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Page No 87:

Question 1:

2x+3y=2,x-2y=8.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-axis and y-axis, respectively.
                                               Graph of 2x + 3y = 2

2x + 3y = 2
⇒ 3y = (2 − 2x)
⇒ 3y = 2(1 − x)
y = 21-x3              ...(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)
y=x-82                     ...(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.

Page No 87:

Question 2:

3x+2y=4,2x-3y=7.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-axis and y-axis, respectively.
                                                 Graph of 3x + 2y = 4

3x + 2y = 4
⇒ 2y = (4 − 3x)
⇒ y = 4-3x2                    ...(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)
y=2x-73 
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.

Page No 87:

Question 3:

2x+3y=8,x-2y+3=0.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                                  Graph of 2x + 3y = 8

2x + 3y = 8
⇒ 3y = (8 − 2x)
y=8-2x3 ...........(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=x+32 ..........(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

Page No 87:

Question 4:

2x-5y+4=0,2x+y-8=0.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                             Graph of 2x − 5y + 4 = 0

2x − 5y + 4 = 0
⇒ 5y = (2x + 4)
y=2x+45 ...........(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

Page No 87:

Question 5:

Solve the following system of equations graphically:
 

                                  3x+2y=125x-2y=4

Answer:

The given equations are:
3x+2y=12                                .....i5x-2y=4                                  .....ii
From (i), write y in terms of x
y=12-3x2                               .....iii
Now, substitute different values of x in (iii) to get different values of y
For = 0, y=12-3x2=12-02=6
For = 2, y=12-3x2=12-62=3
For = 4, y=12-3x2=12-122=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
y=5x-42                                 .....iv
Now, substitute different values of x in (iv) to get different values of y
For = 0, y=5x-42=0-42=-2
For = 2, y=5x-42=10-42=3
For = 4, y=5x-42=20-42=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).

Page No 87:

Question 6:

3x+y+1=0,2x-3y+8=0.

Answer:

n a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x - axis and y - axis respectively.
                                             Graph of 3x + y + 1 = 0

3x + y + 1 = 0
y = (−3x 1) ...........(i)
Putting x = 0, we get y = −1.
Putting x = −1, we get y = 2.
Putting x = 1, we get y = −4.
Thus, we have the following table for the equation 3x + y + 1 = 0.

x 0 −1 1
y −1 2 −4

Now, plot the points A(0, −1), B(−1, 2) and C(1, −4) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + y + 1 = 0.
                    
                                         Graph of 2x − 3y + 8 = 0
2x − 3y + 8 = 0
⇒ 3y = (2x + 8)
y=2x+83
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

Page No 87:

Question 7:

Solve the following system of equation graphically:
 2x+3y+5=03x-2y-12=0

Answer:

From the first equation, write y in terms of x
y=-5+2x3                                          .....i
Substitute different values of x in (i) to get different values of y
For x=-1, y=-5-23=-1For x=2, y=-5+43=-3For x=5, y=-5+103=-5
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
y=3x-122                                           .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=0, y=0-122=-6For x=2, y=6-122=-3For x=4, y=12-122=0
So, the table for the second equation ( 3x  2y  12 = 0 ) is
 
x 0 2 4
y −6  −3 0

Now, plot the points D(0,−6), E(2,−3) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 3x − 2y − 12 = 0.




From the graph it is clear that, the given lines intersect at (2,−3).
Hence, the solution of the given system of equation is (2,−3).

Page No 87:

Question 8:

Solve the following system of equation graphically:
 2x-3y+13=03x-2y+12=0

Answer:

From the first equation, write y in terms of x
y=2x+133                                           .....i
Substitute different values of x in (i) to get different values of y
For x=-5, y=-10+133=1For x=1, y=2+133=5For x=4, y=8+133=7
Thus, the table for the first equation ( 2x  3y + 13 = 0 ) is
 

x −5 1 4
y 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
y=3x+122                                           .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=-4, y=-12+122=0For x=-2, y=-6+122=3For x=0, y=0+122=6
So, the table for the second equation ( 3x  2y + 12 = 0 ) is
 
x −4 −2 0
y 3 6

Now, plot the points D(−4,0), E(2,3) and F(0,6) on the same graph paper and join
D, E and F to get the graph of 3x − 2y + 12 = 0.




From the graph it is clear that, the given lines intersect at (2,3).
Hence, the solution of the given system of equation is (2,3).

Page No 87:

Question 9:

2x+3y=4,3x-y=-5.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                                 Graph of 2x + 3y = 4

2x + 3y = 4
⇒ 3y = (4 − 2x)
y=4-2x3              ...(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.

Page No 87:

Question 10:

x+2y+2=0,3x+2y-2=0.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                                 Graph of x + 2y + 2 = 0

x + 2y + 2 = 0
⇒ 2y = (−2 − x)
y=-2-x2...............(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=2-3x2...............(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

Page No 87:

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:
   x-y+3=02x+3y-4=0

Answer:

From the first equation, write y in terms of x
y=x+3                                                   .....i
Substitute different values of x in (i) to get different values of y
For x=-3, y=-3+3=0For x=-1, y=-1+3=2For x=1, y=1+3=4
Thus, the table for the first equation (x  y + 3 = 0) is
 

x −3 1 1
y 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
y=4-2x3                                             .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=-4, y=4+83=4For x=-1, y=4+23=2For x=2, y=4-43=0
So, the table for the second equation ( 2x + 3y  4 = 0 ) is
 
x −4 −1 2
y 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,
AreaEAF=12×AF×EM                     =12×5×2                     =5 sq. units
Hence, the vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2)
and 
(2,0) and its area is 5 sq. units.

Page No 87:

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:
2x-3y+4=0  x+2y-5=0

Answer:

From the first equation, write y in terms of x
y=2x+43                                              .....i
Substitute different values of x in (i) to get different values of y
For x=-2, y=-4+43=0For x=1, y=2+43=2For x=4, y=8+43=4
Thus, the table for the first equation (2x − 3y + 4 = 0) is
 

x −2 1 4
y 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
y=5-x2                                                .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=-3, y=5+32=4For x=1, y=5-12=2For x=5, y=5-52=0
So, the table for the second equation ( x + 2y  5 = 0 ) is
 
x −3 1 5
y 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,
AreaBAF=12×AF×BM                     =12×7×2                     =7 sq. units
Hence, the vertices of the triangle formed by the given lines and the x-axis are (−2,0), (1,2) and (5,0)
and the area of the triangle is
 7 sq. units.

Page No 87:

Question 13:

Solve the following system of linear equations graphically:
4x-3y+4=0, 4x+3y-20=0.
Find the area of the region bounded by these lines and the x-axis.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                                         Graph of 4x − 3y + 4 = 0
4x − 3y + 4 = 0
⇒ 3y = (4x + 4)
y=4x+43............(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=-4x+203 ............(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 = 12×base×height sq. units
                          = 12×6×4=12 sq. units.

Page No 87:

Question 14:

Solve the following system of linear equations graphically:
x-y+1=0, 3x+2y-12=0.
Calculate the area bounded by these lines and the x-axis.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                                   Graph of 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=-3x+122  ............(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 = 12×base×height sq. units
                          = 12×5×3=7.5 sq. units.

Page No 87:

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=02x+y-6=0

Answer:

From the first equation, write y in terms of x
y=x+22                                                .....i
Substitute different values of x in (i) to get different values of y
For x=-2, y=-2+22=0For x=2, y=2+22=2For x=4, y=4+22=3
Thus, the table for the first equation (x − 2y + 2 = 0) is
 

x −2 2 4
y 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
y=6-2x                                                .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=1, y=6-2=4For x=3, y=0For x=4, y=6-8=-2
So, the table for the second equation (2x + y  6 = 0 ) is
 
x 1 3 4
y 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,
AreaBAE=12×AE×BM                     =12×5×2                     =5 sq. units
Hence, the vertices of the triangle formed by the given lines and the x-axis are 
(−2,0), (2,2) and (3,0) and the area of the triangle is 5 sq. units.

Page No 87:

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:
 2x-3y+6=02x+3y-18=0

Answer:

From the first equation, write y in terms of x
y=2x+63                                                .....i
Substitute different values of x in (i) to get different values of y
For x=-3, y=-6+63=0For x=0, y=0+63=2For x=3, y=6+63=4
Thus, the table for the first equation (2x − 3y + 6 = 0) is
 

x −3 0 3
y 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
y=18-2x3                                                 .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=0, y=18-03=6For x=3, y=18-63=4For x=9, y=18-183=0
So, the table for the second equation (2x + 3y  18 = 0 ) is
 
x 0 3 9
y 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,
AreaEDB=12×BD×EM                     =12×4×3                     =6 sq. units
Hence, the vertices of the triangle formed by the given lines and the y-axis are 
(0,2), (0,6) and (3,4) and its area is 6 sq. units.

