)
x − 2y − 2 = 0 ...(iv)
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y − 36 = 0 ...(v)
3x − 6y − 6 = 0 ...(vi)
On adding (v) and (vi), we get:
7x = 42
⇒ x = 6
On substituting x = 6 in (iii), we get:
2(6) + 3y = 18
⇒ 3y = (18 − 12) = 6
⇒ y = 2
Hence, x = 6 and y = 2 is the required solution.
Page No 128:
Question 4:
The given system of equations are:
⇒
2x + 3y = 18
⇒ 2x + 3y − 18 = 0 ...(i)
and
x − 2y = 2
x − 2y − 2 = 0 ...(ii)
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 2, b1= 3, c1 = −18 and a2 = 1, b2= −2, c2 = −2
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
2x + 3y − 18 = 0 ...(iii)
x − 2y − 2 = 0 ...(iv)
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y − 36 = 0 ...(v)
3x − 6y − 6 = 0 ...(vi)
On adding (v) and (vi), we get:
7x = 42
⇒ x = 6
On substituting x = 6 in (iii), we get:
2(6) + 3y = 18
⇒ 3y = (18 − 12) = 6
⇒ y = 2
Hence, x = 6 and y = 2 is the required solution.
Answer:
The given system of equations are
This system is of the form
where
Now, for the given system of equations to have a unique solution, we must have
Hence, .
Page No 128:
Question 5:
The given system of equations are
This system is of the form
where
Now, for the given system of equations to have a unique solution, we must have
Hence, .
Answer:
The given system of equations are
This system of equations is of the form
where
Now, for the given system of equations to have a unique solution, we must have
Hence, .
Page No 128:
Question 6:
The given system of equations are
This system of equations is of the form
where
Now, for the given system of equations to have a unique solution, we must have
Hence, .
Answer:
The given system of equations is
This system is of the form
where
For the given system of equations to have a unique solution, we must have
Hence, .
Page No 128:
Question 7:
The given system of equations is
This system is of the form
where
For the given system of equations to have a unique solution, we must have
Hence, .
Answer:
The given system of equations is
This system is of the form
where
For the given system of equations to have a unique solution, we must have
Hence, .
Page No 129:
Question 8:
The given system of equations is
This system is of the form
where
For the given system of equations to have a unique solution, we must have
Hence, .
Answer:
The given system of equations:
4x − 5y = k
⇒ 4x − 5y − k = 0 ...(i)
And, 2x − 3y = 12
⇒ 2x − 3y − 12 = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 4, b1= −5, c1 = −k and a2 = 2, b2= −3, c2 = −12
For a unique solution, we must have:
i.e.
Thus, for all real values of k, the given system of equations will have a unique solution.
Page No 129:
Question 9:
The given system of equations:
4x − 5y = k
⇒ 4x − 5y − k = 0 ...(i)
And, 2x − 3y = 12
⇒ 2x − 3y − 12 = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 4, b1= −5, c1 = −k and a2 = 2, b2= −3, c2 = −12
For a unique solution, we must have:
i.e.
Thus, for all real values of k, the given system of equations will have a unique solution.
Answer:
The given system of equations:
kx + 3y = (k − 3)
⇒ kx + 3y − (k − 3) = 0 ....(i)
And, 12x + ky = k
⇒ 12x + ky − k = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −(k − 3) and a2 = 12, b2= k, c2 = −k
For a unique solution, we must have:
i.e.
Thus, for all real values of k other than , the given system of equations will have a unique solution.
Page No 129:
Question 10:
The given system of equations:
kx + 3y = (k − 3)
⇒ kx + 3y − (k − 3) = 0 ....(i)
And, 12x + ky = k
⇒ 12x + ky − k = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −(k − 3) and a2 = 12, b2= k, c2 = −k
For a unique solution, we must have:
i.e.
Thus, for all real values of k other than , the given system of equations will have a unique solution.
Answer:
The given system of equations is:
2x − 3y = 5
⇒ 2x − 3y − 5 = 0 ...(i)
6x − 9y = 15
⇒ 6x − 9y − 15 = 0 ...(ii)
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 2, b1= −3, c1 = −5 and a2 = 6, b2= −9, c2 = −15
and
Thus,
Hence, the given system of equations has an infinite number of solutions.
Page No 129:
Question 11:
The given system of equations is:
2x − 3y = 5
⇒ 2x − 3y − 5 = 0 ...(i)
6x − 9y = 15
⇒ 6x − 9y − 15 = 0 ...(ii)
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 2, b1= −3, c1 = −5 and a2 = 6, b2= −9, c2 = −15
and
Thus,
Hence, the given system of equations has an infinite number of solutions.
Answer:
The given system of equations can be written as
This system is of the form
where
Now,
Since, , therefore the given system has no solution.
Page No 129:
Question 12:
The given system of equations can be written as
This system is of the form
where
Now,
Since, , therefore the given system has no solution.
Answer:
The given system of equations is:
kx + 2y = 5
⇒ kx + 2y − 5= 0 ...(i)
3x − 4y = 10
⇒ 3x − 4y − 10 = 0 ...(ii)
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = k, b1= 2, c1 = −5 and a2 = 3, b2= −4, c2 = −10
(i) For a unique solution, we must have:
, i.e.,
Thus for all real values of k other than , the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
Hence, the required value of k is .
Page No 129:
Question 13:
The given system of equations is:
kx + 2y = 5
⇒ kx + 2y − 5= 0 ...(i)
3x − 4y = 10
⇒ 3x − 4y − 10 = 0 ...(ii)
These equations are of the forms:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = k, b1= 2, c1 = −5 and a2 = 3, b2= −4, c2 = −10
(i) For a unique solution, we must have:
, i.e.,
Thus for all real values of k other than , the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
Hence, the required value of k is .
Answer:
The given system of equations is:
x + 2y = 5
⇒ x + 2y − 5= 0 ...(i)
3x + ky + 15 = 0 ...(ii)
These equations are of the form:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 1, b1= 2, c1 = −5 and a2 = 3, b2= k, c2 = 15
(i) For a unique solution, we must have:
, i.e.,
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
Hence, the required value of k is 6.
Page No 129:
Question 14:
The given system of equations is:
x + 2y = 5
⇒ x + 2y − 5= 0 ...(i)
3x + ky + 15 = 0 ...(ii)
These equations are of the form:
a1x + b1y + c1= 0 and a2x + b2y + c2 = 0
where, a1 = 1, b1= 2, c1 = −5 and a2 = 3, b2= k, c2 = 15
(i) For a unique solution, we must have:
, i.e.,
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
Hence, the required value of k is 6.
Answer:
The given system of equations is:
x + 2y = 3
x + 2y − 3= 0 ....(i)
And, 5x + ky + 7 = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 1, b1= 2, c1 = −3 and a2 = 5, b2= k, c2 = 7
(i) For a unique solution, we must have:
, i.e.
Thus, for all real values of kâ, other than 10, the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
Hence, the required value of k is 10.
There is no value of k for which the given system of equations has an infinite number of solutions.
Page No 129:
Question 15:
The given system of equations is:
x + 2y = 3
x + 2y − 3= 0 ....(i)
And, 5x + ky + 7 = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 1, b1= 2, c1 = −3 and a2 = 5, b2= k, c2 = 7
(i) For a unique solution, we must have:
, i.e.
Thus, for all real values of kâ, other than 10, the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
Hence, the required value of k is 10.
There is no value of k for which the given system of equations has an infinite number of solutions.
Answer:
The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0 ....(i)
And, (k − 1)x + (k + 2)y = 3k
⇒ (k − 1)x + (k + 2)y − 3k = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= 3, c1 = −7 and a2 = (k − 1), b2= (k + 2), c2 = −3k
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 7.
Page No 129:
Question 16:
The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0 ....(i)
And, (k − 1)x + (k + 2)y = 3k
⇒ (k − 1)x + (k + 2)y − 3k = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= 3, c1 = −7 and a2 = (k − 1), b2= (k + 2), c2 = −3k
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 7.
Answer:
The given system of equations:
2x + (k − 2)y = k
⇒ 2x + (k − 2)y − k = 0 ....(i)
And, 6x + (2k − 1)y = (2k + 5)
⇒ 6x + (2k − 1)y − (2k + 5) = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= (k − 2), c1 = −k and a2 = 6, b2= (2k − 1), c2 = −(2k + 5)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 5.
Page No 129:
Question 17:
The given system of equations:
2x + (k − 2)y = k
⇒ 2x + (k − 2)y − k = 0 ....(i)
And, 6x + (2k − 1)y = (2k + 5)
⇒ 6x + (2k − 1)y − (2k + 5) = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= (k − 2), c1 = −k and a2 = 6, b2= (2k − 1), c2 = −(2k + 5)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 5.
Answer:
The given system of equations:
kx + 3y = (2k + 1)
⇒ kx + 3y − (2k + 1) = 0 ...(i)
And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y − (7k + 1) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −(2k + 1) and a2 = 2(k + 1), b2= 9, c2 = −(7k + 1)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 2.
Page No 129:
Question 18:
The given system of equations:
kx + 3y = (2k + 1)
⇒ kx + 3y − (2k + 1) = 0 ...(i)
And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y − (7k + 1) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −(2k + 1) and a2 = 2(k + 1), b2= 9, c2 = −(7k + 1)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 2.
Answer:
The given system of equations:
5x + 2y = 2k
⇒ 5x + 2y − 2k= 0 ...(i)
And, 2(k + 1)x + ky = (3k + 4)
⇒ 2(k + 1)x + ky − (3k + 4) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= 2, c1 = −2k and a2 = 2(k + 1), b2= k, c2 = −(3k + 4)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 4.
Page No 129:
Question 19:
The given system of equations:
5x + 2y = 2k
⇒ 5x + 2y − 2k= 0 ...(i)
And, 2(k + 1)x + ky = (3k + 4)
⇒ 2(k + 1)x + ky − (3k + 4) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= 2, c1 = −2k and a2 = 2(k + 1), b2= k, c2 = −(3k + 4)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 4.
Answer:
The given system of equations:
(k − 1)x − y = 5
⇒ (k − 1)x − y − 5 = 0 ...(i)
And, (k + 1)x + (1 − k)y = (3k + 1)
⇒ (k + 1)x + (1 − k)y − (3k + 1) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = (k − 1), b1= −1, c1 = −5 and a2 = (k + 1), b2= (1 − k), c2 = −(3k + 1)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 3.
Page No 129:
Question 20:
The given system of equations:
(k − 1)x − y = 5
⇒ (k − 1)x − y − 5 = 0 ...(i)
And, (k + 1)x + (1 − k)y = (3k + 1)
⇒ (k + 1)x + (1 − k)y − (3k + 1) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = (k − 1), b1= −1, c1 = −5 and a2 = (k + 1), b2= (1 − k), c2 = −(3k + 1)
For an infinite number of solutions, we must have:
Now, we have the following three cases:
Case I:
Case II:
Case III:
Hence, the given system of equations has an infinite number of solutions when k is equal to 3.
Answer:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have an infinite number of solutions
, we must have
Hence, k = 6.
Page No 129:
Question 21:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have an infinite number of solutions
, we must have
Hence, k = 6.
Answer:
The given system of equations:
(a − 1)x + 3y = 2
⇒ (a − 1)x + 3y − 2 = 0 ...(i)
and 6x + (1 − 2b)y = 6
⇒ 6x + (1 − 2b)y − 6= 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
where, a1 = (a − 1), b1= 3, c1 = −2 and a2 = 6, b2= (1 − 2b), c2 = −6
For an infinite number of solutions, we must have:
⇒ 3a − 3 = 6 and 9 = 1 − 2b
⇒ 3a = 9 and 2b = −8
⇒ a = 3 and b = −4
∴â a = 3 and b = −4
Page No 130:
Question 22:
The given system of equations:
(a − 1)x + 3y = 2
⇒ (a − 1)x + 3y − 2 = 0 ...(i)
and 6x + (1 − 2b)y = 6
⇒ 6x + (1 − 2b)y − 6= 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
where, a1 = (a − 1), b1= 3, c1 = −2 and a2 = 6, b2= (1 − 2b), c2 = −6
For an infinite number of solutions, we must have:
⇒ 3a − 3 = 6 and 9 = 1 − 2b
⇒ 3a = 9 and 2b = −8
⇒ a = 3 and b = −4
∴â a = 3 and b = −4
Answer:
The given system of equations:
(2a − 1)x + 3y = 5
⇒ (2a − 1)x + 3y − 5 = 0 ....(i)
And, 3x + (b − 1)y = 2
⇒ 3x + (b − 1)y − 2 = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = (2a − 1), b1= 3, c1 = −5 and a2 = 3, b2= (b − 1), c2 = −2
For an infinite number of solutions, we must have:
⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒ 4a − 2 = 15 and 6 = 5b − 5
⇒ 4a = 17 and 5b = 11
∴â a = and b =
Page No 130:
Question 23:
The given system of equations:
(2a − 1)x + 3y = 5
⇒ (2a − 1)x + 3y − 5 = 0 ....(i)
And, 3x + (b − 1)y = 2
⇒ 3x + (b − 1)y − 2 = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = (2a − 1), b1= 3, c1 = −5 and a2 = 3, b2= (b − 1), c2 = −2
For an infinite number of solutions, we must have:
⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒ 4a − 2 = 15 and 6 = 5b − 5
⇒ 4a = 17 and 5b = 11
∴â a = and b =
Answer:
The given system of equations:
2x − 3y = 7
⇒ 2x − 3y − 7 = 0 ....(i)
And, (a + b)x − (a + b − 3)y = 4a + b
⇒ (a + b)x − (a + b − 3)y − (4a + b) = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= −3, c1 = −7 and a2 = (a + b), b2= −(a + b − 3), c2 = −(4a + b)
For an infinite number of solutions, we must have:
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b − 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b − 21
⇒ a = 5b ....(iii)
And, 5a = 4b − 21 ....(iv)
On substituting a = 5b in (iv), we get:
25b = 4b − 21
⇒ 21b = −21
⇒ b = −1
On substituting b = −1 in (iii), we get:
a = 5(−1) = −5
∴ a = −5 and b = −1
Page No 130:
Question 24:
The given system of equations:
2x − 3y = 7
⇒ 2x − 3y − 7 = 0 ....(i)
And, (a + b)x − (a + b − 3)y = 4a + b
⇒ (a + b)x − (a + b − 3)y − (4a + b) = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= −3, c1 = −7 and a2 = (a + b), b2= −(a + b − 3), c2 = −(4a + b)
For an infinite number of solutions, we must have:
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b − 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b − 21
⇒ a = 5b ....(iii)
And, 5a = 4b − 21 ....(iv)
On substituting a = 5b in (iv), we get:
25b = 4b − 21
⇒ 21b = −21
⇒ b = −1
On substituting b = −1 in (iii), we get:
a = 5(−1) = −5
∴ a = −5 and b = −1
Answer:
The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0 ....(i)
And, (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
⇒ (a + b + 1)x + (a + 2b + 2)y − [4(a + b) + 1] = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= 3, c1 = −7 and a2 = (a + b + 1), b2= (a + 2b + 2), c2 = −[4(a + b) + 1]
For an infinite number of solutions, we must have:
⇒ 2(a + 2b + 2) = 3(a + b+ 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)
⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14
⇒ a − b − 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
⇒ a − b = 1 and 5a − 2b = 11
a = (b + 1) ....(iii)
5a − 2b = 11 ....(iv)
On substituting a = (b + 1) in (iv), we get:
5(b + 1) − 2b = 11
⇒ 5b + 5 − 2b = 11
⇒ 3b = 6
⇒ b = 2
On substituting b = 2 in (iii), we get:
a = 3
∴â a = 3 and b = 2
Page No 130:
Question 25:
The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0 ....(i)
And, (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
⇒ (a + b + 1)x + (a + 2b + 2)y − [4(a + b) + 1] = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= 3, c1 = −7 and a2 = (a + b + 1), b2= (a + 2b + 2), c2 = −[4(a + b) + 1]
For an infinite number of solutions, we must have:
⇒ 2(a + 2b + 2) = 3(a + b+ 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)
⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14
⇒ a − b − 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
⇒ a − b = 1 and 5a − 2b = 11
a = (b + 1) ....(iii)
5a − 2b = 11 ....(iv)
On substituting a = (b + 1) in (iv), we get:
5(b + 1) − 2b = 11
⇒ 5b + 5 − 2b = 11
⇒ 3b = 6
⇒ b = 2
On substituting b = 2 in (iii), we get:
a = 3
∴â a = 3 and b = 2
Answer:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have an infinite number of solutions, we must have
Adding , we get
Now, substituting a = 5 in a + b = 6, we have
Hence, a = 5 and b = 1.
