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#### Question 1:

$2x+3y=2,\phantom{\rule{0ex}{0ex}}x-2y=8.$

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 = $\frac{2\left(1-x\right)}{3}$              ...(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
Now, plot the points P(0, − 4) and Q(2, − 3). The point C(4, −2) has already been plotted. Join PQ and QC and extend it on both ways.
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.

#### Question 2:

$3x+2y=4,\phantom{\rule{0ex}{0ex}}2x-3y=7.$

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 3x + 2y = 4

3x + 2y = 4
⇒ 2y = (4 − 3x)
⇒ y = $\frac{4-3x}{2}$                    ...(i)
Putting x = 0, we get y = 2
Putting x =  2, we get y = −1
Putting x = −2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4.

 x 0 2 − 2 y 2 − 1 5

Now, plot the points A(0, 2), B( 2, −1) and C(−2, 5) 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 + 2y = 4.

Graph of 2x − 3y = 7
2x − 3y = 7
⇒ 3y = (2x − 7)

Putting x = 2, we get y = −1
Putting x = −1, we get y = −3
Putting x = 5, we get y = 1
Thus, we have the following table for the equation 2x − 3y = 7.
 x 2 −1 5 y −1 −3 1
Now, plot the points P(−1, −3) and Q(5, 1). The point B(2, −1) has already been plotted. Join PB and QB and extend it on both ways.
Thus, PQ is the graph of  2x − 3y = 7. The two graph lines intersect at B(2, − 1).
x = 2 and y = −1 are the solutions of the given system of equations.

#### Question 3:

$2x+3y=8,\phantom{\rule{0ex}{0ex}}x-2y+3=0.\phantom{\rule{0ex}{0ex}}$

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)
$y=\frac{8-2x}{3}$ ...........(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)
$y=\frac{x+3}{2}$ ..........(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
Now, plot the points P(3, 3) and Q(−3, 0). The point A(1, 2) has already been plotted. Join AP and QA and extend it on both ways.
Thus, PQ is the graph of x − 2y + 3 = 0. The two graph lines intersect at A(1, 2).
x = 1 and y = 2

#### Question 4:

$2x-5y+4=0,\phantom{\rule{0ex}{0ex}}2x+y-8=0.$

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)
$y=\frac{2x+4}{5}$ ...........(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
Now, plot the points P(1, 6) and Q(2, 4). The point B(3, 2) has already been plotted. Join PQ and QB and extend it on both ways.
Thus, PB is the graph of 2x + y − 8 = 0. The two graph lines intersect at B(3, 2).
x = 3 and y = 2

#### Question 5:

Solve the following system of equations graphically:

$3x+2y=12\phantom{\rule{0ex}{0ex}}5x-2y=4$

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 = 0, $y=\frac{12-3x}{2}=\frac{12-0}{2}=6$
For = 2, $y=\frac{12-3x}{2}=\frac{12-6}{2}=3$
For = 4, $y=\frac{12-3x}{2}=\frac{12-12}{2}=0$
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 = 0, $y=\frac{5x-4}{2}=\frac{0-4}{2}=-2$
For = 2, $y=\frac{5x-4}{2}=\frac{10-4}{2}=3$
For = 4, $y=\frac{5x-4}{2}=\frac{20-4}{2}=8$
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).

#### Question 6:

$3x+y+1=0,\phantom{\rule{0ex}{0ex}}2x-3y+8=0.$

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)
$y=\frac{2x+8}{3}$
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
Now, plot the points P(2, 4), Q(−4, 0). The point B(−1 , 2) has already been plotted. Join PB and BQ and extend it on both ways.
Thus, PQ is the graph of 2x − 3y + 8 = 0. The two graph lines intersect at B(−1, 2).
x = −1 and y = 2

#### Question 7:

Solve the following system of equation graphically:

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).

#### Question 8:

Solve the following system of equation graphically:

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).

#### Question 9:

$2x+3y=4,\phantom{\rule{0ex}{0ex}}3x-y=-5.$

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 + 3y = 4

2x + 3y = 4
⇒ 3y = (4 − 2x)
$y=\frac{4-2x}{3}$              ...(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 3xy = − 5 = 0.
 x −1 0 −2 y 2 5 −1
Now, plot the points P(0, 5), Q(−2 , −1). The point A(−1 , 2) has already been plotted. Join PA and QA and extend it on both ways.
Thus, PQ is the graph of  3xy = −5. The two graph lines intersect at A(−1 , 2).
x = −1 and y = 2 are the solutions of the given system of equations.

#### Question 10:

$x+2y+2=0,\phantom{\rule{0ex}{0ex}}3x+2y-2=0.$

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)
$y=\frac{-2-x}{2}$...............(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)
$y=\frac{2-3x}{2}$...............(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
Now, plot the points P(0, 1) and Q(4, −5). The point C(2, −2) has already been plotted. Join PC and QC and extend it on both ways.
Thus, PQ is the graph of 3x + 2y − 2 = 0. The two graph lines intersect at A(2, −2).
x = 2 and y = −2

#### Question 11:

Solve the following system of equations graphically and find the vertices and area of the
triangle formed by these lines and the x-axis:

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.

#### Question 12:

Solve the following system of equations graphically and find the vertices and area of the
​triangle formed by these lines and the x-axis:

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 2
x  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.

#### Question 13:

Solve the following system of linear equations graphically:

Find the area of the region bounded by these lines and the x-axis.

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)
$y=\frac{4x+4}{3}$............(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)
$y=\frac{-4x+20}{3}$ ............(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
Now, plot the points P(1, −8) and Q(5 , 0). The point B(2, 4) has already been plotted. Join PB and QB to get the graph line PQ. Extend it on both ways.
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 = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×6×4\right)=12$ sq. units.

#### Question 14:

Solve the following system of linear equations graphically:

Calculate the area bounded by these lines and the x-axis.

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 xy + 1 = 0
xy + 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 xy + 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  xy + 1 = 0.

Graph of 3x + 2y − 12 = 0
3x + 2y − 12 = 0
⇒ 2y = (−3x + 12)
$y=\frac{-3x+12}{2}$  ............(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
Now, plot the points P(0, 6) and Q(4 , 0). The point C(2, 3) has already been plotted. Join PC and CQ to get the graph line PQ. Extend it on both ways.
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 = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×5×3\right)=7.5$ sq. units.

#### Question 15:

Solve the following system of equations graphically and find the vertices and area of the
​triangle formed by these lines and the x-axis:
$x-2y+2=0\phantom{\rule{0ex}{0ex}}2x+y-6=0$

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.

#### Question 16:

Solve the following system of equations graphically and find the vertices and area of the
​triangle formed by these lines and the y-axis:

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 2
x  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 (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.

#### Question 17:

Solve the following system of equations graphically and find the vertices and
area of the triangle formed by these lines and the y-axis.

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 4
x  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 $\frac{14}{3}$ y 7 1 0

Now, plot the points D(0,7), E(4,1) and $F\left(\frac{14}{3},0\right)$ 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  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.

#### Question 18:

Solve the following system of equations graphically and find the vertices
and area of the triangle formed by these lines and the y-axis:

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.

#### Question 19:

Solve the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis.

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 $\frac{4}{5}$ 2

Now, plot the points A(−2,0), $B\left(0,\frac{4}{5}\right)$ and C(3,2) on a graph paper and join
A, B and C to get the graph of 2
x  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), $\left(0,\frac{4}{5}\right)$ and (3,2).
Draw a perpendicular from point C to the y-axis. So,

Hence, the veritices of the triangle are (0,8), $\left(0,\frac{4}{5}\right)$ and (3,2) and its area is $\frac{54}{5}$ sq. units.

#### Question 20:

Solve graphically the system of equations:

Calculate the area bounded by these lines and the y-axis.

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 5xy = 7
5xy = 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 5xy = 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  5xy = 7.

Graph of xy + 1 = 0
xy + 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 xy + 1 = 0.
 x 0 1 2 y 1 2 3
Now, plot the points P(0, 1) and Q(1, 2). The point C(2, 3) has already been plotted. Join PQ and QC to get the graph line PC. Extend it on both ways.
Then, PC is the graph of the equation xy + 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 = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×8×2\right)=8$ sq. units.

#### Question 21:

Solve the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis.
$2x-3y=12\phantom{\rule{0ex}{0ex}}x+3y=6$

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 2
x  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.

#### Question 22:

Show graphically that the following given systems of equations has infinitely many solutions:
$2x+3y=6\phantom{\rule{0ex}{0ex}}4x+6y=12$

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 (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 2
x + 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.

#### Question 23:

Show graphically that the system of equations has infinitely many solutions.

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)
$y=\frac{6x-10}{2}$      ...........(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
These are the same points as obtained for the graph line of equation (i). It is clear from the graph that these two lines coincide.
Hence, the given system of equations has infinitely many solutions.

#### Question 24:

Show graphically that the system of equations has infinitely many solutions.

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)
$y=\frac{18-6x}{3}$...........(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
These are the same points as obtained for the graph line of equation (i). It is clear from the graph that these two lines coincide.
Hence, the given system of equations has an infinite number of solutions.

#### Question 25:

Show graphically that the following given systems of equations has infinitely many solutions:
$x-2y=5\phantom{\rule{0ex}{0ex}}3x-6y=15$

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.

#### Question 26:

Show graphically that the following given systems of equations is inconsistent i.e. has no solution:
$x-2y=6\phantom{\rule{0ex}{0ex}}3x-6y=0$

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.

#### Question 27:

Show graphically that the system of equations is inconsistent.

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)
$y=\frac{-2x+4}{3}$  .........(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)
$y=\frac{-4x+12}{6}$      ...........(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.

#### Question 28:

Show graphically that the following given systems of equations is inconsistent i.e. has no solution:

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 2
x + 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 $\frac{10}{3}$ 5 y $\frac{20}{3}$ 0 $-\frac{10}{3}$

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.

#### Question 29:

Draw the graphs of the following equations on the same graph paper:
$2x+y=2\phantom{\rule{0ex}{0ex}}2x+y=6$
Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.

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 2
x + 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.

#### Question 1:

$x+y=3,\phantom{\rule{0ex}{0ex}}4x-3y=26.$

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

#### Question 2:

Solve for x and y:

The given system of equations is

From (i), write y in terms of x to get
$y=x-3$
Substituting y =  − 3 in (ii), we get
$\frac{x}{3}+\frac{x-3}{2}=6\phantom{\rule{0ex}{0ex}}⇒2x+3\left(x-3\right)=36\phantom{\rule{0ex}{0ex}}⇒2x+3x-9=36\phantom{\rule{0ex}{0ex}}⇒x=\frac{45}{5}=9$
Now, substituting= 9 in (i), we have
$9-y=3\phantom{\rule{0ex}{0ex}}⇒y=9-3=6$
Hence, x = 9 and y = 6.

#### Question 3:

$2x+3y=0,\phantom{\rule{0ex}{0ex}}3x+4y=5.$

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.

#### Question 4:

$2x-3y=13,\phantom{\rule{0ex}{0ex}}7x-2y=20.$

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

#### Question 5:

$3x-5y-19=0,\phantom{\rule{0ex}{0ex}}-7x+3y+1=0.$

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.

#### Question 6:

Solve for x and y:
$2x-y+3=0\phantom{\rule{0ex}{0ex}}3x-7y+10=0$

The given system of equations is

From (i), write y in terms of to get
y = 2x + 3
Substituting y = 2x + 3 in (ii), we get
$3x-7\left(2x+3\right)+10=0\phantom{\rule{0ex}{0ex}}⇒3x-14x-21+10=0\phantom{\rule{0ex}{0ex}}⇒-7x=21-10=11\phantom{\rule{0ex}{0ex}}⇒x=-\frac{11}{7}$
Now, substituting $x=-\frac{11}{7}$ in (i), we have
$-\frac{22}{7}-y+3=0\phantom{\rule{0ex}{0ex}}⇒y=3-\frac{22}{7}=-\frac{1}{7}$
Hence, $x=-\frac{11}{7}$ and $y=-\frac{1}{7}$.

#### Question 7:

Solve for x and y:

117 y=1

The given system of equations can be written as

Multiplying  (i) by 7 and (ii) by 2, we get
$63x+6x=108×7+105×2\phantom{\rule{0ex}{0ex}}⇒69x=966\phantom{\rule{0ex}{0ex}}⇒x=\frac{966}{69}=14$
Now, substituting x = 14 in (1), we have
$9×14-2y=108\phantom{\rule{0ex}{0ex}}⇒2y=126-108\phantom{\rule{0ex}{0ex}}⇒y=\frac{18}{2}=9$
Hence, x = 14 and y = 9.

#### Question 8:

$\frac{x}{3}+\frac{y}{4}=11,\phantom{\rule{0ex}{0ex}}\frac{5x}{6}-\frac{y}{3}+7=0.$

The given equations are:
$\frac{x}{3}+\frac{y}{4}=11$
⇒ 4x + 3y = 132 ........(i)

and $\frac{5x}{6}-\frac{y}{3}+7=0$
⇒ 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.

#### Question 9:

$4x-3y=8,\phantom{\rule{0ex}{0ex}}6x-y=\frac{29}{3}.$

The given system of equation is:
4x − 3y = 8 .........(i)
$6x-y=\frac{29}{3}$ ........(ii)
On multiplying (ii) by 3, we get:
18x − 3y = 29.........(iii)

On subtracting (iii) from (i) we get:
−14x = −21
x$\frac{21}{14}=\frac{3}{2}$
On substituting the value of x$\frac{3}{2}$  in (i), we get:
$4×\frac{3}{2}-3y=8\phantom{\rule{0ex}{0ex}}⇒6-3y=8\phantom{\rule{0ex}{0ex}}⇒3y=6-8=-2\phantom{\rule{0ex}{0ex}}⇒y=\frac{-2}{3}$
Hence, the solution is x$\frac{3}{2}$ and y = $\frac{-2}{3}$.

#### Question 10:

$2x-\frac{3y}{4}=3,\phantom{\rule{0ex}{0ex}}5x=2y+7.$

The given equations are:
$2x-\frac{3y}{4}=3$ ........(i)
5x = 2y + 7 ............(ii)

On multiplying (i) by 2 and (ii) by $\frac{3}{4}$, we get:
$4x-\frac{3}{2}y=6$ .......(iii)
$\frac{15}{4}x=\frac{3}{2}y+\frac{21}{4}$ .......(iv)

On subtracting (iii) from (iv), we get:
$-\frac{1}{4}x=-\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒x=3$

On substituting x = 3 in (i), we get:

Hence, the solution is x = 3 and y = 4.

#### Question 11:

$2x+5y=\frac{8}{3},\phantom{\rule{0ex}{0ex}}3x-2y=\frac{5}{6}.$

The given equations are:
$2x+5y=\frac{8}{3}$ ........(i)
$3x-2y=\frac{5}{6}$..........(ii)

On multiplying (i) by 2 and (ii) by 5, we get:
$4x+10y=\frac{16}{3}$ ........(iii)
$15x-10y=\frac{25}{6}$ ...........(iv)

On adding (iii) and (iv), we get:
$19x=\frac{57}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{57}{6×19}=\frac{3}{6}=\frac{1}{2}$

On substituting x =$\frac{1}{2}$ in (i), we get:
$2×\frac{1}{2}+5y=\frac{8}{3}$
$⇒5y=\left(\frac{8}{3}-1\right)=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒y=\frac{5}{3×5}=\frac{1}{3}$

Hence, the solution is x = $\frac{1}{2}$ and  y = $\frac{1}{3}$.

#### Question 12:

$\frac{7-4x}{3}=y,\phantom{\rule{0ex}{0ex}}2x+3y+1=0.$

The given equations are:
$\frac{7-4x}{3}=y$
⇒ 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.