Page No 87:

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.
 4x-y-4=03x+2y-14=0

Answer:

From the first equation, write y in terms of x
y=4x-4                                                 .....i
Substitute different values of x in (i) to get different values of y
For x=0, y=0-4=-4For x=1, y=4-4=0For x=2, y=8-4=4
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
y=14-3x2                                                 .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=0, y=14-02=7For x=4, y=14-122=1For x=143, y=14-142=0
So, the table for the second equation (3x + 2y  14 = 0 ) is
 
x 0 4 143
y 1 0

Now, plot the points D(0,7), E(4,1) and F143,0 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,
AreaDAC=12×DA×CM                     =12×11×2                     =11 sq. units
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4) and its area is 11 sq. units.

Page No 87:

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:
 x-y-5=03x+5y-15=0

Answer:

From the first equation, write y in terms of x
y=x-5                                                   .....i
Substitute different values of x in (i) to get different values of y
For x=0, y=0-5=-5For x=2, y=2-5=-3For x=5, y=5-5=0
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
y=15-3x5                                                 .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=-5, y=15+155=6For x=0, y=15-05=3For x=5, y=15-155=0
So, the table for the second equation (3x + 5y  15 = 0 ) is
 
x −5 0 5
y 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,
AreaCEA=12×EA×OC                     =12×8×5                     =20 sq. units
Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0) and its area is 20 sq. units.

Page No 87:

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.
2x-5y+4=0 2x+y-8=0

Answer:

From the first equation, write y in terms of x
y=2x+45                                                   .....i
Substitute different values of x in (i) to get different values of y
For x=-2, y=-4+45=0For x=0, y=0+45=45For x=3, y=6+45=2
Thus, the table for the first equation (2x − 5y + 4 = 0) is
 

x −2 0 3
y 45 2

Now, plot the points A(−2,0), B0,45 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
y=8-2x                                                 .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=0, y=8-0=8For x=2, y=8-4=4For x=4, y=8-8=0
So, the table for the second equation (2x + y  8 = 0 ) is
 
x 0 2 4
y 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), 0,45 and (3,2).
Draw a perpendicular from point C to the y-axis. So,
AreaDBC=12×DB×CM                     =12×8-45×3                     =545 sq. units
Hence, the veritices of the triangle are (0,8), 0,45 and (3,2) and its area is 545 sq. units.

Page No 87:

Question 20:

Solve graphically the system of equations:
5x-y=7, x-y+1=0.
Calculate the area bounded by these lines and the y-axis.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                              Graph of 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 = 12×base×height sq. units
                          = 12×8×2=8 sq. units.

Page No 87:

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=12x+3y=6

Answer:

From the first equation, write y in terms of x
y=2x-123                                            .....i
Substitute different values of x in (i) to get different values of y
For x=0, y=0-123=-4For x=3, y=6-123=-2For x=6, y=12-123=0
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
y=6-x3                                                 .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=0, y=6-03=2For x=3, y=6-33=1For x=6, y=6-63=0
So, the table for the second equation (x + 3y = 6 ) is
 
x 0 3 6
y 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,
AreaDAC=12×DA×OC                     =12×6×6                     =18 sq. units
Hence, the veritices of the triangle are (0,2), (6,0) and (0,−4) and its area is 18 sq. units.



Page No 88:

Question 22:

Show graphically that the following given systems of equations has infinitely many solutions:
2x+3y=64x+6y=12

Answer:

From the first equation, write y in terms of x
y=6-2x3                                             .....i
Substitute different values of x in (i) to get different values of y
For x=-3, y=6+63=4For x=3, y=6-63=0For x=6, y=6-123=-2
Thus, the table for the first equation (2x + 3y = 6) is
 

x −3 3 6
y 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
y=12-4x6                                                 .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=-6, y=12+246=6For x=0, y=12-06=2For x=9, y=12-366=-4
So, the table for the second equation (4x + 6y = 12 ) is
 
x −6 0 9
y 2 −4

Now, plot the points D(−6,6), E(0,2) and F(9,−4) on the same graph paper and join
D, E and F to get the graph of 4x + 6y = 12.




From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.

Page No 88:

Question 23:

Show graphically that the system of equations 3x-y=5, 6x-2y=10 has infinitely many solutions.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                               Graph of 3x y = 5
3x y = 5
y = (3x − 5)  .........(i)
Putting x = 1, we get y = −2.
Putting x = 0, we get y = −5.
Putting x = 2, we get y = 1.
Thus, we have the following table for the equation 3x y = 5.

x 1 0 2
y −2 −5 1

Now, plot the points A(1, −2), B(0, −5) and C(2, 1) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x y = 5.
                   
                                                 Graph of 6x − 2y = 10
6x − 2y = 10
⇒ 2y = (6x − 10)
y=6x-102      ...........(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.

Page No 88:

Question 24:

Show graphically that the system of equations 2x+y=6, 6x+3y=18 has infinitely many solutions.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                              Graph of 2x + y = 6
2x + y = 6 ⇒ y = (6 − 2x)  ...(i)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 2x + y = 6.

x 3 1 2
y 0 4 2

Now, plot the points A(3, 0), B(1, 4) and C(2, 2) on the graph paper.
Join AC and CB to get the graph line AB. Extend it on both ways.
Thus, AB is the graph of 2x + y = 6.
                   
                                                 Graph of 6x + 3y = 18
6x + 3y = 18 ⇒ 3y = (18 − 6x)
y=18-6x3...........(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.

Page No 88:

Question 25:

Show graphically that the following given systems of equations has infinitely many solutions:
x-2y=53x-6y=15

Answer:

From the first equation, write y in terms of x
y=x-52                                                .....i
Substitute different values of x in (i) to get different values of y
For x=-5, y=-5-52=-5For x=1, y=1-52=-2For x=3, y=3-52=-1
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
y=3x-156                                                 .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=-3, y=-9-156=-4For x=-1, y=-3-156=-3For x=5, y=15-156=0
So, the table for the second equation (3x  6y = 15 ) is
 
x −3 −1 5
y −4  −3 0

Now, plot the points D(−3,−4), E(−1,−3) and F(5,0) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 15.




From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.

Page No 88:

Question 26:

Show graphically that the following given systems of equations is inconsistent i.e. has no solution:
x-2y=63x-6y=0

Answer:

From the first equation, write y in terms of x
y=x-62                                                .....i
Substitute different values of x in (i) to get different values of y
For x=-2, y=-2-62=-4For x=0, y=0-62=-3For x=2, y=2-62=-2
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
y=12x                                                   .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=-4, y=-42=-2For x=0, y=02=0For x=4, y=42=2
So, the table for the second equation (3x  6y = 0 ) is
 
x −4 0 4
y −2  0 2

Now, plot the points D(−4,−2), O(0,0) and E(4,2) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 0.




From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.

Page No 88:

Question 27:

Show graphically that the system of equations 2x+3y=4, 4x+6y=12 is inconsistent.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
                                             Graph of 2x + 3y = 4
2x + 3y = 4
⇒ 3y = (−2x + 4)
y=-2x+43  .........(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=-4x+126      ...........(ii)
Putting x = 3, we get y = 0.
Putting x = 0, we get y = 2.
Putting x = 6, we get y = −2.
Thus, we have the following table for the equation 4x + 6y = 12.
 x 3 0 6
y 0 2 −2

Now, on the same graph, plot the points P(3, 0), Q(0, 2) and R (6, −2).
Join PQ and PR to get the graph line QR. Extend it on both ways.
Then, QR is the graph of the equation 4x + 6y = 12.

It is clear from the graph that these two lines are parallel and do not intersect when produced.
Hence, the given system of equations is inconsistent.

Page No 88:

Question 28:

Show graphically that the following given systems of equations is inconsistent i.e. has no solution:
 2x+y=66x+3y=20

Answer:

From the first equation, write y in terms of x
y=6-2x                                                .....i
Substitute different values of x in (i) to get different values of y
For x=0, y=6-0=6For x=2, y=6-4=2For x=4, y=6-8=-2
Thus, the table for the first equation (2x y = 6) is
 

x 0 2 4
y 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
y=20-6x3                                                  .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=0, y=20-03=203For x=103, y=20-203=0For x=5, y=20-303=-103
So, the table for the second equation (6x + 3y = 20 ) is
 
x 0 103 5
y 203  0 -103

Now, plot the points D0,203,E103,0 and F5,-103 on the same graph paper and join
D, E and F to get the graph of 6x + 3y = 20.




From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.

Page No 88:

Question 29:

Draw the graphs of the following equations on the same graph paper:
2x+y=22x+y=6
Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.

Answer:

From the first equation, write y in terms of x
y=2-2x                                                .....i
Substitute different values of x in (i) to get different values of y
For x=0, y=2-0=2For x=1, y=2-2=0For x=2, y=2-4=-2
Thus, the table for the first equation (2x y = 2) is
 

x 0 1 2
y 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
y=6-2x                                                  .....ii
Now, substitute different values of x in (ii) to get different values of y
For x=0, y=6-0=6For x=1, y=6-2=4For x=3, y=6-6=0
So, the table for the second equation (2x + y = 6 ) is
 
x 0 1 3
y 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,
AreaTrapezium DABF=AreaDOF-AreaAOB                               =12×3×6-12×1×2                               =9-1                                =8 sq. units
Hence, the area of the rquired trapezium is 8 sq. units.