Page No 130:
Question 26:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have an infinite number of solutions, we must have
Adding , we get
Now, substituting a = 5 in a + b = 6, we have
Hence, a = 5 and b = 1.
Answer:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have an infinite number of solutions, we must have
Substituting a = 4 in a + b = 12, we get
Hence, a = 4 and b = 8.
Page No 130:
Question 27:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have an infinite number of solutions, we must have
Substituting a = 4 in a + b = 12, we get
Hence, a = 4 and b = 8.
Answer:
The given system of equations:
8x + 5y = 9
8x + 5y − 9 = 0 ....(i)
kx + 10y = 15
kx + 10y − 15= 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 8, b1= 5, c1 = −9 and a2 = k, b2= 10, c2 = −15
In order that the given system has no solution, we must have:
and
Hence, the given system of equations has no solution when k is equal to 16.
Page No 130:
Question 28:
The given system of equations:
8x + 5y = 9
8x + 5y − 9 = 0 ....(i)
kx + 10y = 15
kx + 10y − 15= 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 8, b1= 5, c1 = −9 and a2 = k, b2= 10, c2 = −15
In order that the given system has no solution, we must have:
and
Hence, the given system of equations has no solution when k is equal to 16.
Answer:
The given system of equations:
kx + 3y = 3
kx + 3y − 3 = 0 ....(i)
12x + ky = 6
12x + ky − 6= 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −3 and a2 = 12, b2= k, c2 = −6
In order that the given system of equations has no solution, we must have:
i.e.
and
Hence, the given system of equations has no solution when k is equal to −6.
Page No 130:
Question 29:
The given system of equations:
kx + 3y = 3
kx + 3y − 3 = 0 ....(i)
12x + ky = 6
12x + ky − 6= 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −3 and a2 = 12, b2= k, c2 = −6
In order that the given system of equations has no solution, we must have:
i.e.
and
Hence, the given system of equations has no solution when k is equal to −6.
Answer:
The given system of equations:
3x − y − 5 = 0 ...(i)
And, 6x − 2y + k = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 3, b1= −1, c1 = −5 and a2 = 6, b2= −2, c2 = k
In order that the given system of equations has no solution, we must have:
i.e.
⇒
Hence, equations (i) and (ii) will have no solution if .
Page No 130:
Question 30:
The given system of equations:
3x − y − 5 = 0 ...(i)
And, 6x − 2y + k = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 3, b1= −1, c1 = −5 and a2 = 6, b2= −2, c2 = k
In order that the given system of equations has no solution, we must have:
i.e.
⇒
Hence, equations (i) and (ii) will have no solution if .
Answer:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have no solution, we must have
Hence, .
Page No 130:
Question 31:
The given system of equations can be written as
This system is of the form
where
For the given system of linear equations to have no solution, we must have
Hence, .
Answer:
The given system of equations:
5x − 3y = 0 ....(i)
2x + ky = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= −3, c1 = 0 and a2 = 2, b2= k, c2 = 0
For a non-zero solution, we must have:
Hence, the required value of k is .
Page No 151:
Question 1:
The given system of equations:
5x − 3y = 0 ....(i)
2x + ky = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= −3, c1 = 0 and a2 = 2, b2= k, c2 = 0
For a non-zero solution, we must have:
Hence, the required value of k is .
Answer:
Let the cost of a chair be â¹ x and that of a table be â¹ y. Then
Multiplying (i) by 3 and (ii) by 4, we get
Substituting x = 560 in (i), we have
Hence, the cost of a chair and that of a table are respectively â¹ 560 and â¹ 700.
Page No 151:
Question 2:
Let the cost of a chair be â¹ x and that of a table be â¹ y. Then
Multiplying (i) by 3 and (ii) by 4, we get
Substituting x = 560 in (i), we have
Hence, the cost of a chair and that of a table are respectively â¹ 560 and â¹ 700.
Answer:
Let the cost of a spoon be â¹x and that of a fork be â¹y. Then
Adding (i) and (ii), we get
Now, subtracting (ii) from (i), we get
Adding (iii) and (iv), we get
Substituting x = 40 in (iii), we get
Hence, the cost of a spoon that of a fork are â¹40 and â¹50 respectively.
Page No 152:
Question 3:
Let the cost of a spoon be â¹x and that of a fork be â¹y. Then
Adding (i) and (ii), we get
Now, subtracting (ii) from (i), we get
Adding (iii) and (iv), we get
Substituting x = 40 in (iii), we get
Hence, the cost of a spoon that of a fork are â¹40 and â¹50 respectively.
Answer:
Let the x and y be the number of 50-paisa and 25-paisa conis respectively. Then
Multiplying (ii) by 2 and subtracting it from (i), we get
Subtracting y = 22 in (i), we get
Hence, the number of 25-paisa and 50-paisa conis are 22 and 28 respectively.
Page No 152:
Question 4:
Let the x and y be the number of 50-paisa and 25-paisa conis respectively. Then
Multiplying (ii) by 2 and subtracting it from (i), we get
Subtracting y = 22 in (i), we get
Hence, the number of 25-paisa and 50-paisa conis are 22 and 28 respectively.
Answer:
Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137 ...(i)
x − y = 43 ...(ii)
On adding (i) and (ii), we get:
2x = 180 ⇒ x = 90
On substituting x = 90 in (i), we get:
90 + y = 137
⇒ y = (137 − 90) = 47
Hence, the required numbers are 90 and 47.
Page No 152:
Question 5:
Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137 ...(i)
x − y = 43 ...(ii)
On adding (i) and (ii), we get:
2x = 180 ⇒ x = 90
On substituting x = 90 in (i), we get:
90 + y = 137
⇒ y = (137 − 90) = 47
Hence, the required numbers are 90 and 47.
Answer:
Let the first number be x and the second number be y.
Then, we have:
2x + 3y = 92 ....(i)
4x − 7y = 2 ....(ii)
On multiplying (i) by 7 and (ii) by 3, we get:
14x + 21y = 644 ....(iii)
12x − 21y = 6 ....(iv)
On adding (iii) and (iv), we get:
26x = 650
⇒ x = 25
On substituting x = 25 in (i), we get:
2 × 25 + 3y = 92
⇒ 50 + 3y = 92
⇒ 3y = (92 − 50) = 42
⇒ y = 14
Hence, the first number is 25 and the second number is 14.
Page No 152:
Question 6:
Let the first number be x and the second number be y.
Then, we have:
2x + 3y = 92 ....(i)
4x − 7y = 2 ....(ii)
On multiplying (i) by 7 and (ii) by 3, we get:
14x + 21y = 644 ....(iii)
12x − 21y = 6 ....(iv)
On adding (iii) and (iv), we get:
26x = 650
⇒ x = 25
On substituting x = 25 in (i), we get:
2 × 25 + 3y = 92
⇒ 50 + 3y = 92
⇒ 3y = (92 − 50) = 42
⇒ y = 14
Hence, the first number is 25 and the second number is 14.
Answer:
Let the first number be x and the second number be y.
Then, we have:
3x + y = 142 ....(i)
4x − y = 138 ....(ii)
On adding (i) and (ii), we get:
7x = 280
⇒ x = 40
On substituting x = 40 in (i), we get:
3 × 40 + y = 142
⇒ y = (142 − 120) = 22
⇒ y = 22
Hence, the first number is 40 and the second number is 22.
Page No 152:
Question 7:
Let the first number be x and the second number be y.
Then, we have:
3x + y = 142 ....(i)
4x − y = 138 ....(ii)
On adding (i) and (ii), we get:
7x = 280
⇒ x = 40
On substituting x = 40 in (i), we get:
3 × 40 + y = 142
⇒ y = (142 − 120) = 22
⇒ y = 22
Hence, the first number is 40 and the second number is 22.
Answer:
Let the greater number be x and the smaller be y.
Then, we have:
2x − 45 = y or 2x − y = 45 .... (i)
2y − 21 = x or −x + 2y = 21 ....(ii)
On multiplying (i) by 2, we get:
4x − 2y = 90 ....(iii)
On adding (ii) and (iii), we get:
3x = (90 + 21) = 111
⇒ x = 37
On substituting x = 37 in (i), we get:
2 × 37 − y = 45
⇒ 74 − y = 45
⇒ y = (74 − 45) = 29
Hence, the greater number is 37 and the smaller number is 29.
Page No 152:
Question 8:
Let the greater number be x and the smaller be y.
Then, we have:
2x − 45 = y or 2x − y = 45 .... (i)
2y − 21 = x or −x + 2y = 21 ....(ii)
On multiplying (i) by 2, we get:
4x − 2y = 90 ....(iii)
On adding (ii) and (iii), we get:
3x = (90 + 21) = 111
⇒ x = 37
On substituting x = 37 in (i), we get:
2 × 37 − y = 45
⇒ 74 − y = 45
⇒ y = (74 − 45) = 29
Hence, the greater number is 37 and the smaller number is 29.
Answer:
We know:
Dividend = Divisor × Quotient + Remainder
Let the larger number be x and the smaller be y.
Then, we have:
3x = y × 4 + 8 or 3x − 4y = 8 ....(i)
5y = x × 3 + 5 or −3x + 5y = 5 ....(ii)
On adding (i) and (ii), we get:
y = (8 + 5) = 13
On substituting y = 13 in (i) we get:
3x − 4 × 13 = 8
⇒ 3x = (8 + 52) = 60
⇒ x = 20
Hence, the larger number is 20 and the smaller number is 13.
Page No 152:
Question 9:
We know:
Dividend = Divisor × Quotient + Remainder
Let the larger number be x and the smaller be y.
Then, we have:
3x = y × 4 + 8 or 3x − 4y = 8 ....(i)
5y = x × 3 + 5 or −3x + 5y = 5 ....(ii)
On adding (i) and (ii), we get:
y = (8 + 5) = 13
On substituting y = 13 in (i) we get:
3x − 4 × 13 = 8
⇒ 3x = (8 + 52) = 60
⇒ x = 20
Hence, the larger number is 20 and the smaller number is 13.
Answer:
Let the required numbers be x and y.
Now, we have:
By cross multiplication, we get:
2x + 4 = y + 2
⇒ 2x − y = −2 ....(i)
Again, we have:
By cross multiplication, we get:
11x − 44 = 5y − 20
⇒ 11x − 5y = 24 ....(ii)
On multiplying (i) by 5, we get:
10x − 5y = −10 ....(iii)
On subtracting (iii) from (ii), we get:
x = (24 + 10) = 34
On substituting x = 34 in (i), we get:
2 × 34 − y = −2
⇒ 68 − y = −2
⇒ y = (68 + 2) = 70
Hence, the required numbers are 34 and 70.
Page No 152:
Question 10:
Let the required numbers be x and y.
Now, we have:
By cross multiplication, we get:
2x + 4 = y + 2
⇒ 2x − y = −2 ....(i)
Again, we have:
By cross multiplication, we get:
11x − 44 = 5y − 20
⇒ 11x − 5y = 24 ....(ii)
On multiplying (i) by 5, we get:
10x − 5y = −10 ....(iii)
On subtracting (iii) from (ii), we get:
x = (24 + 10) = 34
On substituting x = 34 in (i), we get:
2 × 34 − y = −2
⇒ 68 − y = −2
⇒ y = (68 + 2) = 70
Hence, the required numbers are 34 and 70.
Answer:
Let the larger number be x and the smaller number be y.