#### Question 13:

Solve for x and y:
$0.4x+0.3y=1.7\phantom{\rule{0ex}{0ex}}0.7x-0.2y=0.8$

The given system of equations is

Multiplying  (i) by 0.2 and (ii) by 0.3 and adding them, we get
$0.8x+2.1x=3.4+2.4\phantom{\rule{0ex}{0ex}}⇒2.9x=5.8\phantom{\rule{0ex}{0ex}}⇒x=\frac{5.8}{2.9}=2$
Now, substituting x = 2 in (i), we have
$0.4×2+0.3y=1.7\phantom{\rule{0ex}{0ex}}⇒0.3y=1.7-0.8\phantom{\rule{0ex}{0ex}}⇒y=\frac{0.9}{0.3}=3$
Hence, x = 2 and y = 3.

#### Question 14:

Solve for x and y:
$0.3x+0.5y=0.5\phantom{\rule{0ex}{0ex}}0.5x+0.7y=0.74$

The given system of equations is

Multiplying  (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get
$2.5y-2.1y=2.5-2.22\phantom{\rule{0ex}{0ex}}⇒0.4y=0.28\phantom{\rule{0ex}{0ex}}⇒y=\frac{0.28}{0.4}=0.7$
Now, substituting y = 0.7 in (i), we have
$0.3x+0.5×0.7=0.5\phantom{\rule{0ex}{0ex}}⇒0.3x=0.50-0.35=0.15\phantom{\rule{0ex}{0ex}}⇒x=\frac{0.15}{0.3}=0.5$
Hence, x = 0.5 and y = 0.7.

#### Question 15:

$7\left(y+3\right)-2\left(x+2\right)=14,\phantom{\rule{0ex}{0ex}}4\left(y-2\right)+3\left(x-3\right)=2.$

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.

#### Question 16:

$6x+5y=7x+3y+1=2\left(x+6y-1\right)$

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.

#### Question 17:

$\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x+y-12}{11}$

The given equations are:
$\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x+y-12}{11}$
i.e., $\frac{x+y-8}{2}=\frac{3x+y-12}{11}$
By cross multiplication, we get:
11x + 11y − 88 = 6x + 2y − 24
⇒ 11x − 6x + 11y − 2y = −24 + 88
⇒ 5x + 9y = 64 .........(i)

and $\frac{x+2y-14}{3}=\frac{3x+y-12}{11}$
⇒ 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.

#### Question 18:

$\frac{5}{x}+6y=13,\phantom{\rule{0ex}{0ex}}\frac{3}{x}+4y=7,x\ne 0.$

The given equations are:
$\frac{5}{x}+6y=13$ ............(i)
$\frac{3}{x}+4y=7$ .............(ii)

Putting $\frac{1}{x}=u$, 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
$⇒\frac{1}{x}=5⇒x=\frac{1}{5}$
On substituting $x=\frac{1}{5}$ in (i), we get:
$\frac{5}{1}{5}}+6y=13$
⇒ 25 + 6y = 13
⇒ 6y = (13 − 25) = −12
y =  −2

Hence, the required solution is $x=\frac{1}{5}$ and y = −2.

#### Question 19:

The given equations are:
$x+\frac{6}{y}=6$ ............(i)
$3x-\frac{8}{y}=5$ .............(ii)
Putting $\frac{1}{y}=v$, 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:
$3+\frac{6}{y}=6\phantom{\rule{0ex}{0ex}}$
$⇒\frac{6}{y}=\left(6-3\right)=3⇒3y=6⇒y=2$
Hence, the required solution is x = 3 and y = 2.

#### Question 20:

The given equations are:
$2x-\frac{3}{y}=9$ ............(i)
$3x+\frac{7}{y}=2$ .............(ii)
Putting $\frac{1}{y}=v$, 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:
$2×3-\frac{3}{y}=9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒6-\frac{3}{y}=9⇒\frac{3}{y}=-3⇒y=-1$
Hence, the required solution is x = 3 and y = −1.

#### Question 21:

The given equations are:
$\frac{3}{x}-\frac{1}{y}+9=0$
$\frac{3}{x}-\frac{1}{y}=-9$ ............(i)
$\frac{2}{x}+\frac{3}{y}=5$ .............(ii)
Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
3u − v = −9 .............(iii)
2u + 3v = 5 ...........(iv)

On multiplying (iii) by 3, we get:
9u − 3v = −27 .............(v)

On adding (iv) and (v), we get:
11u = −22 ⇒ u = −2
$⇒\frac{1}{x}=-2⇒x=\frac{-1}{2}$

On substituting $x=\frac{-1}{2}$ in (i), we get:
$\frac{3}{-1}{2}}-\frac{1}{y}=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒-6-\frac{1}{y}=-9⇒\frac{1}{y}=\left(-6+9\right)=3\phantom{\rule{0ex}{0ex}}$
$⇒y=\frac{1}{3}$
Hence, the required solution is $x=\frac{-1}{2}$ and $y=\frac{1}{3}$.

#### Question 22:

$\frac{9}{x}-\frac{4}{y}=8,\phantom{\rule{0ex}{0ex}}\frac{13}{x}+\frac{7}{y}=101\left(x\ne 0,y\ne 0\right).$

The given equations are:
$\frac{9}{x}-\frac{4}{y}=8$ ............(i)
$\frac{13}{x}+\frac{7}{y}=101$ .............(ii)

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, 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
$⇒\frac{1}{x}=4⇒x=\frac{1}{4}$
On substituting $x=\frac{1}{4}$ in (i), we get:
$\frac{9}{1}{4}}-\frac{4}{y}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒36-\frac{4}{y}=8⇒\frac{4}{y}=\left(36-8\right)=28\phantom{\rule{0ex}{0ex}}$
$⇒y=\frac{4}{28}=\frac{1}{7}$
Hence, the required solution is $x=\frac{1}{4}$ and $y=\frac{1}{7}$.

#### Question 23:

The given equations are:
$\frac{5}{x}-\frac{3}{y}=1$  ............(i)
$\frac{3}{2x}+\frac{2}{3y}=5$ .............(ii)

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
5u − 3v = 1 .............(iii)
$\frac{3}{2}u+\frac{2}{3}v=5\phantom{\rule{0ex}{0ex}}$
$⇒\frac{9u+4v}{6}=5\phantom{\rule{0ex}{0ex}}$
$⇒9u+4v=30$ ...............(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
$⇒\frac{1}{x}=2⇒x=\frac{1}{2}$

On substituting $x=\frac{1}{2}$ in (i), we get:
$\frac{5}{1}{2}}-\frac{3}{y}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒10-\frac{3}{y}=1⇒\frac{3}{y}=\left(10-1\right)=9\phantom{\rule{0ex}{0ex}}$
$⇒y=\frac{3}{9}=\frac{1}{3}$
Hence, the required solution is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.

#### Question 24:

Solve for x and y:
$\frac{1}{2x}+\frac{1}{3y}=2\phantom{\rule{0ex}{0ex}}\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

Multiplying equation (i) and (ii) by 6, we get

Multiplying  (i) by 3 and (ii) by 2 and subtracting (ii) from (i), we get
$\frac{9}{x}-\frac{4}{x}=36-26\phantom{\rule{0ex}{0ex}}⇒\frac{5}{x}=10\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{10}=\frac{1}{2}$
Now, substituting $x=\frac{1}{2}$ in (i), we have
$6+\frac{2}{y}=12\phantom{\rule{0ex}{0ex}}⇒\frac{2}{y}=6\phantom{\rule{0ex}{0ex}}⇒y=\frac{1}{3}$
Hence, $x=\frac{1}{2}$ and $y=\frac{1}{3}$.

#### Question 25:

The given equations are:
4x + 6y = 3xy  .......(i)
8x + 9y = 5xy .........(ii)

From equation (i), we have:

$\frac{4x+6y}{xy}=3\phantom{\rule{0ex}{0ex}}$
$⇒\frac{4}{y}+\frac{6}{x}=3$
.............(iii)

For equation (ii), we have:

$\frac{8x+9y}{xy}=5\phantom{\rule{0ex}{0ex}}$
$⇒\frac{8}{y}+\frac{9}{x}=5$
.............(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 $⇒v=\frac{3}{12}=\frac{1}{4}$
$⇒\frac{1}{y}=\frac{1}{4}⇒y=4$

On substituting y = 4 in (iii), we get:
$\frac{4}{4}+\frac{6}{x}=3\phantom{\rule{0ex}{0ex}}$

$⇒2x=6⇒x=\frac{6}{2}=3$
Hence, the required solution is x = 3 and  y = 4.

#### Question 26:

$x+y=5xy,\phantom{\rule{0ex}{0ex}}3x+2y=13xy.$

The given equations are:
x + y = 5xy  .......(i)
3x + 2y = 13xy .........(ii)

From equation (i), we have:

$\frac{x+y}{xy}=5\phantom{\rule{0ex}{0ex}}$
$⇒\frac{1}{y}+\frac{1}{x}=5$
.............(iii)

From equation (ii), we have:

$\frac{3x+2y}{xy}=13\phantom{\rule{0ex}{0ex}}$
$⇒\frac{3}{y}+\frac{2}{x}=13$
.............(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
$⇒\frac{1}{y}=3⇒y=\frac{1}{3}$
On substituting $y=\frac{1}{3}$ in (iii), we get:
$\frac{1}{1}{3}}+\frac{1}{x}=5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒3+\frac{1}{x}=5⇒\frac{1}{x}=2⇒x=\frac{1}{2}$
Hence, the required solution is $x=\frac{1}{2}$ and $y=\frac{1}{3}$ or x = 0 and y = 0.

#### Question 27:

Solve for x and y:
$\frac{5}{x+y}-\frac{2}{x-y}=-1\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}+\frac{7}{x-y}=10$

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
$2x=6⇒x=3$
Substituting x = 3 in (vi), we have
$3+y=5⇒y=5-3=2$
Hence, x = 3 and y = 2.

#### Question 28:

The given equations are:
$\frac{3}{x+y}+\frac{2}{x-y}=2$          ...(i)
$\frac{9}{x+y}-\frac{4}{x-y}=1$           ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, 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
$⇒u=\frac{5}{15}=\frac{1}{3}$
$⇒\frac{1}{x+y}=\frac{1}{3}⇒x+y=3$             ...(vi)
On substituting $u=\frac{1}{3}$ in (iii), we get:
1 + 2v = 2
⇒ 2v = 1
$⇒v=\frac{1}{2}$
$⇒\frac{1}{x-y}=\frac{1}{2}⇒x-y=2$                ...(vii)
On adding (vi) and (vii), we get:
2x = 5
$x=\frac{5}{2}$
On substituting $x=\frac{5}{2}$ in (vi), we get:

Hence, the required solution is .

#### Question 29:

The given equations are:
$\frac{5}{x+1}-\frac{2}{y-1}=\frac{1}{2}$ .............(i)
$\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$ ..............(ii)
Putting $\frac{1}{x+1}=u$ and $\frac{1}{y-1}=v$ , we get:
$5u-2v=\frac{1}{2}$ .................(iii)
$10u+2v=\frac{5}{2}$ ................(iv)
On adding (iii) and (iv), we get:
15u = 3
$u=\frac{3}{15}=\frac{1}{5}$
$\frac{1}{x+1}=\frac{1}{5}⇒x+1=5⇒x=4$
On substituting $u=\frac{1}{5}$ in (iii), we get:
$5×\frac{1}{5}-2v=\frac{1}{2}⇒1-2v=\frac{1}{2}$
$⇒2v=\frac{1}{2}⇒v=\frac{1}{4}$
$⇒\frac{1}{y-1}=\frac{1}{4}⇒y-1=4⇒y=5$
Hence, the required solution is x = 4 and y = 5.

#### Question 30:

$\frac{44}{x+y}+\frac{30}{x-y}=10,\frac{55}{x+y}+\frac{40}{x-y}=13.$

The given equations are:
$\frac{44}{x+y}+\frac{30}{x-y}=10$        ...(i)
$\frac{55}{x+y}+\frac{40}{x-y}=13$         ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ , 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
$⇒u=\frac{1}{11}$
$⇒\frac{1}{x+y}=\frac{1}{11}⇒x+y=11$         ...(vii)
On substituting $u=\frac{1}{11}$ in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
$⇒v=\frac{6}{30}=\frac{1}{5}$
$⇒\frac{1}{x-y}=\frac{1}{5}⇒x-y=5$     ...(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.

#### Question 31:

Solve for x and y:
$\frac{10}{x+y}+\frac{2}{x-y}=4\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}-\frac{9}{x-y}=-2$

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 $u=\frac{4}{15}$ in (iii), we get

Adding (v) and (vi), we get
$2x=\frac{15}{4}+\frac{3}{2}⇒2x=\frac{21}{4}⇒x=\frac{21}{8}$
Substituting $x=\frac{21}{8}$ in (v), we have
$\frac{21}{8}+y=\frac{15}{4}⇒y=\frac{15}{4}-\frac{21}{8}=\frac{9}{8}$
Hence, .

#### Question 32:

$71x+37y=253,\phantom{\rule{0ex}{0ex}}37x+71y=287.$

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(xy) = −34
⇒ (xy) = −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.

#### Question 33:

$217x+131y=913,\phantom{\rule{0ex}{0ex}}131x+217y=827.$

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(xy) = 86
xy = 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.

#### Question 34:

Solve for x and y:

The given equations are

Adding (i) and (ii), we get

Subtracting (i) from (ii), we get

Now, adding equation (iii) and (iv), we get
$2x=6⇒x=3$
Substituting x = 3 in (iv), we have
$3+y=2⇒y=2-3=-1$
Hence, .

#### Question 35:

Solve for x and y:
$\frac{2x+5y}{xy}=6\phantom{\rule{0ex}{0ex}}\frac{4x-5y}{xy}=-3$

The given equations can be written as

Adding (i) and (ii), we get
$\frac{6}{y}=3⇒y=2$
Substituting y = 2 in (i), we have
$\frac{5}{x}+\frac{2}{2}=6⇒x=1$
Hence, x = 1 and y = 2..

#### Question 36:

Solve for x and y:
$\frac{1}{\left(3x+y\right)}+\frac{1}{\left(3x-y\right)}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=\frac{-1}{8}$

The given equations are

Substituting in (i) and (ii), we get

Adding (iii) and (iv), we get

Now, substituting $u=\frac{1}{4}$ in (iii), we get

Adding (v) and (vi), we get
$6x=6⇒x=1$
Substituting x = 1 in (v), we have
$3+y=4⇒y=1$
Hence, x = 1 and y = 1.

#### Question 37:

The given equations are:
$\frac{1}{2\left(x+2y\right)}+\frac{5}{3\left(3x-2y\right)}=-\frac{3}{2}$              ...(i)
$\frac{5}{4\left(x+2y\right)}-\frac{3}{5\left(3x-2y\right)}=\frac{61}{60}$              ...(ii)
Putting $\frac{1}{x+2y}=u$ and $\frac{1}{3x-2y}=v$ , we get:
$\frac{1}{2}u+\frac{5}{3}v=-\frac{3}{2}\phantom{\rule{0ex}{0ex}}$         ...(iii)
$\frac{5}{4}u-\frac{3}{5}v=\frac{61}{60}$          ...(iv)
On multiplying (iii) by 6 and (iv) by 20, we get:
3u + 10v = −9         ...(v)
$25u-12v=\frac{61}{3}$           ...(vi)
On multiplying (v) by 6 and (vi) by 5, we get:
18u + 60v = −54       ...(vii)
$125u-60v=\frac{305}{3}$          ...(viii)
On adding (vii) and (viii), we get:
$143u=\frac{305}{3}-54=\frac{305-162}{3}=\frac{143}{3}\phantom{\rule{0ex}{0ex}}$
$⇒u=\frac{1}{3}=\frac{1}{x+2y}$
x + 2y = 3          ...(ix)
On substituting $u=\frac{1}{3}$ in (v), we get:
1 + 10v = −9
⇒ 10v = −10
v = −1
$⇒\frac{1}{3x-2y}=-1⇒3x-2y=-1$            ...(x)
On adding (ix) and (x), we get:
4x = 2
$x=\frac{1}{2}$
On substituting $x=\frac{1}{2}$ in (x), we get:
$\frac{3}{2}-2y=-1\phantom{\rule{0ex}{0ex}}⇒2y=\left(\frac{3}{2}+1\right)=\frac{5}{2}\phantom{\rule{0ex}{0ex}}⇒y=\frac{5}{4}$
Hence, the required solution is .