Page No 103:

Question 1:

x+y=3,4x-3y=26.

Answer:

The given system of equation is:
x + y = 3 .........(i)
4x − 3y = 26 ........(ii)

On multiplying (i) by 3, we get:
3x + 3y = 9 .......(iii)

On adding (ii) and (iii), we get:
7x = 35
x = 5

On substituting the value of x = 5 in (i), we get:
5 + y = 3
y = (3 − 5) = −2

Hence, the solution is x = 5 and y = −2

Page No 103:

Question 2:

Solve for x and y:
   x-y=3x3+y2=6

Answer:

The given system of equations is
   x-y=3                                                .....ix3+y2=6                                             .....ii
From (i), write y in terms of x to get
y=x-3
Substituting y =  − 3 in (ii), we get
x3+x-32=62x+3x-3=362x+3x-9=36x=455=9
Now, substituting= 9 in (i), we have
9-y=3y=9-3=6
Hence, x = 9 and y = 6. 



Page No 104:

Question 3:

2x+3y=0,3x+4y=5.

Answer:

The given system of equation is:
2x + 3y = 0 .........(i)
3x + 4y = 5 ........(ii)

On multiplying (i) by 4 and (ii) by 3, we get:
8x + 12y = 0 .......(iii)
9x + 12y = 15 ......(iv)

On subtracting (iii) from (iv) we get:
x = 15

On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
⇒ 3y = −30 
y = −10

Hence, the solution is x = 15 and y = −10.

Page No 104:

Question 4:

2x-3y=13,7x-2y=20.

Answer:

The given system of equation is:
2x − 3y = 13 .........(i)
7x − 2y = 20 ........(ii)

On multiplying (i) by 2 and (ii) by 3, we get:
4x − 6y = 26 .......(iii)
21x − 6y = 60 ......(iv)

On subtracting (iii) from (iv) we get:
17x = (60 − 26) = 34
x = 2

On substituting the value of x = 2 in (i), we get:
4 − 3y = 13
⇒ 3y = (4 − 13) = −9
y = −3

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

Page No 104:

Question 5:

3x-5y-19=0,-7x+3y+1=0.

Answer:

The given system of equation is:
3x − 5y − 19 = 0 .........(i)
−7x + 3y + 1 = 0 ........(ii)

On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 57  or .........(iii)
−35x + 15y = −5 ........(iv)

On adding (iii) from (iv) we get:
−26x = (57 − 5)) = 52
x =  −2

On substituting the value of x =  −2 in (i), we get:
−6 − 5y − 19 = 0
⇒ 5y = (− 6 − 19) = −25
y = −5

Hence, the solution is x = −2 and y = −5.

Page No 104:

Question 6:

Solve for x and y:
2x-y+3=03x-7y+10=0

Answer:

The given system of equations is
   2x-y+3=0                                              .....i3x-7y+10=0                                             .....ii
From (i), write y in terms of to get
y = 2x + 3 
Substituting y = 2x + 3 in (ii), we get
3x-72x+3+10=03x-14x-21+10=0-7x=21-10=11x=-117
Now, substituting x=-117 in (i), we have
-227-y+3=0y=3-227=-17
Hence, x=-117 and y=-17.

Page No 104:

Question 7:

Solve for x and y:
x2-y9=6          x7+y3=5       

117 y=1

Answer:

The given system of equations can be written as
9x-2y=108                                         .....i3x+7y=105                                         .....ii
Multiplying  (i) by 7 and (ii) by 2, we get
63x+6x=108×7+105×269x=966x=96669=14
Now, substituting x = 14 in (1), we have
9×14-2y=1082y=126-108y=182=9
Hence, x = 14 and y = 9.

Page No 104:

Question 8:

x3+y4=11, 5x6-y3=-7.

Answer:

The given equations are:
x3+y4=11
⇒ 4x + 3y = 132 ........(i)

and 5x6-y3=-7
⇒ 5x − 2y = −42..........(ii)

On multiplying (i) by 2 and (ii) by 3, we get:
8x + 6y = 264 ........(iii)
15x − 6y = −126 ...........(iv)

On adding (iii) and (iv), we get:
23x = 138
x = 6

On substituting x = 6 in (i), we get:
24 + 3y = 132
⇒ 3y = (132 − 24) = 108
y = 36

Hence, the solution is x = 6 and y = 36.

Page No 104:

Question 9:

4x-3y=8,6x-y=293.

Answer:

The given system of equation is:
4x − 3y = 8 .........(i)
6x-y=293 ........(ii)
On multiplying (ii) by 3, we get:
18x − 3y = 29.........(iii)

On subtracting (iii) from (i) we get:
−14x = −21
x2114=32
On substituting the value of x32  in (i), we get:
4×32-3y=86-3y=83y=6-8=-2y=-23
Hence, the solution is x32 and y = -23.

Page No 104:

Question 10:

2x-3y4=3,5x=2y+7.

Answer:

The given equations are:
2x-3y4=3 ........(i)
5x = 2y + 7 ............(ii)

On multiplying (i) by 2 and (ii) by 34, we get:
4x-32y=6 .......(iii)
154x=32y+214 .......(iv)

On subtracting (iii) from (iv), we get:
-14x=-34x=3

On substituting x = 3 in (i), we get:
2×3-3y4=33y4=(6 -3)=3y=3×43=4

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

Page No 104:

Question 11:

2x+5y=83,3x-2y=56.

Answer:

The given equations are:
2x+5y=83 ........(i)
3x-2y=56..........(ii)

On multiplying (i) by 2 and (ii) by 5, we get:
4x+10y=163 ........(iii)
15x-10y=256 ...........(iv)

On adding (iii) and (iv), we get:
19x=576x=576×19=36=12

On substituting x =12 in (i), we get:
2×12+5y=83
5y=83-1=53y=53×5=13

Hence, the solution is x = 12 and  y = 13.

Page No 104:

Question 12:

7-4x3=y,2x+3y+1=0.

Answer:

The given equations are:
7-4x3=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.

Page No 104:

Question 13:

Solve for x and y:
0.4x+0.3y=1.70.7x-0.2y=0.8

Answer:

The given system of equations is
0.4x+0.3y=1.7                                    .....i0.7x-0.2y=0.8                                    .....ii
Multiplying  (i) by 0.2 and (ii) by 0.3 and adding them, we get
0.8x+2.1x=3.4+2.42.9x=5.8x=5.82.9=2
Now, substituting x = 2 in (i), we have
0.4×2+0.3y=1.70.3y=1.7-0.8y=0.90.3=3
Hence, x = 2 and y = 3.

Page No 104:

Question 14:

Solve for x and y:
0.3x+0.5y=0.50.5x+0.7y=0.74

Answer:

The given system of equations is
0.3x+0.5y=0.5                        .....i0.5x+0.7y=0.74                      .....ii
Multiplying  (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get
2.5y-2.1y=2.5-2.220.4y=0.28y=0.280.4=0.7
Now, substituting y = 0.7 in (i), we have
0.3x+0.5×0.7=0.50.3x=0.50-0.35=0.15x=0.150.3=0.5
Hence, x = 0.5 and y = 0.7.

Page No 104:

Question 15:

7y+3-2x+2=14,4y-2+3x-3=2.

Answer:

The given equations are:
7(y + 3) − 2(x + 2) = 14
⇒ 7y + 21 − 2x − 4 = 14
⇒ −2x + 7y = −3 ..........(i)

and 4(y − 2) + 3(x − 3) = 2
⇒ 4y − 8 + 3x − 9 = 2
⇒ 3x + 4y = 19 .........(ii)

On multiplying (i) by 4 and (ii) by 7, we get:
−8x + 28y = −12 ........(iii)
21x + 28y = 133 ...........(iv)

On subtracting (iii) from (iv), we get:
29x = 145
x = 5

On substituting x = 5 in (i), we get:
−10 + 7y = −3
⇒ 7y = (−3 + 10) = 7
y = 1

Hence, the solution is x = 5 and  y = 1.

Page No 104:

Question 16:

6x+5y=7x+3y+1=2x+6y-1

Answer:

The given equations are:
6x + 5y = 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 6x + 5y = 2(x + 6y − 1)
⇒ 6x + 5y = 2x + 12y − 2
⇒ 6x − 2x + 5y − 12y = − 2
⇒ 4x − 7y = −2 ..........(i)

and 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 7x + 3y + 1 = 2x + 12y − 2
⇒ 7x − 2x + 3y − 12y = −2 − 1
⇒ 5x − 9y = −3  .........(ii)

On multiplying (i) by 9 and (ii) by 7, we get:
36x − 63y = −18 ..........(iii)
35x − 63y = −21.........(iv)

On subtracting (iv) from (iii), we get:
x = (−18 + 21) = 3

On substituting x = 3 in (i), we get:
12 − 7y = −2
⇒ 7y = (2 + 12) = 14
y = 2

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

Page No 104:

Question 17:

x+y-82=x+2y-143=3x+y-1211

Answer:

The given equations are:
x+y-82=x+2y-143=3x+y-1211
i.e., x+y-82=3x+y-1211
By cross multiplication, we get:
11x + 11y − 88 = 6x + 2y − 24
⇒ 11x − 6x + 11y − 2y = −24 + 88
⇒ 5x + 9y = 64 .........(i)

and x+2y-143=3x+y-1211
⇒ 11x + 22y − 154 = 9x + 3y − 36
⇒ 11x − 9x + 22y − 3y = −36 + 154
⇒ 2x + 19y = 118 .........(ii)

On multiplying (i) by 19 and (ii) by 9, we get:
95x + 171y = 1216..........(iii)
18x + 171y = 1062.........(iv)

On subtracting (iv) from (iii), we get:
77x = 154
x = 2

On substituting x = 2 in (i), we get:
10 + 9y = 64
⇒ 9y = (64 − 10) = 54
y = 6

Hence, the solution is x = 2 and y = 6.