Then, we have:
x − y = 14 or x = 14 + y ....(i)
x2 − y2 = 448 ....(ii)
On substituting x = 14 + y in (ii), we get:
(14 + y)2 − y2 = 448
⇒ 196 + y2 + 28y − y2 = 448
⇒ 196 + 28y = 448
⇒ 28y = (448 − 196) = 252
⇒
On substituting y = 9 in (i), we get:
x = 14 + 9 = 23
Hence, the required numbers are 23 and 9.
Page No 152:
Question 11:
Let the larger number be x and the smaller number be y.
Then, we have:
x − y = 14 or x = 14 + y ....(i)
x2 − y2 = 448 ....(ii)
On substituting x = 14 + y in (ii), we get:
(14 + y)2 − y2 = 448
⇒ 196 + y2 + 28y − y2 = 448
⇒ 196 + 28y = 448
⇒ 28y = (448 − 196) = 252
⇒
On substituting y = 9 in (i), we get:
x = 14 + 9 = 23
Hence, the required numbers are 23 and 9.
Answer:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
x + y = 12 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 18
⇒ 10y + x − 10x − y = 18
⇒ 9y − 9x = 18
⇒ y − x = 2 ....(ii)
On adding (i) and (ii), we get:
2y = 14
⇒ y = 7
On substituting y = 7 in (i), we get:
x + 7 = 12
⇒ x = (12 − 7) = 5
Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.
Page No 152:
Question 12:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
x + y = 12 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 18
⇒ 10y + x − 10x − y = 18
⇒ 9y − 9x = 18
⇒ y − x = 2 ....(ii)
On adding (i) and (ii), we get:
2y = 14
⇒ y = 7
On substituting y = 7 in (i), we get:
x + 7 = 12
⇒ x = (12 − 7) = 5
Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.
Answer:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
10x + y = 7(x + y)
10x + y = 7x + 7y or 3x − 6y = 0 ....(i)
Number obtained on reversing its digits = (10y + x)
(10x + y) − 27 = 10y + x
⇒ 10x − x + y − 10y = 27
⇒ 9x − 9y = 27
⇒ 9(x − y) = 27
⇒ x − y = 3 ....(ii)
On multiplying (ii) by 6, we get:
6x − 6y = 18 ....(iii)
On subtracting (i) from (iii), we get:
3x = 18
⇒ x = 6
On substituting x = 6 in (i), we get:
3 × 6 − 6y = 0
⇒ 18 − 6y = 0
⇒ 6y = 18
⇒ y = 3
Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63
Hence, the required number is 63.
Page No 152:
Question 13:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
10x + y = 7(x + y)
10x + y = 7x + 7y or 3x − 6y = 0 ....(i)
Number obtained on reversing its digits = (10y + x)
(10x + y) − 27 = 10y + x
⇒ 10x − x + y − 10y = 27
⇒ 9x − 9y = 27
⇒ 9(x − y) = 27
⇒ x − y = 3 ....(ii)
On multiplying (ii) by 6, we get:
6x − 6y = 18 ....(iii)
On subtracting (i) from (iii), we get:
3x = 18
⇒ x = 6
On substituting x = 6 in (i), we get:
3 × 6 − 6y = 0
⇒ 18 − 6y = 0
⇒ 6y = 18
⇒ y = 3
Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63
Hence, the required number is 63.
Answer:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
x + y = 15 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 9
⇒ 10y + x − 10x − y = 9
⇒ 9y − 9x = 9
⇒ y − x = 1 ....(ii)
On adding (i) and (ii), we get:
2y = 16
⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15
⇒ x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.
Page No 152:
Question 14:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
x + y = 15 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 9
⇒ 10y + x − 10x − y = 9
⇒ 9y − 9x = 9
⇒ y − x = 1 ....(ii)
On adding (i) and (ii), we get:
2y = 16
⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15
⇒ x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.
Answer:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
∴ 10x + y = 4(x + y) + 3
⇒10x + y = 4x + 4y + 3
⇒ 6x − 3y = 3
⇒ 2x − y = 1 ....(i)
Again, we have:
10x + y + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ x − y = −2 ....(ii)
On subtracting (ii) from (i), we get:
x = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = 1
⇒ y = 6 − 1 = 5
Required number = (10x + y) = 10 × 3 + 5 = 30 + 5 = 35
Hence, the required number is 35.
Page No 152:
Question 15:
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
∴ 10x + y = 4(x + y) + 3
⇒10x + y = 4x + 4y + 3
⇒ 6x − 3y = 3
⇒ 2x − y = 1 ....(i)
Again, we have:
10x + y + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ x − y = −2 ....(ii)
On subtracting (ii) from (i), we get:
x = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = 1
⇒ y = 6 − 1 = 5
Required number = (10x + y) = 10 × 3 + 5 = 30 + 5 = 35
Hence, the required number is 35.
Answer:
We know:
Dividend = (Divisor × Quotient) + Remainder
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
∴ 10x + y = (x + y) × 6 + 0
⇒ 10x − 6x + y − 6y = 0
⇒ 4x − 5y = 0 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ 10x + y − 9 = 10y + x
⇒ 9x − 9y = 9
⇒ 9(x − y) = 9
⇒ x − y = 1 ....(ii)
On multiplying (ii) by 5, we get:
5x − 5y = 5 ....(iii)
On subtracting (i) from (iii), we get:
x = 5
On substituting x = 5 in (i), we get:
4 × 5 − 5y = 0
⇒ 20 − 5y = 0
⇒ y = 4
∴ Number = (10x + y) = 10 × 5 + 4 = 50 + 4 = 54
Hence, the required number is 54.
Page No 153:
Question 16:
We know:
Dividend = (Divisor × Quotient) + Remainder
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
∴ 10x + y = (x + y) × 6 + 0
⇒ 10x − 6x + y − 6y = 0
⇒ 4x − 5y = 0 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ 10x + y − 9 = 10y + x
⇒ 9x − 9y = 9
⇒ 9(x − y) = 9
⇒ x − y = 1 ....(ii)
On multiplying (ii) by 5, we get:
5x − 5y = 5 ....(iii)
On subtracting (i) from (iii), we get:
x = 5
On substituting x = 5 in (i), we get:
4 × 5 − 5y = 0
⇒ 20 − 5y = 0
⇒ y = 4
∴ Number = (10x + y) = 10 × 5 + 4 = 50 + 4 = 54
Hence, the required number is 54.
Answer:
Let the tens and the units digits of the required number be x and yâ, respectively.
Then, we have:
xy = 35 ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ 9(y − x) = 18
⇒ y − x = 2 ....(ii)
We know:
(y + x)2 − (y − x)2 = 4xy
⇒
⇒
∴ y + x = 12 .....(iii) (âµ x and y cannot be negative)
On adding (ii) and (iii), we get:
2y = 2 + 12 = 14
⇒ y = 7
On substituting y = 7 in (ii), we get:
7 − x = 2
⇒ x = (7 − 2) = 5
∴ Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.
Page No 153:
Question 17:
Let the tens and the units digits of the required number be x and yâ, respectively.
Then, we have:
xy = 35 ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ 9(y − x) = 18
⇒ y − x = 2 ....(ii)
We know:
(y + x)2 − (y − x)2 = 4xy
⇒
⇒
∴ y + x = 12 .....(iii) (âµ x and y cannot be negative)
On adding (ii) and (iii), we get:
2y = 2 + 12 = 14
⇒ y = 7
On substituting y = 7 in (ii), we get:
7 − x = 2
⇒ x = (7 − 2) = 5
∴ Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.
Answer:
Let the tens and the units digits of the required number be x and yâ, respectively.
Then, we have:
xy = 18 ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) − 63 = 10y + x
⇒ 9x − 9y = 63
⇒ 9(x − y) = 63
⇒ x − y = 7 ....(ii)
We know:
(x + y)2 − (x − y)2 = 4xy
⇒
∴ x + y = 11 ....(iii) (âµ x and y cannot be negative)
On adding (ii) and (iii), we get:
2x = 7 + 11 = 18
⇒ x = 9
On substituting x = 9 in (ii), we get:
9 − y = 7
⇒ y = (9 − 7) = 2
∴ Number = (10x + y) = 10 × 9 + 2 = 90 + 2 = 92
Hence, the required number is 92.
Page No 153:
Question 18:
Let the tens and the units digits of the required number be x and yâ, respectively.
Then, we have:
xy = 18 ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) − 63 = 10y + x
⇒ 9x − 9y = 63
⇒ 9(x − y) = 63
⇒ x − y = 7 ....(ii)
We know:
(x + y)2 − (x − y)2 = 4xy
⇒
∴ x + y = 11 ....(iii) (âµ x and y cannot be negative)
On adding (ii) and (iii), we get:
2x = 7 + 11 = 18
⇒ x = 9
On substituting x = 9 in (ii), we get:
9 − y = 7
⇒ y = (9 − 7) = 2
∴ Number = (10x + y) = 10 × 9 + 2 = 90 + 2 = 92
Hence, the required number is 92.
Answer:
Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per the question
Since the digits differ by 3, so
Adding (i) and (ii), we get
Putting x = 7 in (i), we get
Changing the role of x and y, x = 4 and y = 7
Hence, the two-digit number is 74 or 47.
Page No 153:
Question 19:
Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per the question
Since the digits differ by 3, so
Adding (i) and (ii), we get
Putting x = 7 in (i), we get
Changing the role of x and y, x = 4 and y = 7
Hence, the two-digit number is 74 or 47.
Answer:
Let the required fraction be .
Then, we have:
x + y = 8 ....(i)
And,
⇒ 4(x + 3) = 3(y + 3)
⇒ 4x + 12 = 3y + 9
⇒ 4x − 3y = −3 ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 24 ....(iii)
On adding (ii) and (iii), we get:
7x = 21
⇒ x = 3
On substituting x = 3 in (i), we get:
3 + y = 8
⇒ y = (8 − 3) = 5
∴â âx = 3 and y = 5
Hence, the required fraction is .
Page No 153:
Question 20:
Let the required fraction be .
Then, we have:
x + y = 8 ....(i)
And,
⇒ 4(x + 3) = 3(y + 3)
⇒ 4x + 12 = 3y + 9
⇒ 4x − 3y = −3 ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 24 ....(iii)
On adding (ii) and (iii), we get:
7x = 21
⇒ x = 3
On substituting x = 3 in (i), we get:
3 + y = 8
⇒ y = (8 − 3) = 5
∴â âx = 3 and y = 5
Hence, the required fraction is .
Answer:
Let the required fraction be .
Then, we have:
⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2x − y = −4 .....(i)
Again,
⇒ 3x = 1(y − 1)
⇒ 3x − y = −1 .....(ii)
On subtracting (i) from (ii), we get:
x = (−1 + 4) = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = −4
⇒ 6 − y = −4
⇒ y = (6 + 4) = 10
∴ x = 3 and y = 10
Hence, the required fraction is .
Page No 153:
Question 21:
Let the required fraction be .
Then, we have:
⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2x − y = −4 .....(i)
Again,
⇒ 3x = 1(y − 1)
⇒ 3x − y = −1 .....(ii)
On subtracting (i) from (ii), we get:
x = (−1 + 4) = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = −4
⇒ 6 − y = −4
⇒ y = (6 + 4) = 10
∴ x = 3 and y = 10
Hence, the required fraction is .
Answer:
Let the required fraction be .
Then, we have:
y = x + 11
⇒ y − x = 11 ....(i)
Again,
⇒ 4(x + 8) =3(y + 8)
⇒ 4x + 32 = 3y + 24
⇒ 4x − 3y = −8 ....(ii)
On multiplying (i) by 4, we get:
4y − 4x = 44 ....(iii)
On adding (ii) and (iii), we get:
y = (−8 + 44) = 36
On substituting y = 36 in (i), we get:
36 − x = 11
⇒ x = (36 − 11) = 25
∴â âx = 25 and y = 36
Hence, the required fraction is .
Page No 153:
Question 22:
Let the required fraction be .
Then, we have:
y = x + 11
⇒ y − x = 11 ....(i)
Again,
⇒ 4(x + 8) =3(y + 8)
⇒ 4x + 32 = 3y + 24
⇒ 4x − 3y = −8 ....(ii)
On multiplying (i) by 4, we get:
4y − 4x = 44 ....(iii)
On adding (ii) and (iii), we get:
y = (−8 + 44) = 36
On substituting y = 36 in (i), we get:
36 − x = 11
⇒ x = (36 − 11) = 25
∴â âx = 25 and y = 36
Hence, the required fraction is .
Answer:
Let the required fraction be .
Then, we have:
⇒ 2(x − 1) = 1(y + 2)
⇒ 2x − 2 = y + 2
⇒ 2x − y = 4 ....(i)
Again,
⇒ 3(x − 7) = 1(y − 2)
⇒ 3x − 21 = y − 2
⇒ 3x − y = 19 ....(ii)
On subtracting (i) from (ii), we get:
x = (19 − 4) = 15
On substituting x = 15 in (i), we get:
2 × 15 − y = 4
⇒ 30 − y = 4
⇒ y = 26
∴â âx = 15 and y = 26
Hence, the given fraction is .
Page No 153:
Question 23:
Let the required fraction be .
Then, we have:
⇒ 2(x − 1) = 1(y + 2)
⇒ 2x − 2 = y + 2
⇒ 2x − y = 4 ....(i)
Again,
⇒ 3(x − 7) = 1(y − 2)
⇒ 3x − 21 = y − 2
⇒ 3x − y = 19 ....(ii)
On subtracting (i) from (ii), we get:
x = (19 − 4) = 15
On substituting x = 15 in (i), we get:
2 × 15 − y = 4
⇒ 30 − y = 4
⇒ y = 26
∴â âx = 15 and y = 26
Hence, the given fraction is .
Answer:
Let the fraction be .
As per the question
After changing the numerator and denominator
New numerator = x + 3
New denominator = y + 3
Therefore
Multiplying (i) by 3 and subtracting (ii), we get
Now, putting y = 9 in (i), we get
Hence, the fraction is .
Page No 153:
Question 24:
Let the fraction be .
As per the question
After changing the numerator and denominator
New numerator = x + 3
New denominator = y + 3
Therefore
Multiplying (i) by 3 and subtracting (ii), we get
Now, putting y = 9 in (i), we get
Hence, the fraction is .
Answer:
Let the larger number be x and the smaller number be y.