#### Question 38:

Solve for x and y:
$\frac{2}{\left(3x+2y\right)}+\frac{3}{\left(3x-2y\right)}=\frac{17}{5}\phantom{\rule{0ex}{0ex}}\frac{5}{\left(3x+2y\right)}+\frac{1}{\left(3x-2y\right)}=2$

The given equations are

Substituting in (i) and (ii), we get

Multiplying (iv) by 3 and subtracting from(iii), we get

Now, substituting $u=\frac{1}{5}$ in (iv), we get

Adding (v) and (vi), we get
$6x=6⇒x=1$
Substituting x = 1 in (v), we have
$3+2y=5⇒y=1$
Hence, x = 1 and y = 1.

#### Question 39:

Solve for x and y:
$3\left(2x+y\right)=7xy\phantom{\rule{0ex}{0ex}}3\left(x+3y\right)=11xy$

The given equations can be written as

Multiplying (i) by 3 and subtracting (ii) from it, we get
$\frac{18}{y}-\frac{3}{y}=21-11\phantom{\rule{0ex}{0ex}}⇒\frac{15}{y}=10\phantom{\rule{0ex}{0ex}}⇒y=\frac{15}{10}=\frac{3}{2}$
Substituting $y=\frac{3}{2}$ in (i), we have
$\frac{3}{x}+\frac{6×2}{3}=7\phantom{\rule{0ex}{0ex}}⇒\frac{3}{x}=7-4=3\phantom{\rule{0ex}{0ex}}⇒x=1$
Hence, .

#### Question 40:

Solve for x and y:
$x+y=a+b\phantom{\rule{0ex}{0ex}}ax-by={a}^{2}-{b}^{2}$

The given equations are

Multiplying (i) by b and adding it with (ii), we get
$bx+ax=ab+{b}^{2}+{a}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{ab+{a}^{2}}{a+b}=a$
Substituting x = a in (i), we have
$a+y=a+b\phantom{\rule{0ex}{0ex}}⇒y=b$
Hence, x = a and y = b.

#### Question 41:

$\frac{x}{a}+\frac{y}{b}=2,\phantom{\rule{0ex}{0ex}}ax-by=\left({a}^{2}-{b}^{2}\right).$

The given equations are:
$\frac{x}{a}+\frac{y}{b}=2$

$\frac{bx+ay}{ab}=2$  [Taking LCM]
bx + ay = 2ab .......(i)

Again, axby = (a2b2) ........(ii)

On multiplying (i) by b and (ii) by a, we get:
b2x + bay = 2ab2.........(iii)
a2xbay = a(a2b2) ........(iv)

On adding (iii) from (iv), we get:
(b2 + a2)x = 2a2b + a(a2b2)
⇒ (b2 + a2)x = 2ab2 + a3ab2
⇒ (b2 + a2)x = ab2 + a3
⇒ (b2 + a2)x = a(b2 + a2)
$x=\frac{a\left({b}^{2}+{a}^{2}\right)}{\left({b}^{2}+{a}^{2}\right)}=a$
On substituting x = a in (i), we get:
ba + ay = 2ab
ay = ab
y = b

Hence, the solution is x = a and y = b.

#### Question 42:

Solve for x and y:
$px+qy=p-q\phantom{\rule{0ex}{0ex}}qx-py=p+q$

The given equations are

Multiplying (i) by p and (ii) by q and adding them, we get
${p}^{2}x+{q}^{2}x={p}^{2}-pq+pq+{q}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{p}^{2}+{q}^{2}}{{p}^{2}+{q}^{2}}=1$
Substituting x = 1 in (i), we have
$p+qy=p-q\phantom{\rule{0ex}{0ex}}⇒qy=-q\phantom{\rule{0ex}{0ex}}⇒y=-1$
Hence, x = 1 and $y=-1$.

#### Question 43:

Solve for x and y:
$\frac{x}{a}-\frac{y}{b}=0\phantom{\rule{0ex}{0ex}}ax+by={a}^{2}+{b}^{2}$

The given equations are

From (i)
$y=\frac{bx}{a}$
Substituting $y=\frac{bx}{a}$ in (ii), we get
$ax+\frac{b×bx}{a}={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{\left({a}^{2}+{b}^{2}\right)×a}{{a}^{2}+{b}^{2}}=a$
Now, substitute x = a in (ii) to get
${a}^{2}+by={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒by={b}^{2}\phantom{\rule{0ex}{0ex}}⇒y=b$
Hence, x = a and y = b.

#### Question 44:

$6\left(ax+by\right)=3a+2b,\phantom{\rule{0ex}{0ex}}6\left(bx-ay\right)=3b-2a.$

The given equations are:
6(ax + by) = 3a + 2b
⇒ 6ax + 6by = 3a + 2b ...............(i)

and 6(bxay) = 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)
$⇒x=\frac{3\left({a}^{2}+{b}^{2}\right)}{6\left({a}^{2}+{b}^{2}\right)}=\frac{1}{2}$

On substituting $x=\frac{1}{2}$ in (i), we get:
$6a×\frac{1}{2}+6by=3a+2b$
⇒ 3a + 6by = 3a + 2b
⇒ 6by = 2b
y = $\frac{2b}{6b}=\frac{1}{3}$
Hence, the required solution is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.

#### Question 45:

Solve for x and y:

The given equations are

From (ii)
$y=2a-x$
Substituting $y=2a-x$ in (i), we get
$ax-b\left(2a-x\right)={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒ax-2ab+bx={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{a}^{2}+{b}^{2}+2ab}{a+b}=\frac{{\left(a+b\right)}^{2}}{a+b}=a+b$
Now, substitute x = a + b in (ii) to get
$a+b+y=2a\phantom{\rule{0ex}{0ex}}⇒y=a-b$
Hence, .

#### Question 46:

$\frac{bx}{a}-\frac{ay}{b}+a+b=0,\phantom{\rule{0ex}{0ex}}bx-ay+2ab=0.$

The given equations are:
$\frac{bx}{a}-\frac{ay}{b}+a+b=0$
By taking LCM, we get:
b2xa2y = −a2bb2a .......(i)

and bxay + 2ab = 0
bxay = −2ab ........(ii)

On multiplying (ii) by a, we get:
abxa2y = −2a2b .......(iii)

On subtracting (i) from (iii), we get:
abxb2x = − 2a2b + a2b + b2a = −a2b + b2a
x(abb2) = −ab(a b)
x(ab)b = −ab(a b)
$x=\frac{-ab\left(a-b\right)}{\left(a-b\right)b}=-a$

On substituting x = −a in (i), we get:
b2(−a) − a2y = −a2bb2a
⇒ −b2aa2y = −a2bb2a
⇒ −a2y = −a2b
y = b

Hence, the solution is x = −a and y = b.

#### Question 47:

$\frac{bx}{a}+\frac{ay}{b}={a}^{2}+{b}^{2},\phantom{\rule{0ex}{0ex}}x+y=2ab.$

The given equations are:

$\frac{bx}{a}+\frac{ay}{b}={a}^{2}+{b}^{2}$
By taking LCM, we get:
$\frac{{b}^{2}x+{a}^{2}y}{ab}={a}^{2}+{b}^{2}$
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:
(b2a2)x = a3b + ab3  − 2a3b
⇒ (b2a2)x = −a3b + ab3
⇒ (b2a2)x = ab(b2 a2)
⇒ (b2a2)x = ab(b2a2)
$x=\frac{ab\left({b}^{2}-{a}^{2}\right)}{\left({b}^{2}-{a}^{2}\right)}=ab$

On substituting x = ab in (i), we get:
b2(ab) + a2y = a3b + ab3
a2y = a3b
$\frac{{a}^{3}b}{{a}^{2}}=ab$
Hence, the solution is x = ab and  y = ab.

#### Question 48:

Solve for x and y:
$x+y=a+b\phantom{\rule{0ex}{0ex}}ax-by={a}^{2}-{b}^{2}$

The given equations are

From (i)
$y=a+b-x$
Substituting $y=a+b-x$ in (ii), we get
$ax-b\left(a+b-x\right)={a}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}⇒ax-ab-{b}^{2}+bx={a}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{a}^{2}+ab}{a+b}=a$
Now, substitute x = a in (i) to get
$a+y=a+b\phantom{\rule{0ex}{0ex}}⇒y=b$
Hence, x = a and y = b.

#### Question 49:

Solve for x and y:
${a}^{2}x+{b}^{2}y={c}^{2}\phantom{\rule{0ex}{0ex}}{b}^{2}x+{a}^{2}y={d}^{2}$

The given equations are

Multiplying (i) by aand (ii) by b2 and subtracting, we get
${a}^{4}x-{b}^{4}x={a}^{2}{c}^{2}-{b}^{2}{d}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{a}^{2}{c}^{2}-{b}^{2}{d}^{2}}{{a}^{4}-{b}^{4}}$
Now, multiplying (i) by band (ii) by a2 and subtracting, we get
${b}^{4}y-{a}^{4}y={b}^{2}{c}^{2}-{a}^{2}{d}^{2}\phantom{\rule{0ex}{0ex}}⇒y=\frac{{b}^{2}{c}^{2}-{a}^{2}{d}^{2}}{{b}^{4}-{a}^{4}}$

Hence, .

#### Question 50:

Solve for x and y:
$\frac{x}{a}+\frac{y}{b}=a+b\phantom{\rule{0ex}{0ex}}\frac{x}{{a}^{2}}+\frac{y}{{b}^{2}}=2$

The given equations are

Multiplying (i) by b and (ii) by b2 and subtracting, we get
$\frac{bx}{a}-\frac{{b}^{2}x}{{a}^{2}}=ab+{b}^{2}-2{b}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{ab-{b}^{2}}{{a}^{2}}x=ab-{b}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{\left(ab-{b}^{2}\right){a}^{2}}{ab-{b}^{2}}={a}^{2}$
Now, substituting x = ain (i), we get
$\frac{{a}^{2}}{a}+\frac{y}{b}=a+b\phantom{\rule{0ex}{0ex}}⇒\frac{y}{b}=a+b-a=b\phantom{\rule{0ex}{0ex}}⇒y={b}^{2}$

Hence, .

#### Question 1:

$x+2y+1=0,\phantom{\rule{0ex}{0ex}}2x-3y-12=0.$

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: $\frac{x}{\left[2×\left(-12\right)-1×\left(-3\right)\right]}=\frac{y}{\left[1×2-1×\left(-12\right)\right]}=\frac{1}{\left[1×\left(-3\right)-2×2\right]}$
$\frac{x}{\left(-24+3\right)}=\frac{y}{\left(2+12\right)}=\frac{1}{\left(-3-4\right)}$
$\frac{x}{\left(-21\right)}=\frac{y}{\left(14\right)}=\frac{1}{\left(-7\right)}$

Hence, x = 3 and y = −2 is the required solution.

#### Question 2:

$3x-2y+3=0,\phantom{\rule{0ex}{0ex}}4x+3y-47=0.$

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: $\frac{x}{\left[\left(-2\right)×\left(-47\right)-3×3\right]}=\frac{y}{\left[3×4-\left(-47\right)×3\right]}=\frac{1}{\left[3×3-\left(-2\right)×4\right]}$
$\frac{x}{\left(94-9\right)}=\frac{y}{\left(12+141\right)}=\frac{z}{\left(9+8\right)}$
$\frac{x}{85}=\frac{y}{153}=\frac{1}{17}$

Hence, x = 5 and y = 9 is the required solution.

#### Question 3:

$6x-5y-16=0,\phantom{\rule{0ex}{0ex}}7x-13y+10=0.$

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: $\frac{x}{\left[\left(-5\right)×10-\left(-16\right)×\left(-13\right)\right]}=\frac{y}{\left[\left(-16\right)×7-10×6\right]}=\frac{1}{\left[6×\left(-13\right)-\left(-5\right)×7\right]}$
$\frac{x}{\left(-50-208\right)}=\frac{y}{\left(-112-60\right)}=\frac{z}{\left(-78+35\right)}$
$\frac{x}{\left(-258\right)}=\frac{y}{\left(-172\right)}=\frac{1}{\left(-43\right)}$

Hence, x = 6 and y = 4 is the required solution.

#### Question 4:

$3x+2y+25=0,\phantom{\rule{0ex}{0ex}}2x+y+10=0.$

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: $\frac{x}{\left[2×10-25×1\right]}=\frac{y}{\left[25×2-10×3\right]}=\frac{1}{\left[3×1-2×2\right]}$
$\frac{x}{\left(20-25\right)}=\frac{y}{\left(50-30\right)}=\frac{1}{\left(3-4\right)}$
$\frac{x}{\left(-5\right)}=\frac{y}{20}=\frac{1}{\left(-1\right)}$

Hence, x = 5 and y = −20 is the required solution.

#### Question 5:

$2x+5y=1,\phantom{\rule{0ex}{0ex}}2x+3y=3.$

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: $\frac{x}{\left[5×\left(-3\right)-3×\left(-1\right)\right]}=\frac{y}{\left[\left(-1\right)×2-\left(-3\right)×2\right]}=\frac{1}{\left[2×3-2×5\right]}$
$\frac{x}{\left(-15+3\right)}=\frac{y}{\left(-2+6\right)}=\frac{z}{\left(6-10\right)}$
$\frac{x}{-12}=\frac{y}{4}=\frac{1}{-4}$

Hence, x = 3 and y = −1 is the required solution.

#### Question 6:

$2x+y=35,\phantom{\rule{0ex}{0ex}}3x+4y=65.$

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: $\frac{x}{\left[1×\left(-65\right)-4×\left(-35\right)\right]}=\frac{y}{\left[\left(-35\right)×3-\left(-65\right)×2\right]}=\frac{1}{\left[2×4-3×1\right]}$
$\frac{x}{\left(-65+140\right)}=\frac{y}{\left(-105+130\right)}=\frac{1}{\left(8-3\right)}$
$\frac{x}{75}=\frac{y}{25}=\frac{1}{5}$

Hence, x = 15 and y = 5 is the required solution.

#### Question 7:

$7x-2y=3,\phantom{\rule{0ex}{0ex}}11x-\frac{3}{2}y=8.$

The given equations may be written as:
7x − 2y − 3 = 0        ...(i)
$11x-\frac{3}{2}y-8=0$           ...(ii)
Here, a1 = 7, b1 = −2 , c1 = −3, a2 = 11, b2 = $-\frac{3}{2}$ and c2 = −8
By cross multiplication, we have: $\frac{x}{\left[\left(-2\right)×\left(-8\right)-\left(-\frac{3}{2}\right)×\left(-3\right)\right]}=\frac{y}{\left[\left(-3\right)×11-\left(-8\right)×7\right]}=\frac{1}{\left[7×\left(\frac{-3}{2}\right)-11×\left(-2\right)\right]}$
$\frac{x}{\left(16-\frac{9}{2}\right)}=\frac{y}{\left(-33+56\right)}=\frac{1}{\left(-\frac{21}{2}+22\right)}$
$\frac{x}{\left(\frac{23}{2}\right)}=\frac{y}{23}=\frac{1}{\left(\frac{23}{2}\right)}$

Hence, x = 1 and y = 2 is the required solution.