Page No 104:

Question 18:

5x+6y=13,3x+4y=7,x0.

Answer:

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

Putting 1x=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
1x=5x=15
On substituting x=15 in (i), we get:
515+6y=13
⇒ 25 + 6y = 13
⇒ 6y = (13 − 25) = −12
y =  −2

Hence, the required solution is x=15 and y = −2.

Page No 104:

Question 19:

x+6y=6,3x-8y=5, y0.

Answer:

The given equations are:
x+6y=6 ............(i)
3x-8y=5 .............(ii)
Putting 1y=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+6y=6
6y=6-3=33y=6y=2
Hence, the required solution is x = 3 and y = 2.

Page No 104:

Question 20:

2x-3y=9,3x+7y=2, y0.

Answer:

The given equations are:
2x-3y=9 ............(i)
3x+7y=2 .............(ii)
Putting 1y=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:
2×3-3y=9
6-3y=93y=-3y=-1
Hence, the required solution is x = 3 and y = −1.

Page No 104:

Question 21:

3x-1y+9=0,2x+3y=5x0, y0.

Answer:

The given equations are:
3x-1y+9=0
3x-1y=-9 ............(i)
2x+3y=5 .............(ii)
Putting 1x=u and 1y=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
1x=-2x=-12

On substituting x=-12 in (i), we get:
3-12-1y=-9
-6-1y=-91y=-6+9=3
y=13
Hence, the required solution is x=-12 and y=13.

Page No 104:

Question 22:

9x-4y=8,13x+7y=101x0,y0.

Answer:

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

Putting 1x=u and 1y=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
1x=4x=14
On substituting x=14 in (i), we get:
914-4y=8
36-4y=84y=36-8=28
y=428=17
Hence, the required solution is x=14 and y=17.

Page No 104:

Question 23:

5x-3y=1,32x+23y=5 x0, y0.

Answer:

The given equations are:
5x-3y=1  ............(i)
32x+23y=5 .............(ii)
 
Putting 1x=u and 1y=v, we get:
5u − 3v = 1 .............(iii)
32u+23v=5
9u+4v6=5
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
1x=2x=12

On substituting x=12 in (i), we get:
512-3y=1
10-3y=13y=10-1=9
y=39=13
Hence, the required solution is x=12 and y=13.

Page No 104:

Question 24:

Solve for x and y:
12x+13y=213x+12y=136

Answer:

Multiplying equation (i) and (ii) by 6, we get
3x+2y=12                                           .....i2x+3y=13                                            .....ii
Multiplying  (i) by 3 and (ii) by 2 and subtracting (ii) from (i), we get
9x-4x=36-265x=10x=510=12
Now, substituting x=12 in (i), we have
6+2y=122y=6y=13
Hence, x=12 and y=13.

Page No 104:

Question 25:

4x+6y=3xy,8x+9y=5xy x0, y0.

Answer:

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

From equation (i), we have:

4x+6yxy=3
4y+6x=3
.............(iii)

For equation (ii), we have:

8x+9yxy=5
8y+9x=5
.............(iv)
On substituting 1y=v and 1x=u, 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=312=14
1y=14y=4

On substituting y = 4 in (iii), we get:
44+6x=3
1+6x=3 6x=3-1=2
2x=6x=62=3
Hence, the required solution is x = 3 and  y = 4.

Page No 104:

Question 26:

x+y=5xy,3x+2y=13xy.

Answer:

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

From equation (i), we have:

x+yxy=5
1y+1x=5
.............(iii)

From equation (ii), we have:

3x+2yxy=13
3y+2x=13
.............(iv)
On substituting 1y=v and 1x=u , 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
1y=3y=13
On substituting y=13 in (iii), we get:
113+1x=5
3+1x=51x=2x=12
Hence, the required solution is x=12 and y=13 or x = 0 and y = 0.

Page No 104:

Question 27:

Solve for x and y:
5x+y-2x-y=-115x+y+7x-y=10

Answer:

The given equations are
5x+y-2x-y=-1                             .....i15x+y+7x-y=10                               .....ii
Substituting 1x+y=u and 1x-y=v in (i) and (ii), we get
5u-2v=-1                              .....iii15u+7v=10                             .....iv
Multiplying  (iii) by 3 and subtracting it from (iv), we get
7v+6v=10+313v=13v=1x-y=1           1x-y=v          .....v
Now, substituting v = 1 in (iii), we get
5u-2=-15u=1u=15x+y=5                                                  .....vi
Adding (v) and (vi), we get
2x=6x=3
Substituting x = 3 in (vi), we have
3+y=5y=5-3=2
Hence, x = 3 and y = 2.



Page No 105:

Question 28:

3x+y+2x-y=2,9x+y-4x-y=1, where x+y0 and x-y0.

Answer:

The given equations are:
3x+y+2x-y=2          ...(i)
9x+y-4x-y=1           ...(ii)
Putting 1x+y=u and 1x-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=515=13 
1x+y=13x+y=3             ...(vi)
On substituting u=13 in (iii), we get:
1 + 2v = 2
⇒ 2v = 1
v=12
1x-y=12x-y=2                ...(vii)
On adding (vi) and (vii), we get:
2x = 5
x=52
On substituting x=52 in (vi), we get:
52+y=3 y=3-52=12
Hence, the required solution is x=52 and y=12.

Page No 105:

Question 29:

5x+1-2y-1=12,10x+1+2y-1=52, where x-1, y1.

Answer:

The given equations are:
5x+1-2y-1=12 .............(i)
10x+1+2y-1=52 ..............(ii)
Putting 1x+1=u and 1y-1=v , we get:
5u-2v=12 .................(iii)
10u+2v=52 ................(iv)
On adding (iii) and (iv), we get:
15u = 3
u=315=15
1x+1=15x+1=5x=4
On substituting u=15 in (iii), we get:
5×15-2v=121-2v=12
2v=12v=14
1y-1=14y-1=4y=5
Hence, the required solution is x = 4 and y = 5.

Page No 105:

Question 30:

44x+y+30x-y=10,55x+y+40x-y=13.

Answer:

The given equations are:
44x+y+30x-y=10        ...(i)
55x+y+40x-y=13         ...(ii)
Putting 1x+y=u and 1x-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=111 
1x+y=111x+y=11         ...(vii)
On substituting u=111 in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
v=630=15
1x-y=15x-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.

Page No 105:

Question 31:

Solve for x and y:
10x+y+2x-y=415x+y-9x-y=-2

Answer:

The given equations are
10x+y+2x-y=4                                 .....i15x+y-9x-y=-2                              .....ii
Substituting 1x+y=u and 1x-y=v in (i) and (ii), we get
10u+2v=4                                           .....iii15u-9v=-2                                        .....iv
Multiplying  (iii) by 9 and (iv) by 2 and adding, we get
90u+30u=36-4120u=32u=32120=415x+y=154           1x+y=u          .....v
Now, substituting u=415 in (iii), we get
10×415+2v=483+2v=42v=4-83=43v=23x-y=32         1x-y=v        .....vi
Adding (v) and (vi), we get
2x=154+322x=214x=218
Substituting x=218 in (v), we have
218+y=154y=154-218=98
Hence, x=218 and y=98.

Page No 105:

Question 32:

71x+37y=253,37x+71y=287.

Answer:

The given equations are:
71x + 37y = 253 ..............(i)
37x + 71y = 287 ............(ii)

On adding (i) and (ii), we get:
108x + 108y = 540
⇒ 108(x + y) = 540
⇒ (x + y) = 5................(iii)

On subtracting (ii) from (i), we get:
34x − 34y = −34
⇒ 34(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.

Page No 105:

Question 33:

217x+131y=913,131x+217y=827.

Answer:

The given equations are:
217x + 131y = 913 .............(i)
131x + 217y = 827 .............(ii)

On adding (i) and (ii), we get:
348x + 348y = 1740
⇒ 348(x + y) = 1740
 x + y = 5 ............(iii)

On subtracting (ii) from (i), we get:
86x − 86y = 86
⇒ 86(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.

Page No 105:

Question 34:

Solve for x and y:
23x-29y=98     29x-23y=110 

Answer:

The given equations are
23x-29y=98                                       .....i29x-23y=110                                     .....ii
Adding (i) and (ii), we get
52x-52y=208x-y=4                                              .....iii
Subtracting (i) from (ii), we get
6x+6y=12x+y=2                                              .....iv
Now, adding equation (iii) and (iv), we get
2x=6x=3
Substituting x = 3 in (iv), we have
3+y=2y=2-3=-1
Hence, x=3 and y=-1.