Then, we have:
(x + y) = 16 ....(i)
And, ....(ii)
⇒
⇒ 3(x + y) = xy
⇒ 3 × 16 = xy [Since from (i), we have: x + y = 16]
∴ xy = 48 ....(iii)
We know:
(x − y)2 = (x + y)2 − 4xy
(x − y)2 = (16)2 − 4 × 48 = 256 − 192 = 64
∴ (x − y) =
Since x is larger and y is smaller, we have:
x − y = 8 .....(iv)
On adding (i) and (iv), we get:
2x = 24
⇒ x = 12
On substituting x = 12 in (i), we get:
12 + y = 16 ⇒ y = (16 − 12) = 4
Hence, the required numbers are 12 and 4.
Page No 153:
Question 25:
Let the larger number be x and the smaller number be y.
Then, we have:
(x + y) = 16 ....(i)
And, ....(ii)
⇒
⇒ 3(x + y) = xy
⇒ 3 × 16 = xy [Since from (i), we have: x + y = 16]
∴ xy = 48 ....(iii)
We know:
(x − y)2 = (x + y)2 − 4xy
(x − y)2 = (16)2 − 4 × 48 = 256 − 192 = 64
∴ (x − y) =
Since x is larger and y is smaller, we have:
x − y = 8 .....(iv)
On adding (i) and (iv), we get:
2x = 24
⇒ x = 12
On substituting x = 12 in (i), we get:
12 + y = 16 ⇒ y = (16 − 12) = 4
Hence, the required numbers are 12 and 4.
Answer:
Let the number of students in classroom A be x
Let the number of students in classroom B be y.
If 10 students are transferred from A to B, then we have:
x − 10 = y + 10
⇒ x − y = 20 ...(i)
If 20 students are transferred from B to A, then we have:
2(y − 20) = x + 20
⇒ 2y − 40 = x + 20
⇒ −x + 2y = 60 ...(ii)
On adding (i) and (ii), we get:
y = (20 + 60) = 80
On substituting y = 80 in (i), we get:
x − 80 = 20
⇒ x = (20 + 80) = 100
Hence, the number of students in classroom A is 100 and the number of students in classroom B is 80.
Page No 153:
Question 26:
Let the number of students in classroom A be x
Let the number of students in classroom B be y.
If 10 students are transferred from A to B, then we have:
x − 10 = y + 10
⇒ x − y = 20 ...(i)
If 20 students are transferred from B to A, then we have:
2(y − 20) = x + 20
⇒ 2y − 40 = x + 20
⇒ −x + 2y = 60 ...(ii)
On adding (i) and (ii), we get:
y = (20 + 60) = 80
On substituting y = 80 in (i), we get:
x − 80 = 20
⇒ x = (20 + 80) = 100
Hence, the number of students in classroom A is 100 and the number of students in classroom B is 80.
Answer:
Let fixed charges be â¹x and rate per km be â¹y.
Then as per the question
Subtracting (i) from (ii), we get
Now, putting y = 16, we have
Hence, the fixed charges be â¹50 and the rate per km is â¹16.
Page No 154:
Question 27:
Let fixed charges be â¹x and rate per km be â¹y.
Then as per the question
Subtracting (i) from (ii), we get
Now, putting y = 16, we have
Hence, the fixed charges be â¹50 and the rate per km is â¹16.
Answer:
Let the fixed charges be â¹x and the cost of food per day be â¹y.
Then as per the question
Subtracting (i) from (ii), we get
Now, putting y = 140, we have
Hence, the fixed charges is â¹1000 and the cost of the food per day is â¹140.
Page No 154:
Question 28:
Let the fixed charges be â¹x and the cost of food per day be â¹y.
Then as per the question
Subtracting (i) from (ii), we get
Now, putting y = 140, we have
Hence, the fixed charges is â¹1000 and the cost of the food per day is â¹140.
Answer:
Let the the amounts invested at 10% and 8% be â¹x and â¹y respectively.
Then as per the question
After the amounts interchanged but the rate being the same, we have
Adding (i) and (ii) and dividing by 9, we get
Subtracting (ii) from (i), we get
Now, adding (iii) and (iv), we have
Putting x = 8500 in (iii), we get
Hence, the amounts invested are â¹8,500 at 10% and â¹6,250 at 8%.
Page No 154:
Question 29:
Let the the amounts invested at 10% and 8% be â¹x and â¹y respectively.
Then as per the question
After the amounts interchanged but the rate being the same, we have
Adding (i) and (ii) and dividing by 9, we get
Subtracting (ii) from (i), we get
Now, adding (iii) and (iv), we have
Putting x = 8500 in (iii), we get
Hence, the amounts invested are â¹8,500 at 10% and â¹6,250 at 8%.
Answer:
Let the monthly income of A and B are â¹x and â¹y respectively.
Then as per the question
Since each save â¹9,000, so
Expenditure of A = ââ¹
Expenditure of B = ââ¹
The ratio of expenditures of A and B are in the ratio 7 : 5.
From (i), substitute in (ii) to get
Now, putting x = 30000, we get
Hence, the monthly incomes of A and B are ââ¹30,000 and ââ¹24,000.
Page No 154:
Question 30:
Let the monthly income of A and B are â¹x and â¹y respectively.
Then as per the question
Since each save â¹9,000, so
Expenditure of A = ââ¹
Expenditure of B = ââ¹
The ratio of expenditures of A and B are in the ratio 7 : 5.
From (i), substitute in (ii) to get
Now, putting x = 30000, we get
Hence, the monthly incomes of A and B are ââ¹30,000 and ââ¹24,000.
Answer:
Let the cost price of the chair and table be â¹x and â¹y respectively.
Then as per the question
Selling price of chair + Selling price of table = 1520
When the profit on chair and table are 10% and 25% respectively, then
Solving (i) and (ii) by cross multiplication, we get
Hence, the cost of chair and table are ââ¹600 and ââ¹700 respectively.
Page No 154:
Question 31:
Let the cost price of the chair and table be â¹x and â¹y respectively.
Then as per the question
Selling price of chair + Selling price of table = 1520
When the profit on chair and table are 10% and 25% respectively, then
Solving (i) and (ii) by cross multiplication, we get
Hence, the cost of chair and table are ââ¹600 and ââ¹700 respectively.
Answer:
Let X and Y be the cars starting from points A and B, respectively and let their speeds be x km/h and y km/h, respectively.
Then, we have the following cases:
Case I: When the two cars move in the same direction
In this case, let the two cars meet at point M.
Distance covered by car X in 7 hours = 7x km
Distance covered by car Y in 7 hours = 7y km
∴ AM = (7x) km and BM = (7y) km
⇒ (AM − BM) = AB
⇒ (7x − 7y) = 70
⇒ 7(x − y) = 70
⇒ (x − y) = 10 ....(i)
Case II: When the two cars move in opposite directions
In this case, let the two cars meet at point N.
Distance covered by car X in 1 hour = x km
Distance covered by car Y in 1 hour = y km
∴ AN = x km and BN = y km
⇒ AN + BN = AB
⇒ x + y = 70 ....(ii)
On adding (i) and (ii), we get:
2x = 80
⇒ x = 40
On substituting x = 40 in (i), we get:
40 − y = 10
⇒ y = (40 − 10) = 30
Hence, the speed of car X is 40 km/h and the speed of car Y is 30 km/h.
Page No 154:
Question 32:
Let X and Y be the cars starting from points A and B, respectively and let their speeds be x km/h and y km/h, respectively.
Then, we have the following cases:
Case I: When the two cars move in the same direction
In this case, let the two cars meet at point M.
Distance covered by car X in 7 hours = 7x km
Distance covered by car Y in 7 hours = 7y km
∴ AM = (7x) km and BM = (7y) km
⇒ (AM − BM) = AB
⇒ (7x − 7y) = 70
⇒ 7(x − y) = 70
⇒ (x − y) = 10 ....(i)
Case II: When the two cars move in opposite directions
In this case, let the two cars meet at point N.
Distance covered by car X in 1 hour = x km
Distance covered by car Y in 1 hour = y km
∴ AN = x km and BN = y km
⇒ AN + BN = AB
⇒ x + y = 70 ....(ii)
On adding (i) and (ii), we get:
2x = 80
⇒ x = 40
On substituting x = 40 in (i), we get:
40 − y = 10
⇒ y = (40 − 10) = 30
Hence, the speed of car X is 40 km/h and the speed of car Y is 30 km/h.
Answer:
Let the original speed be x kmph and let the time taken to complete the journey be y hours.
∴ Length of the whole journey = (xy) km
Case I:
When the speed is (x + 5) kmph and the time taken is (y − 3) hrs:
Total journey = (x + 5) (y − 3) km
⇒ (x + 5) (y − 3) = xy
⇒ xy + 5y − 3x − 15 = xy
⇒ 5y − 3x = 15 ...(i)
Case II:
When the speed is (x − 4) kmph and the time taken is (y + 3) hrs:
Total journey = (x − 4) (y + 3) km
⇒ (x − 4) (y + 3) = xy
⇒ xy − 4y + 3x − 12 = xy
⇒ 3x − 4y = 12 ...(ii)
On adding (i) and (ii), we get:
y = 27
On substituting y = 27 in (i), we get:
5 × 27 − 3x = 15
⇒ 135 − 3x = 15
⇒ 3x = 120
⇒ x = 40
∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km
Page No 154:
Question 33:
Let the original speed be x kmph and let the time taken to complete the journey be y hours.
∴ Length of the whole journey = (xy) km
Case I:
When the speed is (x + 5) kmph and the time taken is (y − 3) hrs:
Total journey = (x + 5) (y − 3) km
⇒ (x + 5) (y − 3) = xy
⇒ xy + 5y − 3x − 15 = xy
⇒ 5y − 3x = 15 ...(i)
Case II:
When the speed is (x − 4) kmph and the time taken is (y + 3) hrs:
Total journey = (x − 4) (y + 3) km
⇒ (x − 4) (y + 3) = xy
⇒ xy − 4y + 3x − 12 = xy
⇒ 3x − 4y = 12 ...(ii)
On adding (i) and (ii), we get:
y = 27
On substituting y = 27 in (i), we get:
5 × 27 − 3x = 15
⇒ 135 − 3x = 15
⇒ 3x = 120
⇒ x = 40
∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km
Answer:
Let the speed of the train and taxi be x km/h and y km/h respectively.
Then as per the question
When the speeds of the train and taxi are 260 km and 240 km respectively, then
Multiplying (i) by 6 and subtracting (ii) from it, we get
Putting x = 100 in (i), we have
Hence, the speed of the train and that of the taxi are 100 km/h and 80 km/h respectively.
Page No 154:
Question 34:
Let the speed of the train and taxi be x km/h and y km/h respectively.
Then as per the question
When the speeds of the train and taxi are 260 km and 240 km respectively, then
Multiplying (i) by 6 and subtracting (ii) from it, we get
Putting x = 100 in (i), we have
Hence, the speed of the train and that of the taxi are 100 km/h and 80 km/h respectively.
Answer:
Let the speed of the car A and B be x km/h and y km/h respectively . Let x > y.
Case-1: When they travel in the same direction
From the figure
Case-2: When they travel in opposite direction
From the figure
Adding (i) and (ii), we get
Putting x = 50 in (ii), we have
Hence, the speeds of the cars are 50 km/h and 30 km/h.
Page No 154:
Question 35:
Let the speed of the car A and B be x km/h and y km/h respectively . Let x > y.
Case-1: When they travel in the same direction
From the figure
Case-2: When they travel in opposite direction
From the figure
Adding (i) and (ii), we get
Putting x = 50 in (ii), we have
Hence, the speeds of the cars are 50 km/h and 30 km/h.
Answer:
Let the speed of the sailor in still water be x km/h and that of the current y km/h.
Speed downstream = (x + y) km/h
Speed upstream = (x − y) km/h
As per the question
When the sailor goes upstream, then
Adding (i) and (ii), we get
Putting x = 10 in (i), we have
Hence, the speeds of the sailor in staill water and the current are 10 km/h and 2 km/h respectively.
Page No 155:
Question 36:
Let the speed of the sailor in still water be x km/h and that of the current y km/h.
Speed downstream = (x + y) km/h
Speed upstream = (x − y) km/h
As per the question
When the sailor goes upstream, then
Adding (i) and (ii), we get
Putting x = 10 in (i), we have
Hence, the speeds of the sailor in staill water and the current are 10 km/h and 2 km/h respectively.
Answer:
Let speed of boat in still water be x km/h and speed of stream be y km/h.
Speed Upstream = (x − y) km/h
Speed downstream = (x + y) km/h
According to the question,
Hence, the speed of the stream and that of the boat in still water is 3 km/h and 8 km/h, respectively.
Page No 155:
Question 37:
Let speed of boat in still water be x km/h and speed of stream be y km/h.
Speed Upstream = (x − y) km/h
Speed downstream = (x + y) km/h
According to the question,
Hence, the speed of the stream and that of the boat in still water is 3 km/h and 8 km/h, respectively.
Answer:
Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days.
∴ One man's one day's work =
And, one boy's one day's work =
2 men and 5 boys can finish the work in 4 days.
∴ (2 men's one day's work) + (5 boys' one day's work) =
⇒
⇒ ...(i) Here,
Again, 3 men and 6 boys can finish the work in 3 days.
∴ (3 men's one day's work) + (6 boys' one day's work) =
⇒
⇒ ....(ii) Here,
On multiplying (i) by 6 and (ii) by 5, we get:
....(iii)
....(iv)
On subtracting (iii) from (iv), we get:
⇒
On substituting in (i), we get:
⇒
Hence, one man alone can finish the work in 18 days and one boy alone can finish the work in 36 days.
Page No 155:
Question 38:
Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days.
∴ One man's one day's work =
And, one boy's one day's work =
2 men and 5 boys can finish the work in 4 days.
∴ (2 men's one day's work) + (5 boys' one day's work) =
⇒
⇒ ...(i) Here,
Again, 3 men and 6 boys can finish the work in 3 days.