#### Question 8:

$\frac{x}{6}+\frac{y}{15}=4,\phantom{\rule{0ex}{0ex}}\frac{x}{3}-\frac{y}{12}=\frac{19}{4}.$

The given equations may be written as:
$\frac{x}{6}+\frac{y}{15}-4=0$           ...(i)
$\frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0$       ...(ii)
Here,
By cross multiplication, we have: $\frac{x}{\left[\frac{1}{15}×\left(-\frac{19}{4}\right)-\left(-\frac{1}{12}\right)×\left(-4\right)\right]}=\frac{y}{\left[\left(-4\right)×\frac{1}{3}-\left(\frac{1}{6}\right)×\left(-\frac{19}{4}\right)\right]}=\frac{1}{\left[\frac{1}{6}×\left(-\frac{1}{12}\right)-\frac{1}{3}×\frac{1}{15}\right]}$
$\frac{x}{\left(-\frac{19}{60}-\frac{1}{3}\right)}=\frac{y}{\left(-\frac{4}{3}+\frac{19}{24}\right)}=\frac{1}{\left(-\frac{1}{72}-\frac{1}{45}\right)}$
$\frac{x}{\left(-\frac{39}{60}\right)}=\frac{y}{\left(-\frac{13}{24}\right)}=\frac{1}{\left(-\frac{13}{360}\right)}$

Hence, x = 18 and y = 15 is the required solution.

#### Question 9:

Taking $\frac{1}{x}=u$  and $\frac{1}{y}=v$, 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: $\frac{u}{\left[1×\left(-17\right)-3×\left(-7\right)\right]}=\frac{v}{\left[\left(-7\right)×2-1×\left(-17\right)\right]}=\frac{1}{\left[3-2\right]}$
$\frac{u}{-17+21}=\frac{v}{-14+17}=\frac{1}{1}$
$\frac{u}{4}=\frac{v}{3}=\frac{1}{1}$
$u=\frac{4}{1}=4,v=\frac{3}{1}=3$
$\frac{1}{x}=4,\frac{1}{y}=3$
$x=\frac{1}{4},y=\frac{1}{3}$
Hence, $x=\frac{1}{4}$and $y=\frac{1}{3}$ is the required solution.

#### Question 10:

Taking $\frac{1}{x+y}=u$  and $\frac{1}{x-y}=v$, 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: $\frac{u}{\left[-2×\left(-10\right)-1×7\right]}=\frac{v}{\left[1×15-\left(-10\right)×5\right]}=\frac{1}{\left[35+30\right]}$
$\frac{u}{20-7}=\frac{v}{15+50}=\frac{1}{65}$
$\frac{u}{13}=\frac{v}{65}=\frac{1}{65}$
$u=\frac{13}{65}=\frac{1}{5},v=\frac{65}{65}=1$
$\frac{1}{x+y}=\frac{1}{5},\frac{1}{x-y}=1$
So, (x + y) = 5            ...(iii)
and (xy) = 1           ...(iv)

Again, the above equations (iii) and (iv) may be written as:
x + y − 5 = 0           ...(v)
xy − 1 = 0          ...(vi)
Here, a1 = 1, b1 = 1, c1 = −5, a2 = 1, b2 = −1 and c2 = −1
By cross multiplication, we have: ∴ $\frac{x}{\left[1×\left(-1\right)-\left(-5\right)×\left(-1\right)\right]}=\frac{y}{\left[\left(-5\right)×1-\left(-1\right)×1\right]}=\frac{1}{\left[1×\left(-1\right)-1×1\right]}$

$\frac{x}{\left(-1-5\right)}=\frac{y}{\left(-5+1\right)}=\frac{1}{\left(-1-1\right)}$
$\frac{x}{-6}=\frac{y}{-4}=\frac{1}{-2}$
$x=\frac{-6}{-2}=3,y=\frac{-4}{-2}=2$
Hence, x = 3 and y = 2 is the required solution.

#### Question 11:

$\frac{ax}{b}-\frac{by}{a}=a+b,\phantom{\rule{0ex}{0ex}}ax-by=2ab$

The given equations may be written as:
$\frac{ax}{b}-\frac{by}{a}-\left(a+b\right)=0$         ...(i)
$ax-by-2ab=0$            ...(ii)
Here, a1 = $\frac{a}{b}$, b1 = $\frac{-b}{a}$, c1 = −(a + b), a2 = a, b2 = −b and c2 = −2ab
By cross multiplication, we have: $\frac{x}{\left(-\frac{b}{a}\right)×\left(-2ab\right)-\left(-b\right)×\left(-\left(a+b\right)\right)}=\frac{y}{-\left(a+b\right)×a-\left(-2ab\right)×\frac{a}{b}}=\frac{1}{\frac{a}{b}×\left(-b\right)-a×\left(-\frac{b}{a}\right)}$
$\frac{x}{2{b}^{2}-b\left(a+b\right)}=\frac{y}{-a\left(a+b\right)+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{2{b}^{2}-ab-{b}^{2}}=\frac{y}{-{a}^{2}-ab+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{{b}^{2}-ab}=\frac{y}{{a}^{2}-ab}=\frac{1}{-\left(a-b\right)}$
$\frac{x}{-b\left(a-b\right)}=\frac{y}{a\left(a-b\right)}=\frac{1}{-\left(a-b\right)}$
$x=\frac{-b\left(a-b\right)}{-\left(a-b\right)}=b,y=\frac{a\left(a-b\right)}{-\left(a-b\right)}=-a$
Hence, x = b and y = −a is the required solution.

#### Question 12:

$2ax+3by=\left(a+2b\right),\phantom{\rule{0ex}{0ex}}3ax+2by=\left(2a+b\right).$

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: $\frac{x}{\left[3b×\left(-\left(2a+b\right)\right)-2b×\left(-\left(a+2b\right)\right)\right]}=\frac{y}{\left[-\left(a+2b\right)×3a-2a×\left(-\left(2a+b\right)\right)\right]}=\frac{1}{\left[2a×2b-3a×3b\right]}$
$\frac{x}{\left(-6ab-3{b}^{2}+2ab+4{b}^{2}\right)}=\frac{y}{\left(-3{a}^{2}-6ab+4{a}^{2}+2ab\right)}=\frac{1}{4ab-9ab}$
$\frac{x}{{b}^{2}-4ab}=\frac{y}{{a}^{2}-4ab}=\frac{1}{-5ab}$
$\frac{x}{-b\left(4a-b\right)}=\frac{y}{-a\left(4b-a\right)}=\frac{1}{-5ab}$

Hence, $x=\frac{\left(4a-b\right)}{5a}$ and $y=\frac{\left(4b-a\right)}{5b}$ is the required solution.

#### Question 13:

Solve the following system of equations by using the method of cross-multiplication:

Substituting  in the given equations, we get

Here, .
So, by cross-multiplication, we have
$\frac{u}{{b}_{1}{c}_{2}-{b}_{2}{c}_{1}}=\frac{v}{{c}_{1}{a}_{2}-{c}_{2}{a}_{1}}=\frac{1}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\phantom{\rule{0ex}{0ex}}⇒\frac{u}{\left(-b\right)\left[-\left({a}^{2}+{b}^{2}\right)\right]-\left({a}^{2}b\right)\left(0\right)}=\frac{v}{\left(0\right)\left(a{b}^{2}\right)-\left(-{a}^{2}-{b}^{2}\right)\left(a\right)}=\frac{1}{\left(a\right)\left({a}^{2}b\right)-\left(a{b}^{2}\right)\left(-b\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{u}{b\left({a}^{2}+{b}^{2}\right)}=\frac{v}{a\left({a}^{2}+{b}^{2}\right)}=\frac{1}{ab\left({a}^{2}+{b}^{2}\right)}$

Hence, x = a and y = b.

#### Question 1:

Show that the following system of equations has a unique solution:
$3x+5y=12,5x+3y=4.$
Also, find the solution of the given system of equations.

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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{3}{5}\ne \frac{5}{3}$
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.

#### Question 2:

Show that the following  system of equations has a unique solution and solve it:
$2x-3y=17\phantom{\rule{0ex}{0ex}}4x+y=13$

The system of equations can be written as

The given equations are of the form

where
Now,

Since, $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, therefore the system of equations has unique solution.
Using cross multiplication method, we have
$\frac{x}{{b}_{1}{c}_{2}-{b}_{2}{c}_{1}}=\frac{y}{{c}_{1}{a}_{2}-{c}_{2}{a}_{1}}=\frac{1}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{-3\left(-13\right)-1×\left(-17\right)}=\frac{y}{-17×4-\left(-13\right)×2}=\frac{1}{2×1-4×\left(-3\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{39+17}=\frac{y}{-68+26}=\frac{1}{2+12}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{56}=\frac{y}{-42}=\frac{1}{14}$

Hence, .

#### Question 3:

Show that the following system of equations has a unique solution:
$\frac{x}{3}+\frac{y}{2}=3,x-2y=2.$
Also, find the solution of the given system of equations.

The given system of equations are:
$\frac{x}{3}+\frac{y}{2}=3$
$\frac{2x+3y}{6}=3$
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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{2}{1}\ne \frac{3}{-2}$
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.

#### Question 4:

Find the value of k for which the following equations has a unique solution:
$2x+3y-5=0\phantom{\rule{0ex}{0ex}}kx-6y-8=0$

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
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{k}\ne \frac{3}{-6}\phantom{\rule{0ex}{0ex}}⇒k\ne -4$
Hence, $k\ne -4$.

#### Question 5:

Find the value of k for which the following equations has a unique solution:
$x-ky-2=0\phantom{\rule{0ex}{0ex}}3x+2y+5=0$

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
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{3}\ne \frac{-k}{2}\phantom{\rule{0ex}{0ex}}⇒k\ne -\frac{2}{3}$
Hence, $k\ne -\frac{2}{3}$.

#### Question 6:

Find the value of k for which the following system of equations has a unique solution:
$5x-7y-5=0\phantom{\rule{0ex}{0ex}}2x+ky-1=0$

The given system of equations is

This system is of the form
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
where
For the given system of equations to have a unique solution, we must have
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{2}\ne \frac{-7}{k}\phantom{\rule{0ex}{0ex}}⇒k\ne \frac{-14}{5}$
Hence, $k\ne -\frac{14}{5}$.

#### Question 7:

Find the value of k for which the following system of equations has a unique solution:
$4x+ky+8=0\phantom{\rule{0ex}{0ex}}x+y+1=0$

The given system of equations is

This system is of the form
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
where
For the given system of equations to have a unique solution, we must have
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{1}\ne \frac{k}{1}\phantom{\rule{0ex}{0ex}}⇒k\ne 4$
Hence, $k\ne 4$.

#### Question 8:

Find the value of k for which each of the following systems of equations has a unique solution:

The given system of equations:
4x − 5y = k
⇒ 4x − 5yk = 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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
i.e.  $\frac{4}{2}\ne \frac{-5}{-3}$
$⇒2\ne \frac{5}{3}⇒6\ne 5$
Thus, for all real values of k, the given system of equations will have a unique solution.

#### Question 9:

Find the value of k for which each of the following systems of equations has a unique solution:

The given system of equations:
kx + 3y = (k − 3)
⇒ kx + 3y − (k − 3) = 0            ....(i)
And, 12x + ky = k
⇒ 12x + kyk = 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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
i.e.  $\frac{k}{12}\ne \frac{3}{k}$
$⇒{k}^{2}\ne 36⇒k\ne ±6$
Thus, for all real values of k other than $±6$, the given system of equations will have a unique solution.

#### Question 10:

Show that the system of equations

has an infinite number of solutions.

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

$\therefore \frac{{a}_{1}}{{a}_{2}}=\frac{2}{6}=\frac{1}{3},\frac{{b}_{1}}{b2}=\frac{-3}{-9}=\frac{1}{3}$ and $\frac{{c}_{1}}{{c}_{2}}=\frac{-5}{-15}=\frac{1}{3}$
Thus, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
Hence, the given system of equations has an infinite number of solutions.

#### Question 11:

Show that the system of equations  has no solution.

The given system of equations can be written as

This system is of the form
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
where
Now,
$\frac{{a}_{1}}{{a}_{2}}=\frac{6}{9}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\frac{{b}_{1}}{{b}_{2}}=\frac{5}{\frac{15}{2}}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\frac{{c}_{1}}{{c}_{2}}=\frac{-11}{-21}=\frac{11}{21}$
Since, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$, therefore the given system has no solution.

#### Question 12:

For what value of k, the system of equations

has (i) a unique solution, (ii) no solution?

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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{k}{3}\ne \frac{2}{-4}⇒k\ne \frac{-3}{2}$
Thus for all real values of k other than $\frac{-3}{2}$, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$⇒\frac{k}{3}=\frac{2}{-4}\ne \frac{-5}{-10}$

$⇒k=\frac{-3}{2},k\ne \frac{3}{2}$
Hence, the required value of k is $\frac{-3}{2}$.

#### Question 13:

For what value of k, the system of equations

has (i) a unique solution, (ii) no solution?

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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{1}{3}\ne \frac{2}{k}⇒k\ne 6$
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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$⇒\frac{1}{3}=\frac{2}{k}\ne \frac{-5}{15}$

Hence, the required value of k is 6.

#### Question 14:

For what value of k does the system of equations

has (i) a unique solution, (ii) no solution?
Also, show that there is no value of k for which the given system of equations has infinitely many solutions.

The given system of equations is:
x + 2y = 3
x + 2y − 3= 0                    ....(i)
And, 5xky + 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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e. $\frac{1}{5}\ne \frac{2}{k}⇒k\ne 10$
Thus, for all real values of k​, 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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{1}{5}=\frac{2}{k}\ne \frac{-3}{7}$

$⇒k=10,k\ne \frac{14}{-3}$
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.

#### Question 15:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$2x+3y=7,\phantom{\rule{0ex}{0ex}}\left(k-1\right)x+\left(k+2\right)y=3k.$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(k-1\right)}=\frac{3}{\left(k+2\right)}=\frac{-7}{-3k}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{2}{\left(k-1\right)}=\frac{3}{\left(k+2\right)}=\frac{7}{3k}$

Now, we have the following three cases:
Case I:
$\frac{2}{k-1}=\frac{3}{k+2}$
$⇒2\left(k+2\right)=3\left(k-1\right)⇒2k+4=3k-3⇒k=7$

Case II:
$\frac{3}{k+2}=\frac{7}{3k}$
$⇒7\left(k+2\right)=9k⇒7k+14=9k⇒2k=14⇒k=7$

Case III:
$\frac{2}{k-1}=\frac{7}{3k}$
$⇒7k-7=6k⇒k=7$

Hence, the given system of equations has an infinite number of solutions when k is equal to 7.

#### Question 16:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$2x+\left(k-2\right)y=k,\phantom{\rule{0ex}{0ex}}6+\left(2k-1\right)y=\left(2k+5\right).$

The given system of equations:
2x + (k − 2)y = k
⇒ 2x + (k − 2)yk = 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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{6}=\frac{\left(k-2\right)}{\left(2k-1\right)}=\frac{-k}{-\left(2k+5\right)}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{1}{3}=\frac{\left(k-2\right)}{\left(2k-1\right)}=\frac{k}{\left(2k+5\right)}$

Now, we have the following three cases:
Case I:
$\frac{1}{3}=\frac{k-2}{2k-1}$
$⇒\left(2k-1\right)=3\left(k-2\right)$
$⇒2k-1=3k-6⇒k=5$

Case II:
$\frac{k-2}{2k-1}=\frac{k}{2k+5}$
$⇒\left(k-2\right)\left(2k+5\right)=k\left(2k-1\right)$
$⇒2{k}^{2}+5k-4k-10=2{k}^{2}-k$
$⇒k+k=10⇒2k=10⇒k=5$

Case III:
$\frac{1}{3}=\frac{k}{2k+5}$
$⇒2k+5=3k⇒k=5$

Hence, the given system of equations has an infinite number of solutions when k is equal to 5.