Page No 105:

Question 35:

Solve for x and y:
2x+5yxy=64x-5yxy=-3

Answer:

The given equations can be written as
5x+2y=6                                             .....i-5x+4y=-3                                      .....ii
Adding (i) and (ii), we get
6y=3y=2
Substituting y = 2 in (i), we have
5x+22=6x=1
Hence, x = 1 and y = 2..

Page No 105:

Question 36:

Solve for x and y:
13x+y+13x-y=34123x+y-123x-y=-18

Answer:

The given equations are
13x+y+13x-y=34                                             .....i123x+y-123x-y=-18 13x+y-13x-y=-14  Multiplying by 2         .....ii
Substituting 13x+y=u and 13x-y=v in (i) and (ii), we get
u+v=34                                                                 .....iiiu-v=-14                                                             .....iv
Adding (iii) and (iv), we get
2u=12u=143x+y=4           13x+y=u                        .....v
Now, substituting u=14 in (iii), we get
14+v=34v=34-14v=123x-y=2         13x-y=v                         .....vi
Adding (v) and (vi), we get
6x=6x=1
Substituting x = 1 in (v), we have
3+y=4y=1
Hence, x = 1 and y = 1.

Page No 105:

Question 37:

12(x+2y)+53(3x-2y)=-32,54(x+2y)-35(3x-2y)=6160, where x+2y0 and 3x-2y0.

Answer:

The given equations are:
12x+2y+533x-2y=-32              ...(i)
54x+2y-353x-2y=6160              ...(ii)
Putting 1x+2y=u and 13x-2y=v , we get:
12u+53v=-32         ...(iii)
54u-35v=6160          ...(iv)
On multiplying (iii) by 6 and (iv) by 20, we get:
3u + 10v = −9         ...(v)
25u-12v=613           ...(vi)
On multiplying (v) by 6 and (vi) by 5, we get:
18u + 60v = −54       ...(vii)
125u-60v=3053          ...(viii)
On adding (vii) and (viii), we get:
143u=3053-54=305-1623=1433
u=13=1x+2y
x + 2y = 3          ...(ix)
On substituting u=13 in (v), we get:
1 + 10v = −9
⇒ 10v = −10
v = −1
13x-2y=-13x-2y=-1            ...(x)
On adding (ix) and (x), we get:
4x = 2
x=12
On substituting x=12 in (x), we get:
32-2y=-12y=32+1=52y=54
Hence, the required solution is x=12and y=54.

Page No 105:

Question 38:

Solve for x and y:
23x+2y+33x-2y=17553x+2y+13x-2y=2

Answer:

The given equations are
23x+2y+33x-2y=175                                     .....i53x+2y+13x-2y=2                                          .....ii
Substituting 13x+2y=u and 13x-2y=v in (i) and (ii), we get
2u+3v=175                                                           .....iii5u+v=2                                                                 .....iv
Multiplying (iv) by 3 and subtracting from(iii), we get
2u-15u=175-6-13u=-135u=153x+2y=5           13x+2y=u                   .....v
Now, substituting u=15 in (iv), we get
1+v=2v=13x-2y=1         13x-2y=v                     .....vi
Adding (v) and (vi), we get
6x=6x=1
Substituting x = 1 in (v), we have
3+2y=5y=1
Hence, x = 1 and y = 1.

Page No 105:

Question 39:

Solve for x and y:
32x+y=7xy3x+3y=11xy

Answer:

The given equations can be written as
3x+6y=7                                             .....i9x+3y=11                                            .....ii
Multiplying (i) by 3 and subtracting (ii) from it, we get
18y-3y=21-1115y=10y=1510=32
Substituting y=32 in (i), we have
3x+6×23=73x=7-4=3x=1
Hence, x=1 and y=32.

Page No 105:

Question 40:

Solve for x and y:
x+y=a+bax-by=a2-b2

Answer:

The given equations are
x+y=a+b                                             .....iax-by=a2-b2                                     .....ii
Multiplying (i) by b and adding it with (ii), we get
bx+ax=ab+b2+a2-b2x=ab+a2a+b=a
Substituting x = a in (i), we have
a+y=a+by=b
Hence, x = a and y = b.

Page No 105:

Question 41:

xa+yb=2,ax-by=a2-b2.

Answer:

The given equations are:
xa+yb=2

bx+ayab=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=ab2+a2b2+a2=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.

Page No 105:

Question 42:

Solve for x and y:
px+qy=p-qqx-py=p+q

Answer:

The given equations are
px+qy=p-q                                        .....iqx-py=p+q                                        .....ii
Multiplying (i) by p and (ii) by q and adding them, we get
p2x+q2x=p2-pq+pq+q2x=p2+q2p2+q2=1
Substituting x = 1 in (i), we have
p+qy=p-qqy=-qy=-1
Hence, x = 1 and y=-1.

Page No 105:

Question 43:

Solve for x and y:
xa-yb=0ax+by=a2+b2

Answer:

The given equations are
xa-yb=0                                            .....iax+by=a2+b2                                     .....ii
From (i)
y=bxa
Substituting y=bxa in (ii), we get
ax+b×bxa=a2+b2x=a2+b2×aa2+b2=a
Now, substitute x = a in (ii) to get
a2+by=a2+b2by=b2y=b
Hence, x = a and y = b.

Page No 105:

Question 44:

6ax+by=3a+2b,6bx-ay=3b-2a.

Answer:

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=3a2+b26a2+b2=12

On substituting x=12 in (i), we get:
6a×12+6by=3a+2b
⇒ 3a + 6by = 3a + 2b
⇒ 6by = 2b
y = 2b6b=13
Hence, the required solution is x=12 and y=13.

Page No 105:

Question 45:

Solve for x and y:
ax-by=a2+b2    x+y=2a

Answer:

The given equations are
ax-by=a2+b2                                     .....ix+y=2a                                                 .....ii
From (ii)
y=2a-x
Substituting y=2a-x in (i), we get
ax-b2a-x=a2+b2ax-2ab+bx=a2+b2x=a2+b2+2aba+b=a+b2a+b=a+b
Now, substitute x = a + b in (ii) to get
a+b+y=2ay=a-b
Hence, x=a+b and y=a-b.

Page No 105:

Question 46:

bxa-ayb+a+b=0,bx-ay+2ab=0.

Answer:

The given equations are:
bxa-ayb+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=-aba-ba-bb=-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.

Page No 105:

Question 47:

bxa+ayb=a2+b2,x+y=2ab.

Answer:

The given equations are:

bxa+ayb=a2+b2
By taking LCM, we get:
b2x+a2yab=a2+b2
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=abb2-a2b2-a2=ab

On substituting x = ab in (i), we get:
b2(ab) + a2y = a3b + ab3
a2y = a3b
a3ba2=ab
Hence, the solution is x = ab and  y = ab.

Page No 105:

Question 48:

Solve for x and y:
x+y=a+bax-by=a2-b2

Answer:

The given equations are
x+y=a+b                                             .....iax-by=a2-b2                                      .....ii
From (i)
y=a+b-x
Substituting y=a+b-x in (ii), we get
ax-ba+b-x=a2-b2ax-ab-b2+bx=a2-b2x=a2+aba+b=a
Now, substitute x = a in (i) to get
a+y=a+by=b
Hence, x = a and y = b.

Page No 105:

Question 49:

Solve for x and y:
a2x+b2y=c2b2x+a2y=d2

Answer:

The given equations are
a2x+b2y=c2                                               .....ib2x+a2y=d2                                                .....ii
Multiplying (i) by aand (ii) by b2 and subtracting, we get
a4x-b4x=a2c2-b2d2x=a2c2-b2d2a4-b4
Now, multiplying (i) by band (ii) by a2 and subtracting, we get
b4y-a4y=b2c2-a2d2y=b2c2-a2d2b4-a4

Hence, x=a2c2-b2d2a4-b4 and y=b2c2-a2d2b4-a4.

Page No 105:

Question 50:

Solve for x and y:
xa+yb=a+bxa2+yb2=2

Answer:

The given equations are
xa+yb=a+b                                            .....ixa2+yb2=2                                                .....ii
Multiplying (i) by b and (ii) by b2 and subtracting, we get
bxa-b2xa2=ab+b2-2b2ab-b2a2x=ab-b2x=ab-b2a2ab-b2=a2
Now, substituting x = ain (i), we get
a2a+yb=a+byb=a+b-a=by=b2

Hence, x=a2 and y=b2.



Page No 111:

Question 1:

x+2y+1=0,2x-3y-12=0.

Answer:

The given equations are:
x + 2y + 1 = 0       ...(i)
2x − 3y − 12 = 0      ...(ii)
Here, a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = −3 and c2 = −12
By cross multiplication, we have:

x2×-12-1×-3=y1×2-1×-12=11×-3-2×2
x-24+3=y2+12=1-3-4
x-21=y14=1-7
x=-21-7=3, y=14-7=-2
Hence, x = 3 and y = −2 is the required solution.