∴ (3 men's one day's work) + (6 boys' one day's work) =
⇒
⇒ ....(ii) Here,
On multiplying (i) by 6 and (ii) by 5, we get:
....(iii)
....(iv)
On subtracting (iii) from (iv), we get:
⇒
On substituting in (i), we get:
⇒
Hence, one man alone can finish the work in 18 days and one boy alone can finish the work in 36 days.
Answer:
Let the length of the room be x metres and the breadth of the room be y metres.
Then, we have:
Area of the room = xy
According to the question, we have:
x = y + 3
⇒ x − y = 3 ....(i)
And, (x + 3) (y − 2) = xy
⇒ xy − 2x + 3y − 6 = xy
⇒ 3y − 2x = 6 ....(ii)
On multiplying (i) by 2, we get:
2x − 2y = 6 ....(iii)
On adding (ii) and (iii), we get:
y = (6 + 6) = 12
On substituting y = 12 in (i), we get:
x − 12 = 3
⇒ x = (3 + 12) =15
Hence, the length of the room is 15 metres and its breadth is 12 metres.
Page No 155:
Question 39:
Let the length of the room be x metres and the breadth of the room be y metres.
Then, we have:
Area of the room = xy
According to the question, we have:
x = y + 3
⇒ x − y = 3 ....(i)
And, (x + 3) (y − 2) = xy
⇒ xy − 2x + 3y − 6 = xy
⇒ 3y − 2x = 6 ....(ii)
On multiplying (i) by 2, we get:
2x − 2y = 6 ....(iii)
On adding (ii) and (iii), we get:
y = (6 + 6) = 12
On substituting y = 12 in (i), we get:
x − 12 = 3
⇒ x = (3 + 12) =15
Hence, the length of the room is 15 metres and its breadth is 12 metres.
Answer:
Let the length and the breadth of the rectangle be x m and y m, respectively.
∴ Area of the rectangle = (xy) sq. m
Case 1: When the length is reduced by 5 m and the breadth is increased by 3 m:
New length = (x − 5) m
New breadth = (y + 3) m
∴ New area = (x − 5) (y + 3) sq. m
∴ xy − (x − 5) (y + 3) = 8
⇒ xy − [xy − 5y + 3x − 15] = 8
⇒ xy − xy + 5y − 3x + 15 = 8
⇒ 3x − 5y = 7 .....(i)
Case 2: When the length is increased by 3 m and the breadth is increased by 2 m:
New length = (x + 3) m
New breadth = (y + 2) m
∴ New area = (x + 3) (y + 2) sq. m
∴ (x + 3) (y + 2) − xy = 74
⇒ [xy + 3y + 2x + 6] − xy = 74
⇒ 2x + 3y = 68 .....(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 21 .....(iii)
10x + 15y = 340 .....(iv)
On adding (iii) and (iv), we get:
19x = 361
⇒ x = 19
On substituting x = 19 in (iii), we get:
9 × 19 − 15y = 21
⇒ 171 −15y = 21
⇒ 15y = (171 − 21) = 150
⇒ y = 10
Hence, the length is 19 m and the breadth is 10 m.
Page No 155:
Question 40:
Let the length and the breadth of the rectangle be x m and y m, respectively.
∴ Area of the rectangle = (xy) sq. m
Case 1: When the length is reduced by 5 m and the breadth is increased by 3 m:
New length = (x − 5) m
New breadth = (y + 3) m
∴ New area = (x − 5) (y + 3) sq. m
∴ xy − (x − 5) (y + 3) = 8
⇒ xy − [xy − 5y + 3x − 15] = 8
⇒ xy − xy + 5y − 3x + 15 = 8
⇒ 3x − 5y = 7 .....(i)
Case 2: When the length is increased by 3 m and the breadth is increased by 2 m:
New length = (x + 3) m
New breadth = (y + 2) m
∴ New area = (x + 3) (y + 2) sq. m
∴ (x + 3) (y + 2) − xy = 74
⇒ [xy + 3y + 2x + 6] − xy = 74
⇒ 2x + 3y = 68 .....(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 21 .....(iii)
10x + 15y = 340 .....(iv)
On adding (iii) and (iv), we get:
19x = 361
⇒ x = 19
On substituting x = 19 in (iii), we get:
9 × 19 − 15y = 21
⇒ 171 −15y = 21
⇒ 15y = (171 − 21) = 150
⇒ y = 10
Hence, the length is 19 m and the breadth is 10 m.
Answer:
Let the length and breadth of the rectangle be x m and y m respectively.
Case1: When length is increased by 3 m and breadth is decreased by 4 m
Case2: When length is reduced by 1 m and breadth is increased by 4 m
Subtracting (i) from (ii), we get
Now, putting y = 19 in (ii), we have
Hence, length = 28 m and breadth = 19 m.
Page No 155:
Question 41:
Let the length and breadth of the rectangle be x m and y m respectively.
Case1: When length is increased by 3 m and breadth is decreased by 4 m
Case2: When length is reduced by 1 m and breadth is increased by 4 m
Subtracting (i) from (ii), we get
Now, putting y = 19 in (ii), we have
Hence, length = 28 m and breadth = 19 m.
Answer:
Let the the basic first class full fare be â¹x and the reservation charge be â¹y.
Case 1: One reservation first class full ticket cost â¹4,150
Case 2: One full and one half reserved first class tickets cost â¹6,255
Substituting from (i) in (ii), we get
Now, putting x = 4090 in (i), we have
Hence, cost of basic first class full fare = â¹4,090 and reservation charge = â¹60.
Page No 155:
Question 42:
Let the the basic first class full fare be â¹x and the reservation charge be â¹y.
Case 1: One reservation first class full ticket cost â¹4,150
Case 2: One full and one half reserved first class tickets cost â¹6,255
Substituting from (i) in (ii), we get
Now, putting x = 4090 in (i), we have
Hence, cost of basic first class full fare = â¹4,090 and reservation charge = â¹60.
Answer:
Let the the present age of the man be x years and that of his son be y years.
After 5 years man's age = x + 5
After 5 years ago son's age = y + 5
As per the question
5 years ago man's age = x − 5
5 years ago son's age = y − 5
As per the question
Subtracting (ii) from (i), we have
Putting y = 10 in (i), we get
Hence, man's present age = 40 years and son's present age = 10 years.
Page No 155:
Question 43:
Let the the present age of the man be x years and that of his son be y years.
After 5 years man's age = x + 5
After 5 years ago son's age = y + 5
As per the question
5 years ago man's age = x − 5
5 years ago son's age = y − 5
As per the question
Subtracting (ii) from (i), we have
Putting y = 10 in (i), we get
Hence, man's present age = 40 years and son's present age = 10 years.
Answer:
Let the man's present age be x years.
Let his son's present age be y years.
According to question, we have:
Two years ago:
Age of the man = Five times the age of the son
⇒ (x − 2) = 5(y − 2)
⇒ x − 2 = 5y − 10
⇒ x − 5y = −8 .....(i)
Two years later:
Age of the man = Three times the age of the son + 8
⇒ (x + 2) = 3(y + 2) + 8
⇒ x + 2 = 3y + 6 + 8
⇒ x − 3y = 12 .....(ii)
Subtracting (i) from (ii), we get:
2y = 20
⇒ y = 10
On substituting y = 10 in (i), we get:
x − 5 × 10 = −8
⇒ x − 50 = −8
⇒ x = (−8 + 50) = 42
Hence, the present age of the man is 42 years and the present age of the son is 10 years.
Page No 155:
Question 44:
Let the man's present age be x years.
Let his son's present age be y years.
According to question, we have:
Two years ago:
Age of the man = Five times the age of the son
⇒ (x − 2) = 5(y − 2)
⇒ x − 2 = 5y − 10
⇒ x − 5y = −8 .....(i)
Two years later:
Age of the man = Three times the age of the son + 8
⇒ (x + 2) = 3(y + 2) + 8
⇒ x + 2 = 3y + 6 + 8
⇒ x − 3y = 12 .....(ii)
Subtracting (i) from (ii), we get:
2y = 20
⇒ y = 10
On substituting y = 10 in (i), we get:
x − 5 × 10 = −8
⇒ x − 50 = −8
⇒ x = (−8 + 50) = 42
Hence, the present age of the man is 42 years and the present age of the son is 10 years.
Answer:
Let the father's present age be x years.
Let her son's present age be y years.
Then, we have:
x + 2y = 70 ....(i)
And, 2x + y = 95 ....(ii)
On multiplying (ii) by 2, we get:
4x + 2y = 190 ....(iii)
On subtracting (i) from (iii), we get:
3x = 120
⇒ x = 40
On substituting x = 40 in (i), we get:
40 + 2y = 70
⇒ 2y = (70 − 40) = 30
⇒ y = 15
Hence, the father's present age is 40 years and her son's present age is 15 years.
Page No 155:
Question 45:
Let the father's present age be x years.
Let her son's present age be y years.
Then, we have:
x + 2y = 70 ....(i)
And, 2x + y = 95 ....(ii)
On multiplying (ii) by 2, we get:
4x + 2y = 190 ....(iii)
On subtracting (i) from (iii), we get:
3x = 120
⇒ x = 40
On substituting x = 40 in (i), we get:
40 + 2y = 70
⇒ 2y = (70 − 40) = 30
⇒ y = 15
Hence, the father's present age is 40 years and her son's present age is 15 years.
Answer:
Let the woman's present age be x years.
Let her daughter's present age be y years.
Then, we have:
x = 3y + 3
⇒ x − 3y = 3 ....(i)
After three years, we have:
(x + 3) = 2(y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒ x − 2y = 13 ....(ii)
Subtracting (ii) from (i), we get:
−y = (3 − 13) = −10
⇒ y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 3
⇒ x − 30 = 3
⇒ x = (3 + 30) = 33
Hence, the woman's present age is 33 years and her daughter's present age is 10 years.
Page No 155:
Question 46:
Let the woman's present age be x years.
Let her daughter's present age be y years.
Then, we have:
x = 3y + 3
⇒ x − 3y = 3 ....(i)
After three years, we have:
(x + 3) = 2(y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒ x − 2y = 13 ....(ii)
Subtracting (ii) from (i), we get:
−y = (3 − 13) = −10
⇒ y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 3
⇒ x − 30 = 3
⇒ x = (3 + 30) = 33
Hence, the woman's present age is 33 years and her daughter's present age is 10 years.
Answer:
Let the actual price of the tea and lemon set be â¹x and â¹y respectively.
When gain is â¹7, then
When gain is â¹14, then
Multiplying (i) by 2 and adding with (ii), we have
Putting y = 80 in (ii), we get
Hence, actual price of the tea set and lemon set are â¹100 and â¹80 respectively.
Page No 156:
Question 47:
Let the actual price of the tea and lemon set be â¹x and â¹y respectively.
When gain is â¹7, then
When gain is â¹14, then
Multiplying (i) by 2 and adding with (ii), we have
Putting y = 80 in (ii), we get
Hence, actual price of the tea set and lemon set are â¹100 and â¹80 respectively.
Answer:
Let the fixed charge be â¹x and the charge for each extra day be â¹y.
In case of Mona, as per the question
In case of Tanvy, as per the question
Subtracting (ii) from (i), we get
Now, putting y = 3 in (ii), we have
Hence, the fixed charge be â¹15 and the charge for each extra day is â¹3.
Page No 156:
Question 48:
Let the fixed charge be â¹x and the charge for each extra day be â¹y.
In case of Mona, as per the question
In case of Tanvy, as per the question
Subtracting (ii) from (i), we get
Now, putting y = 3 in (ii), we have
Hence, the fixed charge be â¹15 and the charge for each extra day is â¹3.
Answer:
Let the digit at the tens place be x and digit at the units place be y.
The number is 10x + y.
According to the question,
Page No 156:
Question 49:
Let the digit at the tens place be x and digit at the units place be y.
The number is 10x + y.
According to the question,
Answer:
Let the numerator of the fraction be x and the denominator be y.
According to the question,
Page No 156:
Question 50:
Let the numerator of the fraction be x and the denominator be y.
According to the question,
Answer:
Let the age of first son be x and of second son be y.
Then, the father's present age be 3(x + y).
According to the question,
After 5 years, the father's age will be two times the sum of the ages of his two children.
Hence, the present age of the father is 45 years.
Page No 156:
Question 51:
Let the age of first son be x and of second son be y.
Then, the father's present age be 3(x + y).
According to the question,
After 5 years, the father's age will be two times the sum of the ages of his two children.
Hence, the present age of the father is 45 years.
Answer:
Let the length of the side of one square be x m and the length of the side of another square be y m.
According to the question,
Page No 156:
Question 52:
Let the length of the side of one square be x m and the length of the side of another square be y m.
According to the question,
Answer:
Let the length of the side of one square be x m and the length of the side of another square be y m.
According to the question,
Page No 156:
Question 53:
Let the length of the side of one square be x m and the length of the side of another square be y m.
According to the question,
Answer:
Let x litres and y litres be the amount of acids from 50% and 25% acid solutions respectively.
As per the question
Since, the total volume is 10 litres, so
Subtracting (ii) from (i), we get
Now, putting x = 6 in (ii), we have
Hence, volume of 50% acid solution = 6 litres and volume of 25% acid solution = 4 litres.
Page No 156:
Question 54:
Let x litres and y litres be the amount of acids from 50% and 25% acid solutions respectively.
As per the question
Since, the total volume is 10 litres, so
Subtracting (ii) from (i), we get
Now, putting x = 6 in (ii), we have
Hence, volume of 50% acid solution = 6 litres and volume of 25% acid solution = 4 litres.
Answer:
Let x g and y g be the weight of 18-carat and 12-carat gold respectively.
As per the given condition
And
Multiplying (ii) by 2 and subtracting from (i), we get
Now, putting x = 80 in (ii), we have
Hence, the required weight of 18-carat and 12-carat gold bars are 80 g and 40 g respectively.
Page No 156:
Question 55:
Let x g and y g be the weight of 18-carat and 12-carat gold respectively.
As per the given condition
And
Multiplying (ii) by 2 and subtracting from (i), we get
Now, putting x = 80 in (ii), we have
Hence, the required weight of 18-carat and 12-carat gold bars are 80 g and 40 g respectively.
Answer:
Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions.