#### Question 17:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$kx+3y=\left(2x+1\right),\phantom{\rule{0ex}{0ex}}2\left(k+1\right)x+9y=\left(7k+1\right).$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$

$⇒\frac{k}{2\left(k+1\right)}=\frac{1}{3}=\frac{\left(2k+1\right)}{\left(7k+1\right)}$

Now, we have the following three cases:
Case I:
$\frac{k}{2\left(k+1\right)}=\frac{1}{3}$
$⇒2\left(k+1\right)=3k$
$⇒2k+2=3k$
$⇒k=2$

Case II:
$\frac{1}{3}=\frac{2k+1}{7k+1}$
$⇒\left(7k+1\right)=6k+3$
$⇒k=2$

Case III:
$\frac{k}{2\left(k+1\right)}=\frac{2k+1}{7k+1}$
$⇒k\left(7k+1\right)=\left(2k+1\right)×2\left(k+1\right)$
$⇒7{k}^{2}+k=\left(2k+1\right)\left(2k+2\right)$
$⇒7{k}^{2}+k=4{k}^{2}+4k+2k+2$
$⇒3{k}^{2}-5k-2=0$
$⇒3{k}^{2}-6k+k-2=0$
$⇒3k\left(k-2\right)+1\left(k-2\right)=0$
$⇒\left(3k+1\right)\left(k-2\right)=0$

Hence, the given system of equations has an infinite number of solutions when k is equal to 2.

#### Question 18:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$5x+2y=2k,\phantom{\rule{0ex}{0ex}}2\left(k+1\right)x+ky=\left(3k+4\right).$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{-2k}{-\left(3k+4\right)}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$

Now, we have the following three cases:
Case I:
$\frac{5}{2\left(k+1\right)}=\frac{2}{k}$
$⇒2×2\left(k+1\right)=5k⇒4\left(k+1\right)=5k$
$⇒4k+4=5k⇒k=4$

Case II:
$\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$
$⇒2{k}^{2}=2×\left(3k+4\right)$
$⇒2{k}^{2}=6k+8⇒2{k}^{2}-6k-8=0$
$⇒2\left({k}^{2}-3k-4\right)=0$
$⇒{k}^{2}-4k+k-4=0$
$⇒k\left(k-4\right)+1\left(k-4\right)=0$

Case III:
$\frac{5}{2\left(k+1\right)}=\frac{2k}{3k+4}\phantom{\rule{0ex}{0ex}}$
$⇒15k+20=4{k}^{2}+4k\phantom{\rule{0ex}{0ex}}$
$⇒4{k}^{2}-11k-20=0\phantom{\rule{0ex}{0ex}}$
$⇒4{k}^{2}-16k+5k-20=0\phantom{\rule{0ex}{0ex}}$
$⇒4k\left(k-4\right)+5\left(k-4\right)=0\phantom{\rule{0ex}{0ex}}$

Hence, the given system of equations has an infinite number of solutions when k is equal to 4.

#### Question 19:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$\left(k-1\right)x-y=5,\phantom{\rule{0ex}{0ex}}\left(k+1\right)x+\left(1-k\right)y=\left(3k+1\right).$

The given system of equations:
(k − 1)xy = 5
⇒ (k − 1)xy − 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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$

$⇒\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{1}{\left(k-1\right)}=\frac{5}{\left(3k+1\right)}$

Now, we have the following three cases:
Case I:
$\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{1}{\left(k-1\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒{\left(k-1\right)}^{2}=\left(k+1\right)\phantom{\rule{0ex}{0ex}}$
$⇒{k}^{2}+1-2k=k+1\phantom{\rule{0ex}{0ex}}$

Case II:
$\frac{1}{\left(k-1\right)}=\frac{5}{\left(3k+1\right)}\phantom{\rule{0ex}{0ex}}$
$⇒3k+1=5\left(k-1\right)\phantom{\rule{0ex}{0ex}}$
$⇒3k+1=5k-5\phantom{\rule{0ex}{0ex}}$
$⇒2k=6⇒k=3$

Case III:
$\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{5}{\left(3k+1\right)}\phantom{\rule{0ex}{0ex}}$
$⇒\left(3k+1\right)\left(k-1\right)=5\left(k+1\right)\phantom{\rule{0ex}{0ex}}$
$⇒3{k}^{2}+k-3k-1=5k+5\phantom{\rule{0ex}{0ex}}$
$⇒3{k}^{2}-2k-5k-1-5=0\phantom{\rule{0ex}{0ex}}$
$⇒3{k}^{2}-7k-6=0\phantom{\rule{0ex}{0ex}}$
$⇒3{k}^{2}-9k+2k-6=0\phantom{\rule{0ex}{0ex}}$
$⇒3k\left(k-3\right)+2\left(k-3\right)=0\phantom{\rule{0ex}{0ex}}$
$⇒\left(k-3\right)\left(3k+2\right)=0\phantom{\rule{0ex}{0ex}}$

Hence, the given system of equations has an infinite number of solutions when k is equal to 3.

#### Question 20:

Find the value of k for which the following system of linear equations has an infinite number of solutions:
$\left(k-3\right)x+3y=k\phantom{\rule{0ex}{0ex}}kx+ky=12$

The given system of equations can be written as

This system is of the form
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
where
For the given system of linear equations to have an infinite number of solutions
, we must have

Hence, k = 6.

#### Question 21:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$\left(a-1\right)x+3y=2,\phantom{\rule{0ex}{0ex}}6x+\left(1-2b\right)y=6.$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{\left(a-1\right)}{6}=\frac{3}{\left(1-2b\right)}=\frac{-2}{-6}$
$⇒\frac{a-1}{6}=\frac{3}{\left(1-2b\right)}=\frac{1}{3}$
$⇒\frac{a-1}{6}=\frac{1}{3}\mathrm{and}\frac{3}{\left(1-2\mathrm{b}\right)}=\frac{1}{3}$
⇒ 3a − 3 = 6 and 9 = 1 − 2b
⇒ 3a = 9 and 2b = −8
a = 3 and b = −4
∴​ a = 3 and b = −4

#### Question 22:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$\left(2a-1\right)x+3y=5,\phantom{\rule{0ex}{0ex}}3x+\left(b-1\right)y=2.$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{\left(2a-1\right)}{3}=\frac{3}{\left(b-1\right)}=\frac{-5}{-2}$
$⇒\frac{\left(2a-1\right)}{3}=\frac{3}{\left(b-1\right)}=\frac{5}{2}$

⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒  4a − 2 = 15 and 6 = 5b − 5
⇒  4a = 17 and 5b = 11

∴​ a$\frac{17}{4}$ and b = $\frac{11}{5}$

#### Question 23:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$2x-3y=7,\phantom{\rule{0ex}{0ex}}\left(a+b\right)x-\left(a+b-3\right)y=4a+b.$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{a+b}=\frac{-3}{-\left(a+b-3\right)}=\frac{-7}{-\left(4a+b\right)}$
$⇒\frac{2}{a+b}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}$

⇒ 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

#### Question 24:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$2x+3y=7,\phantom{\rule{0ex}{0ex}}\left(a+b+1\right)x+\left(a+2b+2\right)y=4\left(a+b\right)+1.$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{-7}{-\left[4\left(a+b\right)+1\right]}$
$⇒\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{7}{4\left(a+b\right)+1}$
$⇒\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}\mathrm{and}\frac{3}{\left(\mathrm{a}+2\mathrm{b}+2\right)}=\frac{7}{4\left(a+b\right)+1}$

⇒ 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
ab − 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
ab = 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
∴​ a = 3 and b = 2

#### Question 25:

Find the values of a and b for which the following system of linear equations has an infinite number of solutions:
$2x+3y=7\phantom{\rule{0ex}{0ex}}\left(a+b\right)x+\left(2a-b\right)y=21$                               [CBSE 2001]

The given system of equations can be written as

This system is of the form
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
where
For the given system of linear equations to have an infinite number of solutions, we must have

Adding , we get
$3a=15⇒a=\frac{15}{3}=5$
Now, substituting a = 5 in a + b = 6, we have

Hence, a = 5 and b = 1.

#### Question 26:

Find the values of and b for which the following system of linear equations has an infinite number of solutions:
$2x+3y=7\phantom{\rule{0ex}{0ex}}2ax+\left(a+b\right)y=28$                            [CBSE 2001]

The given system of equations can be written as

This system is of the form
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
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
$4+b=12⇒b=12-4=8$
Hence, a = 4 and b = 8.

#### Question 27:

Find the value of k for which each of the following system of equations has no solution:

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$

$\frac{8}{k}=\frac{1}{2}$ and $\frac{8}{k}\ne \frac{3}{5}$

Hence, the given system of equations has no solution when k is equal to 16.

#### Question 28:

Find the value of k for which each of the following system of equations has no solution:

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
i.e.  $\frac{k}{12}=\frac{3}{k}\ne \frac{-3}{-6}$
$\frac{k}{12}=\frac{3}{k}$ and $\frac{3}{k}\ne \frac{1}{2}$

Hence, the given system of equations has no solution when k is equal to −6.

#### Question 29:

Find the value of k for which each of the following system of equations has no solution:

The given system of equations:
3xy − 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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
i.e. $\frac{3}{6}=\frac{-1}{-2}\ne \frac{-5}{k}$
$\frac{-1}{-2}\ne \frac{-5}{k}⇒k\ne -10$
Hence, equations (i) and (ii) will have no solution if $k\ne -10$.

#### Question 30:

Find the value of for which the following system of linear equations has no solutions:
$kx+3y=k-3\phantom{\rule{0ex}{0ex}}12x+ky=k$

The given system of equations can be written as

This system is of the form
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
where
For the given system of linear equations to have no solution, we must have

Hence, $k=-6$.

#### Question 31:

Find the value of k for which the system of equations

has a nonzero solution.

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}$
$⇒\frac{5}{2}=\frac{-3}{k}\phantom{\rule{0ex}{0ex}}$
$⇒5k=-6⇒k=\frac{-6}{5}$
Hence, the required value of k is $\frac{-6}{5}$.

#### Question 1:

5 chairs and 4 tables together cost ₹ 5,600, while 4 chairs and 3 tables together cost
₹ 4,340. Find the cost of a chair and that of a table.

Let the cost of a chair be ₹ x and that of a table be ₹ y. Then

Multiplying (i) by 3 and (ii) by 4, we get
$15x-16x=16800-17360\phantom{\rule{0ex}{0ex}}⇒-x=-560\phantom{\rule{0ex}{0ex}}⇒x=560$
Substituting x = 560 in (i), we have
$5×560+4y=5600\phantom{\rule{0ex}{0ex}}⇒4y=5600-2800\phantom{\rule{0ex}{0ex}}⇒y=\frac{2800}{4}=700$
Hence, the cost of a chair and that of a table are respectively ₹ 560 and ₹ 700.

#### Question 2:

23 spoons and 17 forks together cost ₹1,770, while 17 spoons and 23 forks together
cost ₹1,830. Find the cost of a spoon and that of a fork.

Let the cost of a spoon be ₹and that of a fork be ₹y. Then

Adding (i) and (ii), we get

Now, subtracting (ii) from (i), we get

Adding (iii) and (iv), we get
$2x=80⇒x=40$
Substituting x = 40 in (iii), we get
$40+y=90⇒y=50$
Hence, the cost of a spoon that of a fork are ₹40 and ₹50 respectively.

#### Question 3:

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all
totalling ₹19.50, how many coins of each kind does she have?

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
$x+22=50\phantom{\rule{0ex}{0ex}}⇒x=50-22=28$
Hence, the number of 25-paisa and 50-paisa conis are 22 and 28 respectively.

#### Question 4:

The sum of two numbers is 137 and their difference is 43. Find the numbers.

Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137                  ...(i)
xy = 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.

#### Question 5:

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.

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.

#### Question 6:

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.

Let the first number be x and the second number be y.
Then, we have:
3x + y = 142                              ....(i)
4xy = 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.

#### Question 7:

If 45 is subtracted from twice the greater of two numbers, it result in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.

Let the greater number be x and the smaller be y.
Then, we have:
2x − 45 = y  or  2xy = 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.

#### Question 8:

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.

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.

#### Question 9:

If 2 is added to each of two given numbers, their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.

Let the required numbers be x and y.
Now, we have:
$\frac{x+2}{y+2}=\frac{1}{2}$
By cross multiplication, we get:
2x + 4 = y + 2
⇒ 2xy  = −2                  ....(i)
Again, we have:
$\frac{x-4}{y-4}=\frac{5}{11}$
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.

#### Question 10:

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.

Let the larger number be x and the smaller number be y.
Then, we have:
xy = 14 or x = 14 + y                    ....(i)
x2y2 = 448                                      ....(ii)
On substituting x = 14 + y in (ii), we get:
(14 + y)2y2 = 448
⇒ 196 + y2 + 28yy2 = 448
⇒ 196 + 28y = 448
⇒ 28y = (448 − 196) = 252
⇒ $y=\frac{252}{28}=9$
On substituting y = 9 in (i), we get:
x = 14 + 9 = 23
Hence, the required numbers are 23 and 9.

#### Question 11:

The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
x + y = 12                 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 18
⇒ 10y + x − 10xy = 18
⇒ 9y − 9x = 18
yx = 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.

#### Question 12:

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.

Let the tens and the units digits of the required number be x and y​, 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
⇒ 10xx + y − 10y = 27
⇒ 9x − 9y = 27
⇒ 9(xy) = 27
xy = 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.

#### Question 13:

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
x + y = 15                  ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 9
⇒ 10y + x − 10xy = 9
⇒ 9y − 9x = 9
yx = 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.

#### Question 14:

A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
∴ 10x + y = 4(x + y) + 3
⇒10x + y = 4x + 4y + 3
⇒ 6x − 3y = 3
⇒ 2xy = 1                   ....(i)

Again, we have:
10x + y + 18 = 10y + x
⇒ 9x − 9y = −18
xy = −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.

#### Question 15:

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.

We know:
Dividend = (Divisor × Quotient) + Remainder

Let the tens and the units digits of the required number be x and y​, 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(xy) = 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.

#### Question 16:

A two-digit number is such that the product of its digit is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Let the tens and the units digits of the required number be x and y​, 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(yx) = 18
yx = 2                 ....(ii)

We know:
(y + x)2 − (yx)2 = 4xy
$\left(y+x\right)=±\sqrt{{\left(y-x\right)}^{2}+4xy}$

y + x = 12              .....(iii) (∵ 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.

#### Question 17:

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Let the tens and the units digits of the required number be x and y​, 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(xy) = 63
xy = 7                ....(ii)

We know:
(x + y)2 − (xy)2 = 4xy
$\left(x+y\right)=±\sqrt{{\left(x-y\right)}^{2}+4xy}$

x + y = 11              ....(iii) (∵ 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.

#### Question 18:

The sum of a two-digit number and the number obtained by reversing the order of its digits
is 121, and the two digits differ by 3. Find the number.

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
$2x=14⇒x=7$
Putting x = 7 in (i), we get
$7+y=11⇒y=4$
Changing the role of x and y, x = 4 and y = 7
Hence, the two-digit number is 74 or 47.

#### Question 19:

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes $\frac{3}{4}$. Find the fraction.

Let the required fraction be $\frac{x}{y}$.
Then, we have:
x + y = 8                        ....(i)
And, $\frac{x+3}{y+3}=\frac{3}{4}$
⇒ 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
∴​ ​x = 3 and y = 5
Hence, the required fraction is $\frac{3}{5}$.

#### Question 20:

If 2 is added to the numerator of a fraction, it reduces to $\left(\frac{1}{2}\right)$ and if 1 is subtracted from the denominator, it reduces to $\left(\frac{1}{3}\right)$. Find the fraction.