Page No 111:

Question 2:

3x-2y+3=0,4x+3y-47=0.

Answer:

The given equations are:
3x − 2y + 3 = 0       ...(i)
4x + 3y − 47 = 0     ...(ii)
Here, a1 = 3, b1 = −2 , c1 = 3, a2 = 4, b2 =  3 and c2 = −47
By cross multiplication, we have:

x-2×-47-3×3=y3×4--47×3=13×3--2×4
x94-9=y12+141=z9+8
x85=y153=117
x=8517=5, y=15317=9
Hence, x = 5 and y = 9 is the required solution.

Page No 111:

Question 3:

6x-5y-16=0,7x-13y+10=0.

Answer:

The given equations are:
6x − 5y − 16 = 0        ...(i)
7x − 13y + 10 = 0      ...(ii)
Here, a1 = 6, b1 = −5 , c1 = −16, a2 = 7, b2 = −13 and c2 = 10
By cross multiplication, we have:

x-5×10--16×-13=y-16×7-10×6=16×-13--5×7
x-50-208=y-112-60=z-78+35
x-258=y-172=1-43
x=-258-43=6, y=-172-43=4
Hence, x = 6 and y = 4 is the required solution.

Page No 111:

Question 4:

3x+2y+25=0,2x+y+10=0.

Answer:

The given equations are:
3x + 2y + 25 = 0        ...(i)
2x + y + 10 = 0          ...(ii)
Here, a1 = 3, b1 = 2 , c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:

x2×10-25×1=y25×2-10×3=13×1-2×2
x20-25=y50-30=13-4
x-5=y20=1-1
x=-5-1=5, y=20-1=-20
Hence, x = 5 and y = −20 is the required solution.

Page No 111:

Question 5:

2x+5y=1,2x+3y=3.

Answer:

The given equations may be written as:
2x + 5y − 1 = 0         ...(i)
2x + 3y − 3 = 0         ...(ii)
Here, a1 = 2, b1 = 5, c1 = −1, a2 = 2, b2 = 3 and c2 = −3
By cross multiplication, we have:

x5×-3-3×-1=y-1×2--3×2=12×3-2×5
x-15+3=y-2+6=z6-10
x-12=y4=1-4
x=-12-4=3, y=4-4=-1
Hence, x = 3 and y = −1 is the required solution.

Page No 111:

Question 6:

2x+y=35,3x+4y=65.

Answer:

The given equations may be written as:
2x + y − 35 = 0          ...(i)
3x + 4y − 65 = 0        ...(ii)
Here, a1 = 2, b1 = 1, c1 = −35, a2 = 3, b2 = 4 and c2 = −65
By cross multiplication, we have:

x1×-65-4×-35=y-35×3--65×2=12×4-3×1
x-65+140=y-105+130=18-3
x75=y25=15
x=755=15, y=255=5
Hence, x = 15 and y = 5 is the required solution.

Page No 111:

Question 7:

7x-2y=3, 22x-3y=16.

Answer:

The given equations may be written as:
7x − 2y − 3 = 0        ...(i)
22x-3y-16=0           ...(ii)
Here, a1 = 7, b1 = −2 , c1 = −3, a2 = 22, b2 = -3 and c2 = −16
By cross multiplication, we have:

x-2×-16--3×-3=y-3×22--16×7=17×-3-22×-2
x32-9=y-66+112=1-21+44
⇒ x23=y46=123
x=2323=1, y=4623=2
Hence, x = 1 and y = 2 is the required solution.

Page No 111:

Question 8:

x6+y15=4,x3-y12=194.

Answer:

The given equations may be written as:
x6+y15-4=0           ...(i)
x3-y12-194=0       ...(ii)
Here, a1=16, b1=115, c1=-4, a2=13, b2=-112 and c2=-194
By cross multiplication, we have:

x115×-194--112×-4=y-4×13-16×-194=116×-112-13×115
x-1960-13=y-43+1924=1-172-145
x-3960=y-1324=1-13360
x=-3960×-36013=18, y=-1324×-36013=15
Hence, x = 18 and y = 15 is the required solution.

Page No 111:

Question 9:

1x+1y=7,2x+3y=17 (x0, y0).

Answer:

Taking 1x=u  and 1y=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:

u1×-17-3×-7=v-7×2-1×-17=13-2
u-17+21=v-14+17=11
u4=v3=11
u=41=4,v=31=3
1x=4,1y=3
x=14,y=13
Hence, x=14and y=13 is the required solution.

Page No 111:

Question 10:

5(x+y)-2(x-y)+1=0,15(x+y)+7(x-y)-10=0                   (xy, x-y).

Answer:

Taking 1x+y=u  and 1x-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:


u-2×-10-1×7=v1×15--10×5=135+30
u20-7=v15+50=165
u13=v65=165
u=1365=15,v=6565=1
1x+y=15,1x-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:

∴ x1×-1--5×-1=y-5×1--1×1=11×-1-1×1

x-1-5=y-5+1=1-1-1
x-6=y-4=1-2
x=-6-2=3,y=-4-2=2
Hence, x = 3 and y = 2 is the required solution.

Page No 111:

Question 11:

axb-bya=a+b,ax-by=2ab

Answer:

The given equations may be written as:
axb-bya-(a+b)=0         ...(i)
ax-by-2ab=0            ...(ii)
Here, a1 = ab, b1 = -ba, c1 = −(a + b), a2 = a, b2 = −b and c2 = −2ab
By cross multiplication, we have:

x-ba×-2ab--b×-a+b=y-a+b×a--2ab×ab=1ab×-b-a×-ba
x2b2-ba+b=y-aa+b+2a2=1-a+b
x2b2-ab-b2=y-a2-ab+2a2=1-a+b
xb2-ab=ya2-ab=1-a-b
x-ba-b=yaa-b=1-a-b
x=-ba-b-a-b=b,y=aa-b-a-b=-a
Hence, x = b and y = −a is the required solution.

Page No 111:

Question 12:

2ax+3by=a+2b,3ax+2by=2a+b.

Answer:

The given equations may be written as:
2ax + 3by − (a + 2b) = 0         ...(i)
3ax + 2by − (2a + b) = 0         ...(ii)
Here, a1 = 2a, b1 = 3b, c1 = −(a + 2b), a2 = 3a, b2 = 2b and c2 = −(2a + b)
By cross multiplication, we have:

x3b×-2a+b-2b×-a+2b=y-a+2b×3a-2a×-2a+b=12a×2b-3a×3b
x-6ab-3b2+2ab+4b2=y-3a2-6ab+4a2+2ab=14ab-9ab
xb2-4ab=ya2-4ab=1-5ab
x-b4a-b=y-a4b-a=1-5ab
x=-b4a-b-5ab=4a-b5a, y=-a4b-a-5ab=4b-a5b
Hence, x=(4a-b)5a and y=4b-a5b is the required solution.

Page No 111:

Question 13:

Solve the following system of equations by using the method of cross-multiplication:
ax-by=0ab2x+a2by=a2+b2, where x0 and y0

Answer:

Substituting 1x=u and 1y=v in the given equations, we get
     au-bv+0=0                                         .....iab2u+a2bv-a2+b2=0                         .....ii
Here, a1=a,b1=-b,c1=0 and a2=ab2,b2=a2b,c2=-a2+b2.
So, by cross-multiplication, we have
ub1c2-b2c1=vc1a2-c2a1=1a1b2-a2b1u-b-a2+b2-a2b0=v0ab2--a2-b2a=1aa2b-ab2-buba2+b2=vaa2+b2=1aba2+b2
u=ba2+b2aba2+b2, v=aa2+b2aba2+b2u=1a, v=1b1x=1a, 1y=1bx=a, y=b
Hence, x = a and y = b.



Page No 122:

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.

Answer:

The given system of equations is:
3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 3, b1= 5, c1 = −12 and a2 = 5, b2 = 3, c2 = −4
For a unique solution, we must have:
a1a2b1b2, i.e., 3553
Hence, the given system of equations has a unique solution.

Again, the given equations are:
3x + 5y = 12            ...(i)
5x + 3y = 4              ...(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36          ...(iii)
25x + 15y = 20        ...(iv)
On subtracting (iii) from (iv), we get:
16x = −16
x = −1
On substituting x = −1 in (i), we get:
3(−1) + 5y = 12
⇒ 5y = (12 + 3) = 15
y = 3
Hence, x = −1 and y = 3 is the required solution.

Page No 122:

Question 2:

Show that the following  system of equations has a unique solution and solve it:
2x-3y=174x+y=13

Answer:

The system of equations can be written as
2x-3y-17=0                                     .....i4x+y-13=0                                       .....ii
The given equations are of the form
a1x+b1y+c1=0 and a2x+b2y+c2=0
where a1=2, b1=-3, c1=-17 and a2=4, b2=1, c2=-13
Now,
a1a2=24=12 and b1b2=-31=-3
Since, a1a2b1b2, therefore the system of equations has unique solution.
Using cross multiplication method, we have
xb1c2-b2c1=yc1a2-c2a1=1a1b2-a2b1x-3-13-1×-17=y-17×4--13×2=12×1-4×-3x39+17=y-68+26=12+12x56=y-42=114
x=5614, y=-4214x=4, y=-3
Hence, x=4 and y=-3.
 