As per the given condition
And
From (ii), subtitute in (i) to get
Now, putting x = 6 in (ii), we have
Hence, the required quantities are 6 litres and 15 litres.
Page No 156:
Question 56:
Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions.
As per the given condition
And
From (ii), subtitute in (i) to get
Now, putting x = 6 in (ii), we have
Hence, the required quantities are 6 litres and 15 litres.
Answer:
Let x and y be the supplementary angles, where x > y.
As per the given condition
And
Adding (i) and (ii), we get
Now, substituting in (ii), we have
Hence, the required angles are and .
Page No 156:
Question 57:
Let x and y be the supplementary angles, where x > y.
As per the given condition
And
Adding (i) and (ii), we get
Now, substituting in (ii), we have
Hence, the required angles are and .
Answer:
The sum of all the angles of a triangle is , therefore
Subtracting (i) from (ii), we have
Now, substituting in (i), we have
Thus
Hence, the angles are .
Page No 156:
Question 58:
The sum of all the angles of a triangle is , therefore
Subtracting (i) from (ii), we have
Now, substituting in (i), we have
Thus
Hence, the angles are .
Answer:
The opposite angles of cyclic quadrilateral are supplementary, so
And
Subtracting (i) from (ii), we have
Now, substituting in (i), we have
Therefore
Hence, .
Page No 161:
Question 1:
The opposite angles of cyclic quadrilateral are supplementary, so
And
Subtracting (i) from (ii), we have
Now, substituting in (i), we have
Therefore
Hence, .
Answer:
The given equations are
Which is of the form , where
Now
Thus, the pair of linear equations are coincident and therefore has infinitely many solutions.
Page No 161:
Question 2:
The given equations are
Which is of the form , where
Now
Thus, the pair of linear equations are coincident and therefore has infinitely many solutions.
Answer:
The given equations are
Which is of the form , where
For the given pair of linear equations to have infinitely many solutions, we must have
Hence, k = 7.
Page No 162:
Question 3:
The given equations are
Which is of the form , where
For the given pair of linear equations to have infinitely many solutions, we must have
Hence, k = 7.
Answer:
The given pair of linear equations are
Which is of the form , where
For the given pair of linear equations to have infinitely many solutions, we must have
Hence, k = 10.
Page No 162:
Question 4:
The given pair of linear equations are
Which is of the form , where
For the given pair of linear equations to have infinitely many solutions, we must have
Hence, k = 10.
Answer:
The given pair of linear equations are
Which is of the form , where
For the given pair of linear equations to have no solution, we must have
Hence, k = 11.
Page No 162:
Question 5:
The given pair of linear equations are
Which is of the form , where
For the given pair of linear equations to have no solution, we must have
Hence, k = 11.
Answer:
The given pair of linear equations are
Which is of the form , where
Now
Thus, the pair of the given linear equations has no solution.
Page No 162:
Question 6:
The given pair of linear equations are
Which is of the form , where
Now
Thus, the pair of the given linear equations has no solution.
Answer:
The given pair of linear equations is
Which is of the form , where
For the system to have a unique solution, we must have
Hence, .
Page No 162:
Question 7:
The given pair of linear equations is
Which is of the form , where
For the system to have a unique solution, we must have
Hence, .
Answer:
Let the numbers be x and y, where x > y.
Then as per the question
Dividing (ii) by (i), we get
Now, adding (i) and (ii), we have
Substituting x = 9 in (iii), we have
Hence, the numbers are 9 and 4.
Page No 162:
Question 8:
Let the numbers be x and y, where x > y.
Then as per the question
Dividing (ii) by (i), we get
Now, adding (i) and (ii), we have
Substituting x = 9 in (iii), we have
Hence, the numbers are 9 and 4.
Answer:
Let the cost of 1 pen and 1 pencil are â¹x and â¹y respectively.
Then as per the question
Adding (i) and (ii), we get
Subtracting (i) from (ii), we get
Now, adding (iii) and (iv), we get
Substituting x = 16 in (iii), we have
Hence, the cost of 1 pen and 1 pencil are respectively â¹16 and â¹5.
Page No 162:
Question 9:
Let the cost of 1 pen and 1 pencil are â¹x and â¹y respectively.
Then as per the question
Adding (i) and (ii), we get
Subtracting (i) from (ii), we get
Now, adding (iii) and (iv), we get
Substituting x = 16 in (iii), we have
Hence, the cost of 1 pen and 1 pencil are respectively â¹16 and â¹5.
Answer:
Let the larger number be x and the smaller number be y.
Then as per the question
Subtracting (ii) from (i), we get
Now, putting y = 15 in (i), we have
Hence, the numbers are 65 and 15.
Page No 162:
Question 10:
Let the larger number be x and the smaller number be y.
Then as per the question
Subtracting (ii) from (i), we get
Now, putting y = 15 in (i), we have
Hence, the numbers are 65 and 15.
Answer:
Let the ones digit and tens digit be x and y respectively.
Then as per the question
Adding (i) and (ii), we get
Now, putting x = 4 in (i), we have
Hence, the number is 64.
Page No 162:
Question 11:
Let the ones digit and tens digit be x and y respectively.
Then as per the question
Adding (i) and (ii), we get
Now, putting x = 4 in (i), we have
Hence, the number is 64.
Answer:
Let the number of stamps of 20 p and 25 p be x and y respectively.
As per the question
From (i), we get
Now, substituting in (ii), we have
Putting x = 35 in (i), we get
Hence, the number of 20 p stamps and 25 p stamps are 35 and 12 respectively.
Page No 162:
Question 12:
Let the number of stamps of 20 p and 25 p be x and y respectively.
As per the question
From (i), we get
Now, substituting in (ii), we have
Putting x = 35 in (i), we get
Hence, the number of 20 p stamps and 25 p stamps are 35 and 12 respectively.
Answer:
Let the number of hens and cow be x and y respectively.
As per the question
Subtracting (i) from (ii), we have
Hence, the number of cows is 22.
Page No 162:
Question 13:
Let the number of hens and cow be x and y respectively.
As per the question
Subtracting (i) from (ii), we have
Hence, the number of cows is 22.
Answer:
The given pair of equation is
Multiplying (i) and (ii) by xy, we have
Now, multiplying (iii) by 2 and subtracting from (iv), we get
Putting x = 1 in (iii), we have
Hence, x = 1 and y = 3.
Page No 162:
Question 14:
The given pair of equation is
Multiplying (i) and (ii) by xy, we have
Now, multiplying (iii) by 2 and subtracting from (iv), we get
Putting x = 1 in (iii), we have
Hence, x = 1 and y = 3.
Answer:
The given pair of equations is
Multiplying (i) by 12 and (ii) by 4, we get
Now, subtracting (iv) from (iii), we get
Putting x = 1 in (iv), we have
Hence, the value of x + y is .
Page No 162:
Question 15:
The given pair of equations is
Multiplying (i) by 12 and (ii) by 4, we get
Now, subtracting (iv) from (iii), we get
Putting x = 1 in (iv), we have
Hence, the value of x + y is .
Answer:
The given pair of equations is
Adding (i) and (ii), we get
Hence, the value of x + y is 4.
Page No 162:
Question 16:
The given pair of equations is
Adding (i) and (ii), we get
Hence, the value of x + y is 4.
Answer:
The given system is
This is a homogeneous system of linear differential equation, so it always has a zero
solution i.e., x = y = 0.
But to have a nonzero solution, it must have infinitely many solutions.
For this, we have
Hence, k = 6.
Page No 162:
Question 17:
The given system is
This is a homogeneous system of linear differential equation, so it always has a zero
solution i.e., x = y = 0.
But to have a nonzero solution, it must have infinitely many solutions.
For this, we have
Hence, k = 6.
Answer:
The given system is
Here, .
For the system, to have a unique solution, we must have
Hence, .
Page No 162:
Question 18:
The given system is
Here, .
For the system, to have a unique solution, we must have
Hence, .
Answer:
The given system is
Here, .
For the system, to have an infinite number of solutions, we must have
Hence, k = 6.
Page No 162:
Question 19:
The given system is
Here, .
For the system, to have an infinite number of solutions, we must have
Hence, k = 6.
Answer:
The given system is
Here, .
Now,
Thus, and therefor the given system has no solution.
Page No 162:
Question 20:
The given system is
Here, .
Now,
Thus, and therefor the given system has no solution.
Answer:
The given system is
Here, .
For the system to be inconsistent, we must have
Hence, k = 10.
Page No 162:
Question 21:
The given system is
Here, .
For the system to be inconsistent, we must have
Hence, k = 10.
Answer:
The given system of equations is
Substituting in (i) and (ii), the given equations are changed to
Multiplying (i) by 2 and adding it with (ii), we get
Multiplying (i) by 3 and subtracting (ii) from it, we get
Therefore
Now, adding (v) and (vi) we have
Substituting in (v), we have
Hence, .
Page No 164:
Question 19:
The given system of equations is
Substituting in (i) and (ii), the given equations are changed to
Multiplying (i) by 2 and adding it with (ii), we get
Multiplying (i) by 3 and subtracting (ii) from it, we get
Therefore
Now, adding (v) and (vi) we have
Substituting in (v), we have
Hence, .
Answer:
Let the required fraction be .
Then, we have:
⇒ 5(x + 1) = 4(y + 1)
⇒ 5x + 5 = 4y + 4
⇒ 5x − 4y = −1 ....(i)
Again, we have:
⇒ 2(x − 5) = 1(y − 5)
⇒ 2x − 10 = y − 5
⇒ 2x − y = 5 ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 20 ....(iii)
On subtracting (i) from (iii), we get:
3x = (20 − (−1)) = 20 + 1 = 21
⇒ 3x = 21
⇒ x = 7
On substituting x = 7 in (i), we get:
5 × 7 − 4y = −1
⇒ 35 − 4y = −1
⇒ 4y = 36
⇒ y = 9
∴ âx = 7 and y = 9
Hence, the required fraction is .
Page No 164:
Question 20:
Let the required fraction be .
Then, we have:
⇒ 5(x + 1) = 4(y + 1)
⇒ 5x + 5 = 4y + 4
⇒ 5x − 4y = −1 ....(i)
Again, we have:
⇒ 2(x − 5) = 1(y − 5)
⇒ 2x − 10 = y − 5
⇒ 2x − y = 5 ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 20 ....(iii)
On subtracting (i) from (iii), we get:
3x = (20 − (−1)) = 20 + 1 = 21
⇒ 3x = 21
⇒ x = 7
On substituting x = 7 in (i), we get:
5 × 7 − 4y = −1
⇒ 35 − 4y = −1
⇒ 4y = 36
⇒ y = 9
∴ âx = 7 and y = 9
Hence, the required fraction is .
Answer:
The given equations may be written as follows:
....(i)
....(ii)
Here, a1= , b1 = , c1 = −(a + b), a2 = a, b2 = −b, c2 = −2ab
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
⇒
Hence, x = b and y = −a is the required solution.
Page No 165:
Question 1:
The given equations may be written as follows:
....(i)
....(ii)
Here, a1= , b1 = , c1 = −(a + b), a2 = a, b2 = −b, c2 = −2ab
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
⇒
Hence, x = b and y = −a is the required solution.
Answer:
The given system of equations is
Multiplying (i) by 2 and (ii) by 3 and then adding, we get
Now, putting x = 3 in (i), we have
Thus, x = 3 and y = 2.
Hence, the correct answer is option (c).
Page No 165:
Question 2:
The given system of equations is
Multiplying (i) by 2 and (ii) by 3 and then adding, we get
Now, putting x = 3 in (i), we have
Thus, x = 3 and y = 2.
Hence, the correct answer is option (c).
Answer:
The given system of equations is
Adding (i) and (ii), we get
Now, putting x = 6 in (ii), we have
Thus, x = 6 and y = 4.
Hence, the correct answer is option (c).
Page No 165:
Question 3:
The given system of equations is
Adding (i) and (ii), we get
Now, putting x = 6 in (ii), we have
Thus, x = 6 and y = 4.
Hence, the correct answer is option (c).
Answer:
The given system of equations is
Multiplying (i) and (ii) by 6, we get
Multiplying (iii) by 4 and (iv) by 3 and adding, we get
Now, putting x = 2 in (iv), we have
Thus, x = 2 and y = 3.
Hence, the correct answer is option (a).
Page No 165:
Question 4:
The given system of equations is
Multiplying (i) and (ii) by 6, we get
Multiplying (iii) by 4 and (iv) by 3 and adding, we get
Now, putting x = 2 in (iv), we have
Thus, x = 2 and y = 3.
Hence, the correct answer is option (a).
Answer:
The given system of equations is
Adding (i) and (ii), we get
Now, putting in (i), we have
Thus, .
Hence, the correct answer is option (d).
Page No 165:
Question 5:
The given system of equations is
Adding (i) and (ii), we get
Now, putting in (i), we have
Thus, .
Hence, the correct answer is option (d).
Answer:
Consider . Now, simplifying these equations, we get
And
Multiplying (ii) by 2 and subtracting it from (i)
Now, putting x = 1 in (ii), we have
Thus, x = 1, y = 1.
Hence, the correct answer is option (a).
Page No 165:
Question 6:
Consider . Now, simplifying these equations, we get
And
Multiplying (ii) by 2 and subtracting it from (i)
Now, putting x = 1 in (ii), we have
Thus, x = 1, y = 1.
Hence, the correct answer is option (a).
Answer:
The given equations are
Substituting in (i) and (ii), the new system becomes
Now, multiplying (iii) by 2 and adding it with (iv), we get
Again, multiplying (iii) by 3 and subtracting (iv) from it, we get
Therefore
Adding (v) and (vi), we get
Substituting , in (v), we have
Thus, .
Hence, the correct answer is option (b).
Page No 165:
Question 7:
The given equations are
Substituting in (i) and (ii), the new system becomes
Now, multiplying (iii) by 2 and adding it with (iv), we get
Again, multiplying (iii) by 3 and subtracting (iv) from it, we get
Therefore
Adding (v) and (vi), we get
Substituting , in (v), we have
Thus, .
Hence, the correct answer is option (b).