Let the required fraction be $\frac{x}{y}$.
Then, we have:
$\frac{x+2}{y}=\frac{1}{2}$
⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2xy = −4                      .....(i)

Again, $\frac{x}{y-1}=\frac{1}{3}$
⇒ 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 $\frac{3}{10}$.

#### Question 21:

The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes $\frac{3}{4}$. Find the fraction.

Let the required fraction be $\frac{x}{y}$.
Then, we have:
y = x + 11
yx = 11                        ....(i)
Again, $\frac{x+8}{y+8}=\frac{3}{4}$
⇒ 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
∴​ ​x = 25 and  y = 36
Hence, the required fraction is $\frac{25}{36}$.

#### Question 22:

Find a fraction which becomes $\left(\frac{1}{2}\right)$ when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes $\left(\frac{1}{3}\right)$ when 7 is subtracted from the numerator and 2 is subtracted from the denominator.

Let the required fraction be $\frac{x}{y}$.
Then, we have:
$\frac{x-1}{y+2}=\frac{1}{2}$
⇒ 2(x − 1) = 1(y + 2)
⇒ 2x − 2 = y + 2
⇒ 2xy = 4                    ....(i)

Again, $\frac{x-7}{y-2}=\frac{1}{3}$
⇒ 3(x − 7) = 1(y − 2)
⇒ 3x − 21 = y − 2
⇒ 3xy = 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
∴​ ​x = 15 and y = 26
Hence, the given fraction is $\frac{15}{26}$.

#### Question 23:

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.      [CBSE 2010]

Let the fraction be $\frac{x}{y}$
As per the question

After changing the numerator and denominator
New numerator = x + 3
New denominator = + 3
Therefore

Multiplying (i) by 3 and subtracting (ii), we get
$3y-2y=12-3\phantom{\rule{0ex}{0ex}}⇒y=9$
Now, putting y = 9 in (i), we get
$9-x=4⇒x=9-4=5$
Hence, the fraction is $\frac{5}{9}$.

#### Question 24:

The sum of two numbers is 16 and the sum of their reciprocals is $\frac{1}{3}$. Find the numbers.

Let the larger number be x and the smaller number be y.
Then, we have:
(x + y) = 16                      ....(i)
And, $\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$                  ....(ii)
$\frac{x+y}{xy}=\frac{1}{3}$
⇒ 3(x + y) = xy
⇒ 3 × 16 = xy      [Since from (i), we have: x + y = 16]
xy = 48                         ....(iii)
We know:
(xy)2 = (x + y)2 − 4xy
(xy)2 = (16)2 4 × 48 = 256 − 192 = 64
∴ (xy) =
Since x is larger and y is smaller, we have:
xy = 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.

#### Question 25:

There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.

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.

#### Question 26:

Taxi charges in a city consist of fixed charges and the remaining depending upon the distance
travelled in kilometres. If a man travels 80 km, he pays ₹1,330, and travelling 90 km, he pays
₹1,490. Find the fixed charges and rate per km.

Let fixed charges be ₹x and rate per km be ₹y.
Then as per the question

Subtracting (i) from (ii), we get
$10y=160⇒y=\frac{160}{10}=16$
Now, putting y = 16, we have
$x+80×16=1330\phantom{\rule{0ex}{0ex}}⇒x=1330-1280=50\phantom{\rule{0ex}{0ex}}$
Hence, the fixed charges be ₹50 and the rate per km is ₹16.

#### Question 27:

A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay ₹4,500, whereas a student B who takes food for 30 days, he has to pay ₹5,200. Find the fixed charges per month and the cost of the food per day.

Let the fixed charges be ₹x and the cost of food per day be ₹y.
Then as per the question

Subtracting (i) from (ii), we get
$5y=700⇒y=\frac{700}{5}=140$
Now, putting = 140, we have
$x+25×140=4500\phantom{\rule{0ex}{0ex}}⇒x=4500-3500=1000\phantom{\rule{0ex}{0ex}}$
Hence, the fixed charges is ₹1000 and the cost of the food per day is ₹140.

#### Question 28:

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received ₹1,350 as annual interest. Had he interchanged the amounts invested, he would have received ₹45 less as interest. What amounts did he invest at different rates?

Let the the amounts invested at 10% and 8% be ₹x and ₹y 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
$4x=34000\phantom{\rule{0ex}{0ex}}⇒x=\frac{34000}{4}=8500$
Putting x = 8500 in (iii), we get
$2×8500+2y=29500\phantom{\rule{0ex}{0ex}}⇒2y=29500-17000=12500\phantom{\rule{0ex}{0ex}}⇒y=\frac{12500}{2}=6250$
Hence, the amounts invested are ₹8,500 at 10% and ₹6,250 at 8%.

#### Question 29:

The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves ₹9,000 per month, find the monthly income of each.

Let the monthly income of A and B are ₹x and ₹y respectively.
Then as per the question

Since each save ₹9,000, so
Expenditure of A = ​₹$\left(x-9000\right)$
Expenditure of B = ​₹$\left(y-9000\right)$
The ratio of expenditures of A and B are in the ratio 7 : 5.

From (i), substitute $y=\frac{4x}{5}$ in (ii) to get
$7×\frac{4x}{5}-5x=18000\phantom{\rule{0ex}{0ex}}⇒28x-25x=90000\phantom{\rule{0ex}{0ex}}⇒3x=90000\phantom{\rule{0ex}{0ex}}⇒x=30000\phantom{\rule{0ex}{0ex}}$
Now, putting x = 30000, we get
$y=\frac{4×30000}{5}=4×6000=24000$
Hence, the monthly incomes of A and B are ​₹30,000 and ​₹24,000.

#### Question 30:

A man sold a chair and a table together for ​₹1,520, thereby making a profit of 25% on chair and 10% on table. By selling them together for ​₹1,535, he would would have made a profit of 10% on the chair and 25% on the table. Find the cost of each.

Let the cost price of the chair and table be ₹x and ₹y 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 ​₹600 and ​₹700 respectively.

#### Question 31:

Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.

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(xy) = 70
⇒ (xy) = 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.

#### Question 32:

A train covered a certain distance at a uniform speed. If the train had been 5 kmph faster, it would have taken 3 hours less than the scheduled time. And, If the train were slower by 4 kmph, it would have taken 3 hours more than the scheduled time. Find the length of the journey.

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

#### Question 33:

Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.

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
$\frac{18}{x}-\frac{13}{x}=\frac{66}{200}-\frac{28}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{x}=\frac{10}{200}⇒x=100\phantom{\rule{0ex}{0ex}}$
Putting x = 100 in (i), we have
$\frac{3}{100}+\frac{2}{y}=\frac{11}{200}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{y}=\frac{11}{200}-\frac{3}{100}=\frac{1}{40}\phantom{\rule{0ex}{0ex}}⇒y=80$
Hence, the speed of the train and that of the taxi are 100 km/h and 80 km/h respectively.

#### Question 34:

Places A and B are 160 km apart on a highway. One car starts from A and another car from B at the same time. If they travel in the same direction , they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.

Let the speed of the car A and B be x km/h and 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 = 50 in (ii), we have

Hence, the speeds of the cars are 50 km/h and 30 km/h.

#### Question 35:

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.

Let the speed of the sailor in still water be x km/h and that of the current km/h.
Speed downstream = (x + y) km/h
Speed upstream = (x − y) km/h
As per the question
$\left(x+y\right)×\frac{40}{60}=8\phantom{\rule{0ex}{0ex}}⇒x+y=12.....\left(i\right)$
When the sailor goes upstream, then
$\left(x-y\right)×1=8\phantom{\rule{0ex}{0ex}}x-y=8.....\left(ii\right)$
Adding (i) and (ii), we get
$2x=20⇒x=10$
Putting x = 10 in (i), we have
$10+y=12⇒y=2$
Hence, the speeds of the sailor in staill water and the current are 10 km/h and 2 km/h respectively.

#### Question 36:

A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
Then, we have:
Speed upstream = (xy)km/hr
Speed downstream = (x + y) km/hr
Time taken to cover 12 km upstream = $\frac{12}{\left(x-y\right)}$ hrs
Time taken to cover 40 km downstream = $\frac{40}{\left(x+y\right)}$ hrs
Total time taken = 8 hrs
$\frac{12}{\left(x-y\right)}$ + $\frac{40}{\left(x+y\right)}$ = 8        ....(i)

Again, we have:
Time taken to cover 16 km upstream = $\frac{16}{\left(x-y\right)}$ hrs
Time taken to cover 32 km downstream = $\frac{32}{\left(x+y\right)}$ hrs
Total time taken = 8 hrs

$\frac{16}{\left(x-y\right)}$ + $\frac{32}{\left(x+y\right)}$ = 8        ....(ii)
Putting $\frac{1}{\left(x-y\right)}=u$ and $\frac{1}{\left(x+y\right)}=v$ in (i) and (ii), we get:
12u + 40v = 8
3u + 10v = 2            ....(a)
And,  16u + 32v = 8
⇒ 2u + 4v = 1          ....(b)
On multiplying (a) by 4 and (b) by 10, we get:
12u + 40v = 8                    ....(iii)
And, 20u + 40v = 10            ....(iv)
On subtracting (iii) from (iv), we get:
8u = 2
$u=\frac{2}{8}=\frac{1}{4}$
On substituting $u=\frac{1}{4}$ in (iii), we get:
40v = 5
$v=\frac{5}{40}=\frac{1}{8}$
Now, we have:
$u=\frac{1}{4}$
$\frac{1}{x-y}=\frac{1}{4}⇒x-y=4$             ....(v)
$v=\frac{1}{8}$
$\frac{1}{x+y}=\frac{1}{8}⇒x+y=8$            ....(vi)
On adding (v) and (vi), we get:
2x = 12
x = 6
On substituting x = 6 in (v), we get:
6 − y = 4
y = (6 − 4) = 2
∴ Speed of the boat in still water = 6 km/h
And, speed of the stream = 2 km/h

#### Question 37:

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.

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 = $\frac{1}{x}$
And, one boy's one day's work = $\frac{1}{y}$
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) = $\frac{1}{4}$
$\frac{2}{x}+\frac{5}{y}=\frac{1}{4}$
$2u+5v=\frac{1}{4}$              ...(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) = $\frac{1}{3}$
$\frac{3}{x}+\frac{6}{y}=\frac{1}{3}$
$3u+6v=\frac{1}{3}$             ....(ii)           Here,
On multiplying (i) by 6 and (ii) by 5, we get:
$12u+30v=\frac{6}{4}$             ....(iii)
$15u+30v=\frac{5}{3}$             ....(iv)
On subtracting (iii) from (iv), we get:
$3u=\left(\frac{5}{3}-\frac{6}{4}\right)=\frac{2}{12}=\frac{1}{6}$
$u=\frac{1}{6×3}=\frac{1}{18}⇒\frac{1}{x}=\frac{1}{18}⇒x=18$
On substituting $u=\frac{1}{18}$ in (i), we get:
$2×\frac{1}{18}+5v=\frac{1}{4}⇒5v=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$
$v=\left(\frac{5}{36}×\frac{1}{5}\right)=\frac{1}{36}⇒\frac{1}{y}=\frac{1}{36}⇒y=36$
Hence, one man alone can finish the work in 18 days and one boy alone can finish the work in 36 days.

#### Question 38:

The length of a room exceeds its breadth by 3 metres. If the length in increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.

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
xy = 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.

#### Question 39:

The area of a rectangle gets reduced by 8 m2, when its length is reduced by 5 m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m2. Find the length and the breadth of the rectangle.

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
xyxy + 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.

#### Question 40:

The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m.
If the length is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.

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
$2y=38⇒y=19$
Now, putting y = 19 in (ii), we have
$4x-19=93\phantom{\rule{0ex}{0ex}}⇒4x=93+19=112\phantom{\rule{0ex}{0ex}}⇒x=28$
Hence, length = 28 m and breadth = 19 m.

#### Question 41:

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket.
One reserved first class ticket from Mumbai to Delhi costs ₹4,150 while one full and one half reserved first class
tickets cost ₹6,255. What is the basic first class full fare and what is the reservation charge?

Let the the basic first class full fare be ₹x and the reservation charge be ₹y.
Case 1: One reservation first class full ticket cost ₹4,150

Case 2: One full and one half reserved first class tickets cost ₹6,255

Substituting $y=4150-x$ from (i) in (ii), we get
$3x+4\left(4150-x\right)=12510\phantom{\rule{0ex}{0ex}}⇒3x-4x+16600=12510\phantom{\rule{0ex}{0ex}}⇒x=16600-12510=4090$
Now, putting x = 4090 in (i), we have
$4090+y=4150\phantom{\rule{0ex}{0ex}}⇒y=4150-4090=60$
Hence, cost of basic first class full fare = ₹4,090 and reservation charge = ₹60.

#### Question 42:

Five years hence, a man's age will be three times the age of his son. Five years ago, the man was seven times as old as his son.
Find their present ages.

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
$4y=40⇒y=10$
Putting y = 10 in (i), we get
$x-3×10=10\phantom{\rule{0ex}{0ex}}⇒x=10+30=40$
Hence, man's present age = 40 years and son's present age = 10 years.

#### Question 43:

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the ago of the son. Find the present ages of the man and his son.

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.

#### Question 44:

If twice the son's age in years is added to the mother's age, the sum is 70 years. But, if twice the mother's age is added to the son's age, the sum is 95 years. Find the age of the mother and that of the son.

Let the mother'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 mother's present age is 40 years and her son's present age is 15 years.

#### Question 45:

The present age of a woman is 3 years more than three times the ages of her daughter. Three years hence, the woman's age will be 10 years more than twice the age of her daughter. Find their present ages.

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.

#### Question 46:

On selling a tea at 5% loss and a lemon set at 15% gain, a crockery seller gains ₹7. If he sells the tea set at 5% gain and the
lemon set at 10% gain, he gains ₹13. Find the actual price of each of the tea set and the lemon set.

Let the actual price of the tea and lemon set be ₹x and ₹y respectively.
When gain is ₹7, then

When gain is ₹14, then

Multiplying (i) by 2 and adding with (ii), we have
$7y=280+280\phantom{\rule{0ex}{0ex}}⇒y=\frac{560}{7}=80$
Putting y = 80 in (ii), we get
$80+2x=280\phantom{\rule{0ex}{0ex}}⇒x=\frac{200}{2}=100$
Hence, actual price of the tea set and lemon set are ₹100 and ₹80 respectively.

#### Question 47:

A lending library  has a fixed charge for the first three days and an additional charge for each day thereafter.
Mona paid ₹27 for a book kept for 7 days, while Tanvy paid ₹21 for the book she kept for 5 days.
Find the fixed charge and the charge for each extra day.

Let the fixed charge be ₹x and the charge for each extra day be ₹y.
In case of  Mona, as per the question

In case of Tanvy, as per the question

Subtracting (ii) from (i), we get
$2y=6⇒y=3$
Now, putting y = 3 in (ii), we have
$x+2×3=21\phantom{\rule{0ex}{0ex}}⇒x=21-6=15$
Hence, the fixed charge be ₹15 and the charge for each extra day is ₹3.

#### Question 48:

A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each
should be used to make 10 litres of a 40% acid solution?

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
$x=6$
Now, putting x = 6 in (ii), we have
$6+y=10⇒y=4$
Hence, volume of 50% acid solution = 6 litres and volume of 25% acid solution = 4 litres.

#### Question 49:

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a
bar of 16-carat gold, weighing 120 g?

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
$x=320-240=80$
Now, putting x = 80 in (ii), we have
$80+y=120⇒y=40$
Hence, the required weight of 18-carat and 12-carat gold bars are 80 g and 40 g respectively.