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Question 3:

Show that the following system of equations has a unique solution:
x3+y2=3,x-2y=2.
Also, find the solution of the given system of equations.

Answer:

The given system of equations are:
x3+y2=3
2x+3y6=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:
a1a2b1b2, i.e., 213-2
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.

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Question 4:

Find the value of k for which the following equations has a unique solution:
2x+3y-5=0kx-6y-8=0

Answer:

The given system of equations are
2x+3y-5=0                                        .....ikx-6y-8=0                                        .....ii
This system is of the form
a1x+b1y+c1=0 and a2x+b2y+c2=0
where a1=2, b1=3, c1=-5 and a2=k, b2=-6, c2=-8
Now, for the given system of equations to have a unique solution, we must have
a1a2b1b22k3-6k-4
Hence, k-4.

Page No 122:

Question 5:

Find the value of k for which the following equations has a unique solution:
x-ky-2=03x+2y+5=0

Answer:

The given system of equations are
x-ky-2=0                                           .....i3x+2y+5=0                                         .....ii
This system of equations is of the form
a1x+b1y+c1=0 and a2x+b2y+c2=0
where a1=1, b1=-k, c1=-2 and a2=3, b2=2, c2=5
Now, for the given system of equations to have a unique solution, we must have
a1a2b1b213-k2k-23
Hence, k-23.

Page No 122:

Question 6:

Find the value of k for which the following system of equations has a unique solution:
5x-7y-5=02x+ky-1=0

Answer:

The given system of equations is 
5x-7y-5=0                                               .....i2x+ky-1=0                                               .....ii
This system is of the form
a1x+b1y+c1=0a2x+b2y+c2=0
where a1=5, b1=-7, c1=-5 and a2=2, b2=k, c2=-1
For the given system of equations to have a unique solution, we must have
a1a2b1b252-7kk-145
Hence, k-145.

Page No 122:

Question 7:

Find the value of k for which the following system of equations has a unique solution:
4x+ky+8=0x+y+1=0

Answer:

The given system of equations is 
4x+ky+8=0                                               .....ix+y+1=0                                                   .....ii
This system is of the form
a1x+b1y+c1=0a2x+b2y+c2=0
where a1=4, b1=k, c1=8 and a2=1, b2=1, c2=1
For the given system of equations to have a unique solution, we must have
a1a2b1b241k1k4
Hence, k4.



Page No 123:

Question 8:

Find the value of k for which each of the following systems of equations has a unique solution:
4x-5y=k, 2x-3y=12.

Answer:

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:
a1a2b1b2
i.e.  42-5-3
25365
Thus, for all real values of k, the given system of equations will have a unique solution.

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Question 9:

Find the value of k for which each of the following systems of equations has a unique solution:
kx+3y=(k-3), 12x+ky=k.

Answer:

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:
a1a2b1b2
i.e.  k123k
k236k±6
Thus, for all real values of k other than ±6, the given system of equations will have a unique solution.

Page No 123:

Question 10:

Show that the system of equations
2x-3y=5, 6x-9y=15
has an infinite number of solutions.

Answer:

The given system of equations is:
2x − 3y = 5
⇒ 2x − 3y − 5 = 0            ...(i)
6x − 9y = 15
⇒ 6x − 9y − 15 = 0          ...(ii)
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 2, b1= −3, c1 = −5 and a2 = 6, b2 = −9, c2 = −15

a1a2=26=13,b1b2=-3-9=13 and c1c2=-5-15=13
Thus, a1a2=b1b2=c1c2
Hence, the given system of equations has an infinite number of solutions.

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Question 11:

Show that the system of equations 6x+5y=11, 9x+152y=21 has no solution.

Answer:

The given system of equations can be written as 
6x+5y-11=                                               .....i9x+152y-21=0                                       .....ii
This system is of the form
a1x+b1y+c1=0a2x+b2y+c2=0
where a1=6, b1=5, c1=-11 and a2=9, b2=152, c2=-21
Now,
a1a2=69=23b1b2=5152=23c1c2=-11-21=1121
Since, a1a2=b1b2c1c2, therefore the given system has no solution.

Page No 123:

Question 12:

For what value of k, the system of equations
kx+2y=5, 3x-4y=10
has (i) a unique solution, (ii) no solution?

Answer:

The given system of equations is:
kx + 2y = 5
kx + 2y − 5= 0                ...(i)
3x − 4y = 10
⇒ 3x − 4y − 10 = 0             ...(ii)
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = k, b1= 2, c1 = −5 and a2 = 3, b2 = −4, c2 = −10
(i) For a unique solution, we must have:
a1a2b1b2, i.e., k32-4k-32
Thus for all real values of k other than -32, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
a1a2=b1b2c1c2
k3=2-4-5-10
k3=2-4 and k312
k=-32,k32
Hence, the required value of k is -32.

Page No 123:

Question 13:

For what value of k, the system of equations
x+2y=5, 3x+ky+15=0
has (i) a unique solution, (ii) no solution?

Answer:

The given system of equations is:
x + 2y = 5
x + 2y − 5= 0                      ...(i)
3x + ky + 15 = 0          ...(ii)
These equations are of the form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 1, b1= 2, c1 = −5 and a2 = 3, b2 = k, c2 = 15
(i) For a unique solution, we must have:
a1a2b1b2, i.e., 132kk6
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:
a1a2=b1b2c1c2
13=2k-515
13=2k and 2k-515
k=6, k-6
Hence, the required value of k is 6.

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Question 14:

For what value of k does the system of equations
x+2y=3, 5x+ky+7=0
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.

Answer:

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:
a1a2b1b2, i.e. 152kk10
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:
a1a2=b1b2c1c2
15=2k-37
15=2k and 2k-37
k=10,k14-3
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.

Page No 123:

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,(k-1)x+(k+2)y=3k.

Answer:

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0                            ....(i)
And, (k − 1)x + (k + 2)y = 3k
⇒ (k − 1)x + (k + 2)y − 3k = 0         ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= 3, c1 = −7 and a2 = (k − 1), b2 = (k + 2), c2 = −3k
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
2k-1=3k+2=-7-3k
2k-1=3k+2=73k

Now, we have the following three cases:
Case I:
2k-1=3k+2
2k+2=3k-12k+4=3k-3k=7

Case II:
3k+2=73k
7k+2=9k7k+14=9k2k=14k=7

Case III:
2k-1=73k
7k-7=6kk=7

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

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Question 16:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
2x+(k-2)y=k,6+(2k-1)y=(2k+5).

Answer:

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:
a1a2=b1b2=c1c2
26=k-22k-1=-k-2k+5
13=k-22k-1=k2k+5

Now, we have the following three cases:
Case I:
13=k-22k-1
(2k-1)=3(k-2)
2k-1=3k-6k=5

Case II:
k-22k-1=k2k+5
k-22k+5=k2k-1
2k2+5k-4k-10=2k2-k
k+k=102k=10k=5

Case III:
13=k2k+5
2k+5=3kk=5

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

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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=(2x+1),2(k+1)x+9y=(7k+1).

Answer:

The given system of equations:
kx + 3y = (2k + 1)
kx + 3y − (2k + 1) = 0                  ...(i)
And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y − (7k + 1) = 0        ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −(2k + 1) and a2 = 2(k + 1), b2 = 9, c2 = −(7k + 1)
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
i.e. k2k+1=39=-2k+1-7k+1
k2k+1=13=2k+17k+1

Now, we have the following three cases:
Case I:
k2k+1=13
2k+1=3k
2k+2=3k
k=2

Case II:
13=2k+17k+1
7k+1=6k+3
k=2

Case III:
k2k+1=2k+17k+1
k7k+1=2k+1×2k+1
7k2+k=2k+12k+2
7k2+k=4k2+4k+2k+2
3k2-5k-2=0
3k2-6k+k-2=0
3kk-2+1k-2=0
3k+1k-2=0
k=2 or k=-13
Hence, the given system of equations has an infinite number of solutions when k is equal to 2.

Page No 123:

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,2(k+1)x+ky=(3k+4).

Answer:

The given system of equations:
5x + 2y = 2k
⇒ 5x + 2y − 2k= 0                          ...(i)
And, 2(k + 1)x + ky = (3k + 4)
⇒ 2(k + 1)x + ky − (3k + 4) = 0       ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= 2, c1 = −2k and a2 = 2(k + 1), b2 = k, c2 = −(3k + 4)
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
52k+1=2k=-2k-3k+4
52k+1=2k=2k3k+4

Now, we have the following three cases:
Case I:
52k+1=2k
2×2k+1=5k4k+1=5k
4k+4=5kk=4

Case II:
2k=2k3k+4
2k2=2×3k+4
2k2=6k+82k2-6k-8=0
2k2-3k-4=0
k2-4k+k-4=0
k(k-4)+1(k-4)=0
k+1 k-4=0
k+1=0  or k-4=0
k=-1or k=4

Case III:
52k+1=2k3k+4
15k+20=4k2+4k
4k2-11k-20=0
4k2-16k+5k-20=0
4kk-4+5k-4=0
k-4 4k+5=0
k=4 or k=-54

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

Page No 123:

Question 19:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
(k-1)x-y=5,(k+1)x+(1-k)y=(3k+1).