Answer:
The given equations are
Dividing (i) and (ii) by xy, we get
Multiplying (iii) by 2 and subtracting (iv) from it, we get
Substituting x = 3 in (iii), we get
Thus, x = 3 and y = 4.
Hence, the correct answer is option (c).
Page No 165:
Question 8:
The given equations are
Dividing (i) and (ii) by xy, we get
Multiplying (iii) by 2 and subtracting (iv) from it, we get
Substituting x = 3 in (iii), we get
Thus, x = 3 and y = 4.
Hence, the correct answer is option (c).
Answer:
The given equations are
Adding (i) and (ii), we get
Subtracting (i) from (ii), we get
Adding (iii) and (iv), we get
Substituting x = 1 in (iii), we have
Thus, x = 1 and y = 2.
Hence, the correct answer is option (a).
Page No 165:
Question 9:
The given equations are
Adding (i) and (ii), we get
Subtracting (i) from (ii), we get
Adding (iii) and (iv), we get
Substituting x = 1 in (iii), we have
Thus, x = 1 and y = 2.
Hence, the correct answer is option (a).
Answer:
Hence, the correct answer is option (c).
Page No 165:
Question 10:
Hence, the correct answer is option (c).
Answer:
The given equations are
Multiplying (ii) by 2 and subtracting it from (ii), we get
Substituting y = 1 in (ii), we get
Hence, the correct answer is option (b).
Page No 165:
Question 11:
The given equations are
Multiplying (ii) by 2 and subtracting it from (ii), we get
Substituting y = 1 in (ii), we get
Hence, the correct answer is option (b).
Answer:
The given equations are
Here,
For the given system to have a unique solution, we must have
Hence, the correct answer is option (d).
Page No 165:
Question 12:
The given equations are
Here,
For the given system to have a unique solution, we must have
Hence, the correct answer is option (d).
Answer:
The correct option is (b).
The given system of equations can be written as follows:
x − 2y − 3 = 0 and 3x + ky − 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −2, c1 = −3 and a2 = 3, b2 = k and c2 = −1
∴
These graph lines will intersect at a unique point when we have:
⇒
Hence, k has all real values other than −6.
Page No 165:
Question 13:
The correct option is (b).
The given system of equations can be written as follows:
x − 2y − 3 = 0 and 3x + ky − 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −2, c1 = −3 and a2 = 3, b2 = k and c2 = −1
∴
These graph lines will intersect at a unique point when we have:
⇒
Hence, k has all real values other than −6.
Answer:
The correct option is (a).
The given system of equations can be written as follows:
x + 2y − 3 = 0 and 5x + ky + 7 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 5, b2 = k and c2 = 7
∴
For the system of equations to have no solution, we must have:
∴
Page No 166:
Question 14:
The correct option is (a).
The given system of equations can be written as follows:
x + 2y − 3 = 0 and 5x + ky + 7 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 5, b2 = k and c2 = 7
∴
For the system of equations to have no solution, we must have:
∴
Answer:
The correct option is (d).
The given system of equations can be written as follows:
3x + 2ky − 2 = 0 and 2x + 5y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 3, b1 = 2k, c1 = −2 and a2 = 2, b2 = 5 and c2 = 1
∴
For parallel lines, we have:
∴
⇒
Page No 166:
Question 15:
The correct option is (d).
The given system of equations can be written as follows:
3x + 2ky − 2 = 0 and 2x + 5y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 3, b1 = 2k, c1 = −2 and a2 = 2, b2 = 5 and c2 = 1
∴
For parallel lines, we have:
∴
⇒
Answer:
The correct option is (d).
The given system of equations can be written as follows:
kx − 2y − 3 = 0 and 3x + y − 5 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = −2, c1 = −3 and a2 = 3, b2 = 1 and c2 = −5
∴
Thus, for these graph lines to intersect at a unique point, we must have:
⇒
Hence, the graph lines will intersect at all real values of k except −6.
Page No 166:
Question 16:
The correct option is (d).
The given system of equations can be written as follows:
kx − 2y − 3 = 0 and 3x + y − 5 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = −2, c1 = −3 and a2 = 3, b2 = 1 and c2 = −5
∴
Thus, for these graph lines to intersect at a unique point, we must have:
⇒
Hence, the graph lines will intersect at all real values of k except −6.
Answer:
The correct option is (d).
The given system of equations can be written as:
x + 2y + 5 = 0 and −3x − 6y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = 5 and a2 = −3, b2 = −6 and c2 = 1
∴
∴
Hence, the given system has no solution.
Page No 166:
Question 17:
The correct option is (d).
The given system of equations can be written as:
x + 2y + 5 = 0 and −3x − 6y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = 5 and a2 = −3, b2 = −6 and c2 = 1
∴
∴
Hence, the given system has no solution.
Answer:
The correct option is (d).
The given system of equations can be written as follows:
2x + 3y − 5 = 0 and 4x + 6y − 15 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = −5 and a2 = 4, b2 = 6 and c2 = −15
∴
∴ â
Hence, the given system has no solution.
Page No 166:
Question 18:
The correct option is (d).
The given system of equations can be written as follows:
2x + 3y − 5 = 0 and 4x + 6y − 15 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = −5 and a2 = 4, b2 = 6 and c2 = −15
∴
∴ â
Hence, the given system has no solution.
Answer:
The correct option is (d).
If a pair of linear equations is consistent, then the two graph lines either intersect at a point or coincide.
Page No 166:
Question 19:
The correct option is (d).
If a pair of linear equations is consistent, then the two graph lines either intersect at a point or coincide.
Answer:
The correct option is (a).
If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since there is no common solution, their graph lines do not intersect. Hence, they are parallel.
Page No 166:
Question 20:
The correct option is (a).
If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since there is no common solution, their graph lines do not intersect. Hence, they are parallel.
Answer:
The correct option is (b).
Let
∴
Now,
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ...(i)
Also,
⇒ 3y = 2(x + y)
⇒ 2x − y = 0 ...(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0 ...(iii)
On adding (i) and (iii) we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
∴ x = 20 and y = 40
∴â
Page No 166:
Question 21:
The correct option is (b).
Let
∴
Now,
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ...(i)
Also,
⇒ 3y = 2(x + y)
⇒ 2x − y = 0 ...(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0 ...(iii)
On adding (i) and (iii) we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
∴ x = 20 and y = 40
∴â
Answer:
The correct option is (b).
In a cyclic quadrilateral ABCD:
We have:
and [Since ABCD is a cyclic quadrilateral]
Now,
⇒ 2x + 2y − 20 = 180
⇒ x + y − 10 = 90
⇒ x + y = 100 ....(i)
Also,
⇒ x + 2y + 20 = 180
⇒ x + 2y = 160 ....(ii)
On subtracting (i) from (ii), we get:
y = (160 − 100) = 60
On substituting y = 60 in (i), we get:
x + 60 = 100 ⇒ x = (100 − 60) = 40
∴â
Page No 166:
Question 22:
The correct option is (b).
In a cyclic quadrilateral ABCD:
We have:
and [Since ABCD is a cyclic quadrilateral]
Now,
⇒ 2x + 2y − 20 = 180
⇒ x + y − 10 = 90
⇒ x + y = 100 ....(i)
Also,
⇒ x + 2y + 20 = 180
⇒ x + 2y = 160 ....(ii)
On subtracting (i) from (ii), we get:
y = (160 − 100) = 60
On substituting y = 60 in (i), we get:
x + 60 = 100 ⇒ x = (100 − 60) = 40
∴â
Answer:
The correct option is (d).
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
According to the question, we have:
x + y = 15 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) = (10x + y) + 9
⇒ 10y + x − 10x − y = 9
⇒ 9y − 9x = 9
⇒ y − x = 1 ....(ii)
On adding (i) and (ii), we get:
2y = 16 ⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15 ⇒ x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.
Page No 166:
Question 23:
The correct option is (d).
Let the tens and the units digits of the required number be x and yâ, respectively.
Required number = (10x + y)
According to the question, we have:
x + y = 15 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) = (10x + y) + 9
⇒ 10y + x − 10x − y = 9
⇒ 9y − 9x = 9
⇒ y − x = 1 ....(ii)
On adding (i) and (ii), we get:
2y = 16 ⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15 ⇒ x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.
Answer:
Let the fraction be
It is given that
Also,
Subtract (ii) from (i), we get
Put the value of x in equation (i), we get;
Therefore, the fraction is
Hence, the correct answer is option (b)
Page No 166:
Question 24:
Let the fraction be
It is given that
Also,
Subtract (ii) from (i), we get
Put the value of x in equation (i), we get;
Therefore, the fraction is
Hence, the correct answer is option (b)
Answer:
The correct option is (d).
Let the man's present age be x years.
Let his son's present age be y years.
Five years later:
(x + 5) = 3(y + 5)
⇒ x + 5 = 3y + 15
⇒ x − 3y = 10 ....(i)
Five years ago:
(x − 5) = 7(y − 5)
⇒ x − 5 = 7y − 35
⇒ x − 7y = −30 ....(ii)
On subtracting (i) from (ii), we get:
−4y = −40 ⇒ y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 10 ⇒ x − 30 = 10 ⇒ x = (10 + 30) = 40 years
Hence, the man's present age is 40 years.
Page No 167:
Question 25:
The correct option is (d).
Let the man's present age be x years.
Let his son's present age be y years.
Five years later:
(x + 5) = 3(y + 5)
⇒ x + 5 = 3y + 15
⇒ x − 3y = 10 ....(i)
Five years ago:
(x − 5) = 7(y − 5)
⇒ x − 5 = 7y − 35
⇒ x − 7y = −30 ....(ii)
On subtracting (i) from (ii), we get:
−4y = −40 ⇒ y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 10 ⇒ x − 30 = 10 ⇒ x = (10 + 30) = 40 years
Hence, the man's present age is 40 years.
Answer:
The correct option is (b).
The given equations are as follows:
They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 6, b1 = −2, c1 = 9 and a2 = 3, b2 = −1 and c2 = 12
∴
∴
The given system has no solution.
Hence, the lines are parallel.
Page No 167:
Question 26:
The correct option is (b).
The given equations are as follows:
They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 6, b1 = −2, c1 = 9 and a2 = 3, b2 = −1 and c2 = 12
∴
∴
The given system has no solution.
Hence, the lines are parallel.
Answer:
The correct option is (c).
The given equations are as follows:
They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = −2 and a2 = 1, b2 = −2 and c2 = −8
∴
∴
The given system has a unique solution.
Hence, the lines intersect exactly at one point.
Page No 167:
Question 27:
The correct option is (c).
The given equations are as follows:
They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = −2 and a2 = 1, b2 = −2 and c2 = −8
∴
∴
The given system has a unique solution.
Hence, the lines intersect exactly at one point.
Answer:
The correct option is (a).
The given system of equations can be written as follows:
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 5, b1 = −15, c1 = −8 and a2 = 3, b2 = −9 and c2 =
∴
∴
The given system of equations will have an infinite number of solutions.
Hence, the lines are coincident.
Page No 169:
Question 1:
The correct option is (a).
The given system of equations can be written as follows:
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 5, b1 = −15, c1 = −8 and a2 = 3, b2 = −9 and c2 =
∴
∴
The given system of equations will have an infinite number of solutions.
Hence, the lines are coincident.
Answer:
The correct option is (a).
The given system of equations can be written as follows:
x + 2y − 3 = 0 and 2x + 4y + 7 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 2, b2 = 4 and c2 = 7
∴
∴
So, the given system has no solution.
Hence, the lines are parallel.
Page No 169:
Question 2:
The correct option is (a).
The given system of equations can be written as follows:
x + 2y − 3 = 0 and 2x + 4y + 7 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 2, b2 = 4 and c2 = 7
∴
∴
So, the given system has no solution.
Hence, the lines are parallel.
Answer:
The correct option is (d).
The given system of equations can be written as follows:
2x − 3y − 7 = 0 and (a + b)x − (a + b − 3)y − (4a + b) = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = −3, c1 = −7 and a2 = (a + b), b2 = −(a + b − 3) and c2 = −(4a + b)
∴
For an infinite number of solutions, we must have:
∴
Now, we have:
⇒ a + b + 6 = 0 ...(i)
Again, we have:
⇒ 5a − 4b + 21 = 0 ...(ii)
On multiplying (i) by 4, we get:
4a + 4b + 24 = 0 ...(iii)
On adding (ii) and (iii), we get:
9a = −45 ⇒ a = −5
On substituting a = −5 in (i), we get:
−5 + b + 6 = 0 ⇒ b = −1
∴ âa = −5 and b = −1
Page No 170:
Question 3:
The correct option is (d).
The given system of equations can be written as follows:
2x − 3y − 7 = 0 and (a + b)x − (a + b − 3)y − (4a + b) = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = −3, c1 = −7 and a2 = (a + b), b2 = −(a + b − 3) and c2 = −(4a + b)
∴
For an infinite number of solutions, we must have:
∴
Now, we have:
⇒ a + b + 6 = 0 ...(i)
Again, we have:
⇒ 5a − 4b + 21 = 0 ...(ii)
On multiplying (i) by 4, we get:
4a + 4b + 24 = 0 ...(iii)
On adding (ii) and (iii), we get:
9a = −45 ⇒ a = −5
On substituting a = −5 in (i), we get:
−5 + b + 6 = 0 ⇒ b = −1
∴ âa = −5 and b = −1
Answer:
The correct option is (a).
The given system of equations can be written as follows:
2x + y − 5 = 0 and 3x + 2y − 8 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 1, c1 = −5 and a2 = 3, b2 = 2 and c2 = −8
∴
∴
The given system has a unique solution.
Hence, the lines intersect at one point.
Page No 170:
Question 4:
The correct option is (a).
The given system of equations can be written as follows:
2x + y − 5 = 0 and 3x + 2y − 8 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 1, c1 = −5 and a2 = 3, b2 = 2 and c2 = −8
∴
∴
The given system has a unique solution.
Hence, the lines intersect at one point.
Answer:
The correct option is (d).
Given:
x = −y and y > 0
Now, we have:
(i) x2y
On substituting x = −y, we get:
(−y)2y = y3 > 0 (âµ y > 0)
This is true.