#### Question 50:

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution.
Find the quantity of each type of acids to be mixed to form the mixture.

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 $y=21-x$ in (i) to get
$0.90x+0.97\left(21-x\right)=21×0.95\phantom{\rule{0ex}{0ex}}⇒0.90x+0.97×21-0.97x=21×0.95\phantom{\rule{0ex}{0ex}}⇒0.07x=0.97×21-21×0.95\phantom{\rule{0ex}{0ex}}⇒x=\frac{21×0.02}{0.07}=6$
Now, putting x = 6 in (ii), we have
$6+y=21⇒y=15$
Hence, the required quantities are 6 litres and 15 litres.

#### Question 51:

The larger of the two supplementary angles exceeds the smaller by ${18}^{\circ }$. Find them.

Let x and y be the supplementary angles, where x > y.
As per the given condition

And

Adding (i) and (ii), we get
$2x={198}^{\circ }⇒x={99}^{\circ }$
Now, substituting $x={99}^{\circ }$ in (ii), we have
${99}^{\circ }-y={18}^{\circ }⇒x={99}^{\circ }-{18}^{\circ }={81}^{\circ }$
Hence, the required angles are ${99}^{\circ }$ and ${81}^{\circ }$.

#### Question 52:

In Find the three angles.

The sum of all the angles of a triangle is ${180}^{\circ }$, therefore

Subtracting (i) from (ii), we have
$7{x}^{\circ }={182}^{\circ }-{7}^{\circ }={175}^{\circ }\phantom{\rule{0ex}{0ex}}⇒{x}^{\circ }={25}^{\circ }$
Now, substituting ${x}^{\circ }={25}^{\circ }$ in (i), we have
${y}^{\circ }=3{x}^{\circ }+{7}^{\circ }=3×{25}^{\circ }+{7}^{\circ }={82}^{\circ }$
Thus
$\angle A={x}^{\circ }={25}^{\circ }\phantom{\rule{0ex}{0ex}}\angle B={\left(3x-2\right)}^{\circ }={75}^{\circ }-{2}^{\circ }={73}^{\circ }\phantom{\rule{0ex}{0ex}}\angle C={y}^{\circ }={82}^{\circ }$
Hence, the angles are .

#### Question 53:

In a cyclic quadrilateral ABCD, it is given that  .
Find the four angles.

The opposite angles of cyclic quadrilateral are supplementary, so

And

Subtracting (i) from (ii), we have
$3x=99⇒x={33}^{\circ }$
Now, substituting $x={33}^{\circ }$ in (i), we have
${33}^{\circ }+y={83}^{\circ }⇒y={83}^{\circ }-{33}^{\circ }={50}^{\circ }$
Therefore
$\angle A={\left(2x+4\right)}^{\circ }={\left(2×33+4\right)}^{\circ }={70}^{\circ }\phantom{\rule{0ex}{0ex}}\angle B={\left(y+3\right)}^{\circ }={\left(50+3\right)}^{\circ }={53}^{\circ }\phantom{\rule{0ex}{0ex}}\angle C={\left(2y+10\right)}^{\circ }={\left(2×50+10\right)}^{\circ }={110}^{\circ }\phantom{\rule{0ex}{0ex}}\angle D={\left(4x-5\right)}^{\circ }={\left(4×33-5\right)}^{\circ }=132°-{5}^{\circ }={127}^{\circ }$
Hence, .

#### Question 1:

Write the number of solutions of the following pair of linear equations:

The given equations are

Which is of the form , where
Now
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{{b}_{1}}{{b}_{2}}=\frac{2}{4}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{{c}_{1}}{{c}_{2}}=\frac{-8}{-16}=\frac{1}{2}$
$⇒\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}=\frac{1}{2}$
Thus, the pair of linear equations are coincident and therefore has infinitely many solutions.

#### Question 2:

Find the value of k for which the following pair of linear equations have infinitely many solutions:

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.

#### Question 3:

For what value of k does the following pair of linear equations have infinitely many solutions?:

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{10}{20}=\frac{5}{10}=\frac{-\left(k-5\right)}{-k}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{k-5}{k}\phantom{\rule{0ex}{0ex}}⇒2k-10=k⇒k=10$
Hence, k = 10.

#### Question 4:

For what value of k will the following pair of linear equations have no solution?:

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.

#### Question 5:

Write the number of solutions of the following pair of linear equations:

The given pair of linear equations are

Which is of the form , where
Now
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{{b}_{1}}{{b}_{2}}=\frac{3}{6}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{{c}_{1}}{{c}_{2}}=\frac{-4}{-7}=\frac{4}{7}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Thus, the pair of the given linear equations has no solution.

#### Question 6:

Write the value of k for which the system of equations has a unique solution.

The given pair of linear equations is

Which is of the form , where
For the system to have a unique solution, we must have
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{2}\ne \frac{k}{-1}\phantom{\rule{0ex}{0ex}}⇒k\ne -\frac{3}{2}$
Hence, $k\ne -\frac{3}{2}$.

#### Question 7:

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.

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
$2x=18⇒x=9$
Substituting x = 9 in (iii), we have
$9+y=13⇒y=4$
Hence, the numbers are 9 and 4.

#### Question 8:

The cost of 5 pens and 8 pencils is ₹120, while the cost of 8 pens and 5 pencils is ₹153.
Find the cost of 1 pen and that of  1 pencil.

Let the cost of 1 pen and 1 pencil are ₹x and ₹y 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
$2x=32⇒x=16$
Substituting x = 16 in (iii), we have
$16+y=21⇒y=5$
Hence, the cost of 1 pen and 1 pencil are respectively ₹16 and ₹5.

#### Question 9:

The sum of two numbers is 80. The larger number exceeds four times the
smaller one by 5. Find the numbers.

Let the larger number be x and the smaller number be y.
Then as per the question

Subtracting (ii) from (i), we get
$5y=75⇒y=15$
Now, putting y = 15 in (i), we have
$x+15=80⇒x=65$
Hence, the numbers are 65 and 15.

#### Question 10:

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its
digits are reversed. Find the number.

Let the ones digit and tens digit be x and y respectively.
Then as per the question

Adding (i) and (ii), we get
$2x=8⇒x=4$
Now, putting x = 4 in (i), we have
$4+y=10⇒y=6$
Hence, the number is 64.

#### Question 11:

A man purchased 47 stamps of 20 p and 25 p for ₹10. Find the number of each type of stamps.

Let the number of stamps of 20 p and 25 p be x and y respectively.
As per the question

From (i), we get
$y=47-x$
Now, substituting $y=47-x$ in (ii), we have
$4x+5\left(47-x\right)=200\phantom{\rule{0ex}{0ex}}⇒4x-5x+235=200\phantom{\rule{0ex}{0ex}}⇒x=235-200=35$
Putting x = 35 in (i), we get
$35+y=47\phantom{\rule{0ex}{0ex}}⇒y=47-35=12$
Hence, the number of 20 p stamps and 25 p stamps are 35 and 12 respectively.

#### Question 12:

A man has some hens and cows. If the number of heads be 48 and the number of feet
be 140, how many cows are there?

Let the number of hens and cow be x and y respectively.
As per the question

Subtracting (i) from (ii), we have
$y=22$
Hence, the number of cows is 22.

#### Question 13:

If , find the values of x and y.

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
$9x-6x=21-18⇒x=\frac{3}{3}=1\phantom{\rule{0ex}{0ex}}$
Putting x = 1 in (iii), we have
$3×1+2y=9⇒y=\frac{9-3}{2}=3$
Hence, x = 1 and y = 3.

#### Question 14:

If , then find the value of (x + y).

The given pair of equations is

Multiplying (i) by 12 and (ii) by 4, we get

Now, subtracting (iv) from (iii), we get
$x=1$
Putting x = 1 in (iv), we have
$2+4y=4\phantom{\rule{0ex}{0ex}}⇒4y=2\phantom{\rule{0ex}{0ex}}⇒y=\frac{1}{2}$
$\therefore x+y=1+\frac{1}{2}=\frac{3}{2}$
Hence, the value of x + y is $\frac{3}{2}$.

#### Question 15:

If , then find the value of (x + y).

The given pair of equations is

Adding (i) and (ii), we get

Hence, the value of x + y is 4.

#### Question 16:

Find the value of k for which the system 3x + 5y = 0 and kx + 10y  = 0 has a nonzero solution.

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{k}=\frac{5}{10}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k=6$
Hence, k = 6.

#### Question 17:

Find k for which the system kx − y = 2 and 6x − 2y  = 3 has a unique solution.

The given system is

Here, .
For the system, to have a unique solution, we must have
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{k}{6}\ne \frac{-1}{-2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k\ne 3$
Hence, $k\ne 3$.

#### Question 18:

Find k for which the system 2x + 3y − 5 = 0 and 4x + ky − 10 = 0 has an infinite number of solutions.

The given system is

Here, .
For the system, to have an infinite number of solutions, we must have
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{4}=\frac{3}{k}=\frac{-5}{-10}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{3}{k}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k=6$
Hence, k = 6.

#### Question 19:

Show that the system 2x + 3y − 1 = 0, 4x + 6y − 4 = 0 has no solution.

The given system is

Here, .
Now,
$\frac{{a}_{1}}{{a}_{2}}=\frac{2}{4}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{{b}_{1}}{{b}_{2}}=\frac{3}{6}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{{c}_{1}}{{c}_{2}}=\frac{-1}{-4}=\frac{1}{4}$
Thus, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ and therefor the given system has no solution.

#### Question 20:

Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.

The given system is

Here, .
For the system to be inconsistent, we must have

Hence, k = 10.

#### Question 21:

Solve: .

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
$15u=4+1⇒u=\frac{1}{3}$
Multiplying (i) by 3 and subtracting (ii) from it, we get
$6v+4v=6-1⇒u=\frac{5}{10}=\frac{1}{2}$
Therefore

Now, adding (v) and (vi) we have
$2x=5⇒x=\frac{5}{2}$
Substituting $x=\frac{5}{2}$ in (v), we have
$\frac{5}{2}+y=3⇒y=3-\frac{5}{2}=\frac{1}{2}$
Hence, .

#### Question 1:

If 2x + 3y = 12 and 3x − 2y = 5 then
(a) x = 2, y = 3          (b) x = 2, y = − 3          (c) x = 3, y = 2          (d) x = 3, y = − 2

The given system of equations is

Multiplying (i) by 2 and (ii) by 3 and then adding, we get
$4x+9x=24+15\phantom{\rule{0ex}{0ex}}⇒x=\frac{39}{13}=3$
Now, putting x = 3 in (i), we have
$2×3+3y=12⇒y=\frac{12-6}{3}=2$
Thus, x = 3 and y = 2.
Hence, the correct answer is option (c).

#### Question 2:

If then
(a) x = 4, y = 2          (b) x = 5, y = 3          (c) x = 6, y = 4          (d) x = 7, y = 5

The given system of equations is

Adding (i) and (ii), we get
$2x=12⇒x=6$
Now, putting x = 6 in (ii), we have
$6+y=10⇒y=10-6=4$
Thus, x = 6 and y = 4.
Hence, the correct answer is option (c).

#### Question 3:

If then
(a) x = 2, y = 3          (b) x = − 2, y = 3          (c) x = 2, y = − 3          (d) x = − 2, y = − 3

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
$16x+9x=-4+54\phantom{\rule{0ex}{0ex}}⇒x=\frac{50}{25}=2$
Now, putting x = 2 in (iv), we have
$3×2+4y=18⇒y=\frac{18-6}{4}=3$
Thus, x = 2 and y = 3.
Hence, the correct answer is option (a).

#### Question 4:

If then
(a) x = 2, y = 3          (b) x = − 2, y = 3          (c) $x=-\frac{1}{2},y=3$          (d)  $x=-\frac{1}{2},y=\frac{1}{3}$

The given system of equations is

Adding (i) and (ii), we get
$\frac{2}{y}+\frac{3}{y}=15\phantom{\rule{0ex}{0ex}}⇒\frac{5}{y}=15⇒y=\frac{5}{15}=\frac{1}{3}$
Now, putting $y=\frac{1}{3}$ in (i), we have
$\frac{1}{x}+2×3=4⇒\frac{1}{x}=4-6⇒x=-\frac{1}{2}$
Thus, $x=-\frac{1}{2},y=\frac{1}{3}$.
Hence, the correct answer is option (d).

#### Question 5:

If $\frac{2x+y+2}{5}=\frac{3x-y+1}{3}=\frac{3x+2y+1}{6}$ then
(a) x = 1, y = 1          (b) x = − 1, y = − 1          (c) x = 1, y = 2          (d) x = 2, y = 1

Consider . Now, simplifying these equations, we get

And

Multiplying (ii) by 2 and subtracting it from (i)
$9x-6x=1+2⇒x=1$
Now, putting x = 1 in (ii), we have
$3×1-4y=-1⇒y=\frac{3+1}{4}=1$
Thus, x = 1, y = 1.
Hence, the correct answer is option (a).

#### Question 6:

If then
(a)          (b)           (c)           (d)

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
$6u+9u=4+1⇒u=\frac{5}{15}=\frac{1}{3}$
Again, multiplying (iii) by 3 and subtracting (iv) from it, we get
$6v+4v=6-1⇒v=\frac{5}{10}=\frac{1}{2}$
Therefore

Adding (v) and (vi), we get
$2x=3+2⇒x=\frac{5}{2}$
Substituting $x=\frac{5}{2}$, in (v), we have
$\frac{5}{2}+y=3⇒y=3-\frac{5}{2}=\frac{1}{2}$
Thus, .
Hence, the correct answer is option (b).

#### Question 7:

If then
(a) x = 2, y = 3          (b) x = 1, y = 2          (c) x = 3, y = 4          (d) x = 1, y = − 1

The given equations are

Dividing (i) and (ii) by xy, we get

Multiplying (iii) by 2 and subtracting (iv) from it, we get
$\frac{12}{x}-\frac{9}{x}=6-5⇒\frac{3}{x}=1⇒x=3$
Substituting x = 3 in (iii), we get
$\frac{6}{3}+\frac{4}{y}=3⇒\frac{4}{y}=1⇒y=4$
Thus, x = 3 and y = 4.
Hence, the correct answer is option (c).

#### Question 8:

If then
(a) x = 1, y = 2          (b) x = 2, y = 1          (c) x = 3, y = 2          (d) x = 2, y = 3

The given equations are

Adding (i) and (ii), we get

Subtracting (i) from (ii), we get

Adding (iii) and (iv), we get
$2x=2⇒x=1$
Substituting x = 1 in (iii), we have
$1+y=3⇒y=2$
Thus, x = 1 and y = 2.
Hence, the correct answer is option (a).

#### Question 9:

If ${2}^{x+y}={2}^{x-y}=\sqrt{8}$ then the value of y is
(a) $\frac{1}{2}$            (b) $\frac{3}{2}$            (c) 0            (d) none of these

$\because {2}^{x+y}={2}^{x-y}=\sqrt{8}\phantom{\rule{0ex}{0ex}}\therefore x+y=x-y\phantom{\rule{0ex}{0ex}}⇒y=0$
Hence, the correct answer is option (c).

#### Question 10:

If then
(a)             (b)             (c)             (d)

The given equations are

Multiplying (ii) by 2 and subtracting it from (ii), we get
$\frac{3}{y}-\frac{1}{y}=6-4\phantom{\rule{0ex}{0ex}}⇒\frac{2}{y}=2\phantom{\rule{0ex}{0ex}}⇒y=1$
Substituting y = 1 in (ii), we get
$\frac{1}{x}+\frac{1}{2}=2\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=2-\frac{1}{2}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{3}$
Hence, the correct answer is option (b).

#### Question 11:

The system has a unique solution only when
(a) k = 0            (b) $k\ne 0$            (c) k = 3            (d) $k\ne 3$

The given equations are

Here,
For the given system to have a unique solution, we must have
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{k}{6}\ne \frac{-1}{-2}\phantom{\rule{0ex}{0ex}}⇒k\ne 3$
Hence, the correct answer is option (d).

#### Question 12:

The system has a unique solution only when

(a) k = −6
(b) k ≠ −6
(c) k = 0
(d) k ≠ 0

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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$$\frac{1}{3}\ne \frac{-2}{k}⇒k\ne -6$
Hence, k has all real values other than −6.

#### Question 13:

The system has no solution, when

(a) k = 10
(b) $k\ne 10$
(c) $k=\frac{-7}{3}$
(d) k = −21

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{5},\frac{{b}_{1}}{{b}_{2}}=\frac{2}{k}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-3}{7}$
For the system of equations to have no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{1}{5}=\frac{2}{k}\ne \frac{-3}{7}⇒k=10$

#### Question 14:

If the lines given by  are parallel, then the value of k is

(a) $\frac{-5}{4}$
(b) $\frac{2}{5}$
(c) $\frac{3}{2}$
(d) $\frac{15}{4}$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{3}{2}=\frac{2k}{5}\ne \frac{-2}{1}$
$k=\frac{15}{4}$

#### Question 15:

For what value of k do the equations  represent two lines intersecting at a unique point?

(a) k = 3
(b) k = −3
(c) k = 6
(d) all real values except −6

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{k}{3},\frac{{b}_{1}}{{b}_{2}}=\frac{-2}{1}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-3}{-5}=\frac{3}{5}$
Thus, for these graph lines to intersect at a unique point, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
$\frac{k}{3}\ne \frac{-2}{1}⇒k\ne -6$
Hence, the graph lines will intersect at all real values of k except −6.

#### Question 16:

The pair of equations  has

(a) a unique solutions
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{-3},\frac{{b}_{1}}{{b}_{2}}=\frac{2}{-6}=\frac{1}{-3}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{5}{1}$
∴ $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Hence, the given system has no solution.

#### Question 17:

The pair of equations  has

(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{2}{4}=\frac{1}{2},\frac{{b}_{1}}{{b}_{2}}=\frac{3}{6}=\frac{1}{2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-5}{-15}=\frac{1}{3}$
∴ ​$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Hence, the given system has no solution.

#### Question 18:

If a pair of linear equations is consistent, then their graph lines will be

(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident

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.

#### Question 19:

If a pair of linear equations is inconsistent, then their graph lines will be

(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident

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.

#### Question 20:

In a
(a) 20°
(b) 40°
(c) 60°
(d) 80°

The correct option is (b).

Let
$\angle C=3\angle B=\left(3y\right)°$
Now, $\angle A+\angle B+\angle C=180°$
x + y + 3y = 180
x + 4y = 180              ...(i)
Also, $\angle C=2\left(\angle A+\angle B\right)$
⇒ 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
∴​ $\angle B=y°=40°$

#### Question 21:

In a cyclic quadrilateral ABCD, it is being given that
(a) 70°
(b) 80°
(c) 100°
(d) 110°

The correct option is (b).
In a cyclic quadrilateral ABCD:
$\angle A=\left(x+y+10\right)°$
$\angle B=\left(y+20°\right)$
$\angle C=\left(x+y-30\right)°$
$\angle D=\left(x+y\right)°$
We have:
$\angle A+\angle C=180°$ and $\angle B+\angle D=180°$      [Since ABCD is a cyclic quadrilateral]
Now, $\angle A+\angle C=\left(x+y+10\right)°+\left(x+y-30\right)°=180°$
⇒ 2x + 2y − 20 = 180
x + y − 10 = 90
x + y = 100                   ....(i)
Also, $\angle B+\angle D=\left(y+20\right)°+\left(x+y\right)°=180°$
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
∴​ $\angle B=\left(y+20°\right)=\left(60+20\right)°=80°$

#### Question 22:

The sum of the digits of a two digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is

(a) 96
(b) 69
(c) 87
(d) 78

The correct option is (d).

Let the tens and the units digits of the required number be x and y​, 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 − 10xy = 9
⇒ 9y − 9x = 9
yx = 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.

#### Question 23:

Assertion (A)
The system of equations has a unique solutions.
Reason (R)
The system of equations  has a unique solution when $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Option (c) is the correct answer.
Clearly, Reason (R) is false.
On solving x + y = 8 and xy = 2, we get:
x = 5 and y = 3
Thus, the given system has a unique solution. So, Assertion (A) is true.
∴ Assertion (A) is true and Reason (R) is false.

#### Question 24:

5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is
(a) 45 years
(b) 50 years
(c) 47 years
(d) 40 years

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.

#### Question 25:

The graphs of the equations  are two lines which are

(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

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

$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
The given system has no solution.
Hence, the lines are parallel.

#### Question 26:

The graphs of the equations  are two which are

(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{2}{1},\frac{{b}_{1}}{{b}_{2}}=\frac{3}{-2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-2}{-8}=\frac{1}{4}$
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
The given system has a unique solution.
Hence, the lines intersect exactly at one point.

#### Question 27:

The graphs of the equations  are two which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

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 = $-\frac{24}{5}$

$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
The given system of equations will have an infinite number of solutions.
Hence, the lines are coincident.

#### Question 1:

The graphic representation of the equations  gives a pair of
(a) parallel lines
(b) intersecting lines
(c) coincident lines
(d) none of these

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{2},\frac{{b}_{1}}{{b}_{2}}=\frac{2}{4}=\frac{1}{2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-3}{7}$
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
So, the given system has no solution.
Hence, the lines are parallel.

#### Question 2:

If  have an infinite number of solutions, then
(a) a = 5, b = 1
(b) a = −5, b = 1
(c) a = 5, b = −1
(d) a = −5, b = −1

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}$
Now, we have:
$\frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}⇒2a+2b-6=3a+3b$
a + b + 6 = 0                                  ...(i)
Again, we have:
$\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}⇒12a+3b=7a+7b-21\phantom{\rule{0ex}{0ex}}$
⇒ 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
∴ ​a = −5 and b = −1

#### Question 3:

The pair of equations  has
(a) a unique solution
(b) two solutions
(c) no solution
(d) infinitely many solutions

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

$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
The given system has a unique solution.
Hence, the lines intersect at one point.

#### Question 4:

If x = −y and y > 0, which of the following is wrong?
(a) ${x}^{2}y>0$
(b) $x+y=0$
(c) $xy<0$
(d) $\frac{1}{x}-\frac{1}{y}=0$

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 (∵ 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 (∵ y > 0)
This is again true.

(iv) $\frac{1}{x}-\frac{1}{y}=0\phantom{\rule{0ex}{0ex}}$
$⇒\frac{y-x}{xy}=0$
On substituting x = −y, we get:
$\frac{y-\left(-y\right)}{\left(-y\right)y}=0⇒\frac{2y}{-{y}^{2}}=0⇒2y=0⇒y=0$
Hence, from the above equation, we get y = 0, which is wrong.

#### Question 5:

Show that the system of equations has a unique solution.

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 = $\frac{1}{2}$, b2$-\frac{1}{4}$ and c2 = −1

$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
The given system has a unique solution.
Hence, the lines intersect at one point.

#### Question 6:

For what values of k is the system of equations inconsistent?

The given system of equations can be written as follows:
kx + 3y − (k − 2) = 0 and 12x + kyk = 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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{k}{12}=\frac{3}{k}\ne \frac{\left(k-2\right)}{k}⇒{k}^{2}=\left(3×12\right)=36$
$k=\sqrt{36}=±6$
Hence, the pair of equations is inconsistent if $k=±6$.

#### Question 7:

Show that the equations $9x-10y=21,\frac{3x}{2}-\frac{5y}{3}=\frac{7}{2}$ have infinitely many solutions.

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 = $\frac{3}{2}$, b2$\frac{-5}{3}$ and c2 = $\frac{-7}{2}$

$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
This shows that the given system equations has an infinite number of solutions.

#### Question 8:

Solve the system of equations:

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 = 2yy = 2
Hence, the required solution is x = 4 and y = 2.

#### Question 9:

Show that the paths represented by the equations  are parallel.

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
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{-2}=\frac{-1}{2},\frac{{b}_{1}}{{b}_{2}}=\frac{-3}{6}=\frac{-1}{2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-2}{-5}=\frac{2}{5}$
∴ ​$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Thus, the given system of equations has no solution.
Hence, the paths represented by the equations are parallel.

#### Question 10:

The difference between two numbers is 26 and one number is three times the other. Find the numbers.

Let the larger number be x and the smaller number be y.
Then, we have:
xy = 26                           ...(i)
x = 3y                                 ...(ii)
On substituting x = 3y in (i), we get:
3yy = 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.

#### Question 11:

Solve:

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
xy = 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.

#### Question 12:

Solve:

The given equations are as follows:
6x + 3y = 7xy                         ....(i)
3x + 9y = 11xy                       ....(ii)

For equation (i), we have:

$\frac{6x+3y}{xy}=7\phantom{\rule{0ex}{0ex}}$
$⇒\frac{6x}{xy}+\frac{3y}{xy}=7⇒\frac{6}{y}+\frac{3}{x}=7$        ....(iii)

For equation (ii), we have:

$\frac{3x+9y}{xy}=11\phantom{\rule{0ex}{0ex}}$
$⇒\frac{3x}{xy}+\frac{9y}{xy}=11⇒\frac{3}{y}+\frac{9}{x}=11$      ....(iv)
On substituting $\frac{1}{y}=v$ and $\frac{1}{x}=u$ 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 = $\frac{10}{15}=\frac{2}{3}$
$\frac{1}{y}=\frac{2}{3}⇒y=\frac{3}{2}$
On substituting $y=\frac{3}{2}$ in (iii), we get:
$\frac{6}{\left(3}{2}\right)}+\frac{3}{x}=7\phantom{\rule{0ex}{0ex}}$
$⇒4+\frac{3}{x}=7⇒\frac{3}{x}=3⇒3x=3\phantom{\rule{0ex}{0ex}}$
$⇒x=1$
Hence, the required solution is x = 1 and $y=\frac{3}{2}$.

#### Question 13:

Find the value of k for which the system of equations  has (i) a unique solution, (ii) no solution.

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:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$ i.e. $\frac{3}{k}\ne \frac{1}{2}⇒k\ne 6$
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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{3}{k}=\frac{1}{2}\ne \frac{-1}{-5}$

Thus, for k = 6, the given system of equations will have no solution.

#### Question 14:

In $∆ABC,\angle C=3\angle B=2\left(\angle A+\angle B\right),$ find the measure of each one of .

Let
Then, $\angle C=3\angle B=\left(3y\right)°$
Now, we have:
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$
x + y + 3y = 180
x + 4y = 180              ....(i)
Also, $\angle C=2\left(\angle A+\angle B\right)$
⇒ 3y = 2(x + y)
⇒ 2xy = 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
∴ ​

#### Question 15:

5 pencils and 7 pens together cost Rs 195 while 7 pencils and 5 pens together cost Rs 153. Find the cost of each one of the pencil and the pen.

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(xy) = −42
xy = −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.

#### Question 16:

Solve the following system of equations graphically:

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)
$y=\frac{2x-1}{3}$                           ...(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)
$y=\frac{4x+1}{3}$                           ...(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
Now, plots the points P(2, 3) and Q(5, 7). The point A(−1, −1) has already been plotted. Join PA and QP to get the graph line AQ. Extend it on both the sides.
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.

#### Question 17:

Find the angles of a cyclic quadrilateral ABCD in which

Given:
In a cyclic quadrilateral ABCD, we have:
$\angle A=\left(4x+20\right)°$
$\angle B=\left(3x-5\right)°$
$\angle C=\left(4y\right)°$
$\angle D=\left(7y+5\right)°$
$\angle A+\angle C=180°$ and $\angle \mathrm{B}+\angle \mathrm{D}=180°$      [Since ABCD is a cyclic quadrilateral]
Now, $\angle A+\angle C=\left(4x+20\right)°+\left(4y\right)°=180°$
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y  = 180 − 20 = 160
x + y = 40                       ....(i)
Also, $\angle B+\angle D=\left(3x-5\right)°+\left(7y+5\right)°=180°$
⇒ 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:
$\angle A=\left(4x+20\right)°=\left(4×25+20\right)°=120°$
$\angle B=\left(3x-5\right)°=\left(3×25-5\right)°=70°$
$\angle C=\left(4y\right)°=\left(4×15\right)°=60°$
$\angle D=\left(7y+5\right)°=\left(7×15+5\right)°=\left(105+5\right)°=110°$

#### Question 18:

Solve for x and y:

We have:

Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$:
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: $\frac{u}{-518+665}=\frac{v}{-266+1295}=\frac{1}{1225-196}$
$\frac{u}{147}=\frac{v}{1029}=\frac{1}{1029}$
$u=\frac{147}{1029}=\frac{1}{7},v=\frac{1029}{1029}=1$
$\frac{1}{x+y}=\frac{1}{7},\frac{1}{x-y}=1$
∴ (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: $\frac{x}{\left[1×\left(-1\right)-\left(-1\right)×\left(-7\right)\right]}=\frac{y}{\left[\left(-7\right)×1-\left(-1\right)×1\right]}=\frac{1}{\left[1×\left(-1\right)-1×1\right]}$

$\frac{x}{-1-7}=\frac{y}{-7+1}=\frac{1}{-1-1}$
$\frac{x}{-8}=\frac{y}{-6}=\frac{1}{-2}$
$x=\frac{-8}{-2}=4,y=\frac{-6}{-2}=3$
Hence, x = 4 and y = 3 is the required solution.

#### Question 19:

If 1 is added to both of the numerator and denominator of a fraction, it becomes $\frac{4}{5}$. If however, 5 is subtracted from both numerator and denominator, the fraction $\frac{1}{2}$. Find the fraction.

Let the required fraction be $\frac{x}{y}$.
Then, we have:
$\frac{x+1}{y+1}=\frac{4}{5}$
⇒ 5(x + 1) = 4(y + 1)
⇒ 5x + 5 = 4y + 4
⇒ 5x − 4y = −1                       ....(i)

Again, we have:
$\frac{x-5}{y-5}=\frac{1}{2}$
⇒ 2(x − 5) = 1(y − 5)
⇒ 2x − 10 = y  − 5
⇒ 2xy = 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
∴ ​x = 7 and  y = 9
Hence, the required fraction is $\frac{7}{9}$.

#### Question 20:

Solve:

The given equations may be written as follows:
$\frac{ax}{b}-\frac{by}{a}-\left(a+b\right)=0$                       ....(i)
$ax-by-2ab=0$                               ....(ii)
Here, a1 = $\frac{a}{b}$, b1 = $\frac{-b}{a}$, c1 = −(a + b), a2 = a, b2 = −b, c2 = −2ab
By cross multiplication, we have: $\frac{x}{\left(-\frac{b}{a}\right)×\left(-2ab\right)-\left(-b\right)×\left(-\left(a+b\right)\right)}=\frac{y}{-\left(a+b\right)×a-\left(-2ab\right)×\frac{a}{b}}=\frac{1}{\frac{a}{b}×\left(-b\right)-a×\left(-\frac{b}{a}\right)}$
$\frac{x}{2{b}^{2}-b\left(a+b\right)}=\frac{y}{-a\left(a+b\right)+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{2{b}^{2}-ab-{b}^{2}}=\frac{y}{-{a}^{2}-ab+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{{b}^{2}-ab}=\frac{y}{{a}^{2}-ab}=\frac{1}{-\left(a-b\right)}$
$\frac{x}{-b\left(a-b\right)}=\frac{y}{a\left(a-b\right)}=\frac{1}{-\left(a-b\right)}$

Hence, x = b and y = −a is the required solution.

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