(ii) x + y
On substituting x = −y, we get:
(−y) + y = 0
This is also true.
(iii) xy
On substituting x = −y, we get:
(−y) y = −y2 < 0 (âµ y > 0)
This is again true.
(iv)
On substituting x = −y, we get:
Hence, from the above equation, we get y = 0, which is wrong.
Page No 170:
Question 5:
The correct option is (d).
Given:
x = −y and y > 0
Now, we have:
(i) x2y
On substituting x = −y, we get:
(−y)2y = y3 > 0 (âµ y > 0)
This is true.
(ii) x + y
On substituting x = −y, we get:
(−y) + y = 0
This is also true.
(iii) xy
On substituting x = −y, we get:
(−y) y = −y2 < 0 (âµ y > 0)
This is again true.
(iv)
On substituting x = −y, we get:
Hence, from the above equation, we get y = 0, which is wrong.
Answer:
The given system of equations:
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = −1, b1 = 2, c1 = 2 and a2 = , b2 = and c2 = −1
∴
∴
The given system has a unique solution.
Hence, the lines intersect at one point.
Page No 170:
Question 6:
The given system of equations:
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = −1, b1 = 2, c1 = 2 and a2 = , b2 = and c2 = −1
∴
∴
The given system has a unique solution.
Hence, the lines intersect at one point.
Answer:
The given system of equations can be written as follows:
kx + 3y − (k − 2) = 0 and 12x + ky − k = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = 3, c1 = − (k − 2) and a2 = 12, b2 = k and c2 = − k
∴
For inconsistency, we must have:
⇒
⇒
Hence, the pair of equations is inconsistent if .
Page No 170:
Question 7:
The given system of equations can be written as follows:
kx + 3y − (k − 2) = 0 and 12x + ky − k = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = 3, c1 = − (k − 2) and a2 = 12, b2 = k and c2 = − k
∴
For inconsistency, we must have:
⇒
⇒
Hence, the pair of equations is inconsistent if .
Answer:
The given system of equations can be written as follows:
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 9, b1 = −10, c1 = −21 and a2 = , b2 = and c2 =
∴
∴
This shows that the given system equations has an infinite number of solutions.
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Question 8:
The given system of equations can be written as follows:
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 9, b1 = −10, c1 = −21 and a2 = , b2 = and c2 =
∴
∴
This shows that the given system equations has an infinite number of solutions.
Answer:
The given equations are as follows:
x − 2y = 0 ....(i)
3x + 4y = 20 ....(ii)
On multiplying (i) by 2, we get:
2x − 4y = 0 ....(iii)
On adding (ii) and (iii), we get:
5x = 20 ⇒ x = 4
On substituting x = 4 in (i), we get:
4 − 2y = 0 ⇒ 4 = 2y ⇒ y = 2
Hence, the required solution is x = 4 and y = 2.
Page No 170:
Question 9:
The given equations are as follows:
x − 2y = 0 ....(i)
3x + 4y = 20 ....(ii)
On multiplying (i) by 2, we get:
2x − 4y = 0 ....(iii)
On adding (ii) and (iii), we get:
5x = 20 ⇒ x = 4
On substituting x = 4 in (i), we get:
4 − 2y = 0 ⇒ 4 = 2y ⇒ y = 2
Hence, the required solution is x = 4 and y = 2.
Answer:
The given system of equations can be written as follows:
x − 3y − 2 = 0 and −2x + 6y − 5= 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −3, c1 = −2 and a2 = −2, b2 = 6 and c2 = −5
∴
∴ â
Thus, the given system of equations has no solution.
Hence, the paths represented by the equations are parallel.
Page No 170:
Question 10:
The given system of equations can be written as follows:
x − 3y − 2 = 0 and −2x + 6y − 5= 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −3, c1 = −2 and a2 = −2, b2 = 6 and c2 = −5
∴
∴ â
Thus, the given system of equations has no solution.
Hence, the paths represented by the equations are parallel.
Answer:
Let the larger number be x and the smaller number be y.
Then, we have:
x − y = 26 ...(i)
x = 3y ...(ii)
On substituting x = 3y in (i), we get:
3y − y = 26 ⇒ 2y = 26 ⇒ y = 13
On substituting y = 13 in (i), we get:
x − 13 = 26 ⇒ x = 26 + 13 = 39
Hence, the required numbers are 39 and 13.
Page No 170:
Question 11:
Let the larger number be x and the smaller number be y.
Then, we have:
x − y = 26 ...(i)
x = 3y ...(ii)
On substituting x = 3y in (i), we get:
3y − y = 26 ⇒ 2y = 26 ⇒ y = 13
On substituting y = 13 in (i), we get:
x − 13 = 26 ⇒ x = 26 + 13 = 39
Hence, the required numbers are 39 and 13.
Answer:
The given equations are as follows:
23x + 29y = 98 ....(i)
29x + 23y = 110 ....(ii)
On adding (i) and (ii), we get:
52x + 52y = 208
⇒ x + y = 4 ....(iii)
On subtracting (i) from (ii), we get:
6x − 6y = 12
⇒ x − y = 2 ....(iv)
On adding (iii) and (iv), we get:
2x = 6 ⇒ x = 3
On substituting x = 3 in (iii), we get:
3 + y = 4
⇒ y = 4 − 3 = 1
Hence, the required solution is x = 3 and y = 1.
Page No 170:
Question 12:
The given equations are as follows:
23x + 29y = 98 ....(i)
29x + 23y = 110 ....(ii)
On adding (i) and (ii), we get:
52x + 52y = 208
⇒ x + y = 4 ....(iii)
On subtracting (i) from (ii), we get:
6x − 6y = 12
⇒ x − y = 2 ....(iv)
On adding (iii) and (iv), we get:
2x = 6 ⇒ x = 3
On substituting x = 3 in (iii), we get:
3 + y = 4
⇒ y = 4 − 3 = 1
Hence, the required solution is x = 3 and y = 1.
Answer:
The given equations are as follows:
6x + 3y = 7xy ....(i)
3x + 9y = 11xy ....(ii)
For equation (i), we have:
....(iii)
For equation (ii), we have:
....(iv)
On substituting and in (iii) and (iv), we get:
6v + 3u = 7 ....(v)
3v + 9u = 11 ....(vi)
On multiplying (v) by 3, we get:
18v + 9u = 21 ....(vii)
On subtracting (vi) from (vii), we get:
15v = 10 ⇒ v =
⇒
On substituting in (iii), we get:
Hence, the required solution is x = 1 and .
Page No 170:
Question 13:
The given equations are as follows:
6x + 3y = 7xy ....(i)
3x + 9y = 11xy ....(ii)
For equation (i), we have:
....(iii)
For equation (ii), we have:
....(iv)
On substituting and in (iii) and (iv), we get:
6v + 3u = 7 ....(v)
3v + 9u = 11 ....(vi)
On multiplying (v) by 3, we get:
18v + 9u = 21 ....(vii)
On subtracting (vi) from (vii), we get:
15v = 10 ⇒ v =
⇒
On substituting in (iii), we get:
Hence, the required solution is x = 1 and .
Answer:
The given system of equations:
3x + y = 1
⇒ 3x + y − 1= 0 ....(i)
kx + 2y = 5
⇒ kx + 2y − 5 = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 3, b1= 1, c1 = −1 and a2 = k, b2= 2, c2 = −5
(i) For a unique solution, we must have:
i.e.
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) In order that the given equations have no solution, we must have:
Thus, for k = 6, the given system of equations will have no solution.
Page No 170:
Question 14:
The given system of equations:
3x + y = 1
⇒ 3x + y − 1= 0 ....(i)
kx + 2y = 5
⇒ kx + 2y − 5 = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1= 0, a2x + b2y + c2 = 0
Here, a1 = 3, b1= 1, c1 = −1 and a2 = k, b2= 2, c2 = −5
(i) For a unique solution, we must have:
i.e.
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) In order that the given equations have no solution, we must have:
Thus, for k = 6, the given system of equations will have no solution.
Answer:
Let
Then,
Now, we have:
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ....(i)
Also,
⇒ 3y = 2(x + y)
⇒ 2x − y = 0 ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0 ....(iii)
On adding (i) and (iii), we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
∴ x = 20 and y = 40
∴ â
Page No 170:
Question 15:
Let
Then,
Now, we have:
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ....(i)
Also,
⇒ 3y = 2(x + y)
⇒ 2x − y = 0 ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0 ....(iii)
On adding (i) and (iii), we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
∴ x = 20 and y = 40
∴ â
Answer:
Let the cost of each pencil be Rs. x and that of each pen be Rs. y.
Then, we have:
5x + 7y = 195 ....(i)
7x + 5y = 153 ....(ii)
Adding (i) and (ii), we get:
12x + 12y = 348
⇒ 12(x + y) = 348
⇒ x + y = 29 ....(iii)
Subtracting (i) from (ii), we get:
2x − 2y = −42
⇒ 2(x − y) = −42
⇒ x − y = −21 ....(iv)
On adding (iii) and (iv), we get:
2x = 8 ⇒ x = 4
On substituting x = 4 in (iii), we get:
4 + y = 29 ⇒ y = (29 − 4) = 25
Hence, the cost of each pencil is Rs. 4 and the cost of each pen is Rs. 25.
Page No 170:
Question 16:
Let the cost of each pencil be Rs. x and that of each pen be Rs. y.
Then, we have:
5x + 7y = 195 ....(i)
7x + 5y = 153 ....(ii)
Adding (i) and (ii), we get:
12x + 12y = 348
⇒ 12(x + y) = 348
⇒ x + y = 29 ....(iii)
Subtracting (i) from (ii), we get:
2x − 2y = −42
⇒ 2(x − y) = −42
⇒ x − y = −21 ....(iv)
On adding (iii) and (iv), we get:
2x = 8 ⇒ x = 4
On substituting x = 4 in (iii), we get:
4 + y = 29 ⇒ y = (29 − 4) = 25
Hence, the cost of each pencil is Rs. 4 and the cost of each pen is Rs. 25.
Answer:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and the y-axis, respectively.
Graph of 2x − 3y = 1
2x − 3y = 1
⇒ 3y = (2x − 1)
∴ ...(i)
Putting x = −1, we get:
y = −1
Putting x = 2, we get:
y = 1
Putting x = 5, we get:
y = 3
Thus, we have the following table for the equation 2x − 3y = 1.
x | −1 | 2 | 5 |
y | −1 | 1 | 3 |
Now, plots the points A(−1, −1), B(2, 1) and C(5, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both the sides.
Thus, the line AC is the graph of 2x − 3y = 1.
Graph of 4x − 3y + 1 = 0
4x − 3y + 1 = 0
⇒ 3y = ( 4x + 1)
∴ ...(ii)
Putting x = −1, we get:
y = −1
Putting x = 2, we get:
y = 3
Putting x = 5, we get:
y = 7
Thus, we have the following table for the equation 4x − 3y + 1 = 0.
x | −1 | 2 | 5 |
y | −1 | 3 | 7 |
Thus, the line AQ is the graph of the equation 4x − 3y + 1 = 0.
The two lines intersect at A(−1, −1).
Thus, x = −1 and y = −1 is the solution of the given system of equations.
Page No 170:
Question 17:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and the y-axis, respectively.
Graph of 2x − 3y = 1
2x − 3y = 1
⇒ 3y = (2x − 1)
∴ ...(i)
Putting x = −1, we get:
y = −1
Putting x = 2, we get:
y = 1
Putting x = 5, we get:
y = 3
Thus, we have the following table for the equation 2x − 3y = 1.
x | −1 | 2 | 5 |
y | −1 | 1 | 3 |
Now, plots the points A(−1, −1), B(2, 1) and C(5, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both the sides.
Thus, the line AC is the graph of 2x − 3y = 1.
Graph of 4x − 3y + 1 = 0
4x − 3y + 1 = 0
⇒ 3y = ( 4x + 1)
∴ ...(ii)
Putting x = −1, we get:
y = −1
Putting x = 2, we get:
y = 3
Putting x = 5, we get:
y = 7
Thus, we have the following table for the equation 4x − 3y + 1 = 0.
x | −1 | 2 | 5 |
y | −1 | 3 | 7 |
Thus, the line AQ is the graph of the equation 4x − 3y + 1 = 0.
The two lines intersect at A(−1, −1).
Thus, x = −1 and y = −1 is the solution of the given system of equations.
Answer:
Given:
In a cyclic quadrilateral ABCD, we have:
and [Since ABCD is a cyclic quadrilateral]
Now,
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y = 180 − 20 = 160
⇒ x + y = 40 ....(i)
Also,
⇒ 3x + 7y = 180 ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 120 ....(iii)
On subtracting (iii) from (ii), we get:
4y = 60 ⇒ y = 15
On substituting y = 15 in (1), we get:
x + 15 = 40 ⇒ x = (40 − 15) = 25
Therefore, we have:
Page No 170:
Question 18:
Given:
In a cyclic quadrilateral ABCD, we have:
and [Since ABCD is a cyclic quadrilateral]
Now,
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y = 180 − 20 = 160
⇒ x + y = 40 ....(i)
Also,
⇒ 3x + 7y = 180 ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 120 ....(iii)
On subtracting (iii) from (ii), we get:
4y = 60 ⇒ y = 15
On substituting y = 15 in (1), we get:
x + 15 = 40 ⇒ x = (40 − 15) = 25
Therefore, we have:
Answer:
We have:
Taking and :
35u + 14v − 19 = 0 ....(i)
14u + 35v − 37 = 0 ....(ii)
Here, a1= 35, b1 = 14, c1 = −19, a2 = 14, b2 = 35, c2 = −37
By cross multiplication, we have:
∴
⇒
⇒
⇒
⇒
∴ (x + y) = 7 ....(iii)
And, (x − y) = 1 ....(iv)
Again, the equations (iii) and (iv) can be written as follows:
x + y − 7 = 0 ....(v)
x − y − 1 = 0 ....(vi)
Here, a1= 1, b1 = 1, c1 = −7 , a2 = 1 , b2 = −1 , c2 = −1
By cross multiplication, we have:
⇒
⇒
⇒
⇒
Hence, x = 4 and y = 3 is the required solution.
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