Answer:

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:
a1a2=b1b2=c1c2
i.e. k-1k+1=-1-k-1=-5-3k+1
k-1k+1=1k-1=53k+1

Now, we have the following three cases:
Case I:
k-1k+1=1k-1
k-12=k+1
k2+1-2k=k+1
k2-3k=0k(k-3)=0k=0 or k=3

Case II:
1k-1=53k+1
3k+1=5k-1
3k+1=5k-5
2k=6k=3

Case III:
k-1k+1=53k+1
3k+1k-1=5k+1
3k2+k-3k-1=5k+5
3k2-2k-5k-1-5=0
3k2-7k-6=0
3k2-9k+2k-6=0
3kk-3+2k-3=0
k-33k+2=0
k-3=0 or 3k+2=0
k=3 or k=-23

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

Page No 123:

Question 20:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
k-3x+3y=k, kx+ky=12.

Answer:

The given system of equations can be written as 
k-3x+3y-k=0                                      .....ikx+ky-12=0                                             .....ii
This system is of the form
a1x+b1y+c1=0a2x+b2y+c2=0
where a1=k-3, b1=3, c1=-k and a2=k, b2=k, c2=-12
For the given system of linear equations to have an infinite number of solutions
, we must have
a1a2=b1b2=c1c2k-3k=3k=-k-12k-3k=3k  and  3k=-k-12k-3=3  and  k2=36
k=6  and  k=±6k=6
Hence, k = 6.

Page No 123:

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:
(a-1)x+3y=2,6x+(1-2b)y=6.

Answer:

The given system of equations:
(a − 1)x + 3y = 2
⇒ (a − 1)x + 3y − 2 = 0        ...(i)
and 6x + (1 − 2b)y = 6
⇒ 6x + (1 − 2b)y − 6= 0       ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
where, a1 = (a − 1), b1= 3, c1 = −2 and a2 = 6, b2 = (1 − 2b), c2 = −6
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
a-16=31-2b=-2-6
a-16=31-2b=13
a-16=13and31-2b=13
⇒ 3a − 3 = 6 and 9 = 1 − 2b
⇒ 3a = 9 and 2b = −8
a = 3 and b = −4
∴​ a = 3 and b = −4



Page No 124:

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:
(2a-1)x+3y=5,3x+(b-1)y=2.

Answer:

The given system of equations:
(2a − 1)x + 3y = 5
⇒ (2a − 1)x + 3y − 5 = 0          ....(i)
And, 3x + (b − 1)y = 2
⇒ 3x + (b − 1)y − 2 = 0            ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = (2a − 1), b1= 3, c1 = −5 and a2 = 3, b2 = (b − 1), c2 = −2
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
2a-13=3b-1=-5-2
2a-13=3b-1=52
2a-13=52 and 3b-1=52
⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒  4a − 2 = 15 and 6 = 5b − 5
⇒  4a = 17 and 5b = 11
a=174and b=115
∴​ a174 and b = 115

Page No 124:

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,(a+b)x-(a+b-3)y=4a+b.

Answer:

The given system of equations:
2x − 3y = 7
⇒ 2x − 3y − 7 = 0                                ....(i)
And, (a + b)x − (a + b − 3)y = 4a + b
⇒ (a + b)x − (a + b − 3)y − (4a + b) = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= −3, c1 = −7 and a2 = (a + b), b2 = −(a + b − 3), c2 = −(4a + b)
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
2a+b=-3-a+b-3=-7-4a+b
2a+b=3a+b-3=74a+b
2a+b=74a+b and 3a+b-3=74a+b
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b − 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b − 21
a = 5b                  ....(iii)
And, 5a = 4b − 21     ....(iv)
On substituting a = 5b in (iv), we get:
25b = 4b − 21
⇒ 21b = −21
b = −1
On substituting b = −1 in (iii), we get:
a = 5(−1) = −5
a = −5 and b = −1

Page No 124:

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,(a+b+1)x+(a+2b+2)y=4(a+b)+1.

Answer:

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0                                   ....(i)
And, (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
⇒ (a + b + 1)x + (a + 2b + 2)y − [4(a + b) + 1] = 0    ....(ii)          
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1=  3, c1 = −7 and a2 = (a + b + 1), b2 = (a + 2b + 2), c2 = −[4(a + b) + 1]
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
2a+b+1=3a+2b+2=-7-4a+b+1
2a+b+1=3a+2b+2=74a+b+1
2a+b+1=3a+2b+2and3a+2b+2=74a+b+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

Page No 124:

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=7a+bx+2a-by=21                               [CBSE 2001]

Answer:

The given system of equations can be written as 
2x+3y-7=0                                         .....ia+bx+2a-by-21=0                   .....ii
This system is of the form
a1x+b1y+c1=0a2x+b2y+c2=0
where a1=2, b1=3, c1=-7 and a2=a+b, b2=2a-b, c2=-21
For the given system of linear equations to have an infinite number of solutions, we must have
a1a2=b1b2=c1c22a+b=32a-b=-7-212a+b=-7-21=13  and  32a-b=-7-21=13a+b=6  and  2a-b=9
Adding a+b=6 and 2a-b=9, we get
3a=15a=153=5
Now, substituting a = 5 in a + b = 6, we have
5+b=6  b=6-5=1
Hence, a = 5 and b = 1.

Page No 124:

Question 26:

Find the values of and b for which the following system of linear equations has an infinite number of solutions:
2x+3y=72ax+a+by=28                            [CBSE 2001]

Answer:

The given system of equations can be written as 
2x+3y-7=0                                         .....i2ax+a+by-28=0                             .....ii
This system is of the form
a1x+b1y+c1=0a2x+b2y+c2=0
where a1=2, b1=3, c1=-7 and a2=2a, b2=a+b, c2=-28
For the given system of linear equations to have an infinite number of solutions, we must have
a1a2=b1b2=c1c222a=3a+b=-7-2822a=-7-28=14  and  3a+b=-7-28=14a=4  and  a+b=12
Substituting a = 4 in a + b = 12, we get
4+b=12b=12-4=8
Hence, a = 4 and b = 8.

Page No 124:

Question 27:

Find the value of k for which each of the following system of equations has no solution:8x+5y=9, kx+10y=15.

Answer:

The given system of equations:
8x + 5y = 9
8x + 5y − 9 = 0                    ....(i)
kx + 10y = 15
kx + 10y − 15= 0                 ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 8, b1= 5, c1 = −9 and a2 = k, b2 = 10, c2 = −15
In order that the given system has no solution, we must have:
a1a2=b1b2c1c2
 i.e. 8k=510-9-15i.e. 8k=1235
8k=12 and 8k35
k=16 and k403
Hence, the given system of equations has no solution when k is equal to 16.

Page No 124:

Question 28:

Find the value of k for which each of the following system of equations has no solution:
kx+3y=3, 12x+ky=6.

Answer:

The given system of equations:
kx + 3y = 3
kx + 3y − 3 = 0                  ....(i)
12x + ky = 6
12x + ky − 6= 0                 ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −3 and a2 = 12, b2 = k, c2 = −6
In order that the given system of equations has no solution, we must have:
a1a2=b1b2c1c2
i.e.  k12=3k-3-6
k12=3k and 3k12
k2=36 and k6
k=±6 and k  6
Hence, the given system of equations has no solution when k is equal to −6.

Page No 124:

Question 29:

Find the value of k for which each of the following system of equations has no solution:
3x-y-5=0, 6x-2y+k=0,(k0).

Answer:

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:
a1a2=b1b2c1c2
i.e. 36=-1-2-5k
-1-2-5kk-10
Hence, equations (i) and (ii) will have no solution if k-10.

Page No 124:

Question 30:

Find the value of for which the following system of linear equations has no solutions:
kx+3y=k-312x+ky=k

Answer:

The given system of equations can be written as 
kx+3y+3-k=0                                         .....i12x+ky-k=0                                              .....ii
This system is of the form
a1x+b1y+c1=0a2x+b2y+c2=0
where a1=k, b1=3, c1=3-k and a2=12, b2=k, c2=-k
For the given system of linear equations to have no solution, we must have
a1a2=b1b2c1c2k12=3k3-k-k k12=3k and  3k3-k-kk2=36  and  -33-k
k=±6  and  k6k=-6
Hence, k=-6.

Page No 124:

Question 31:

Find the value of k for which the system of equations
5x-3y=0, 2x+ky=0
has a nonzero solution.

Answer:

The given system of equations:
5x − 3y = 0         ....(i)
2x + ky = 0         ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= −3, c1 = 0 and a2 = 2, b2 = k, c2 = 0
For a non-zero solution, we must have:
a1a2=b1b2
52=-3k
5k=-6k=-65
Hence, the required value of k is -65.



Page No 145:

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.

Answer:

Let the cost of a chair be ₹ x and that of a table be ₹ y. Then
5x+4y=5600                                              .....i4x+3y=4340                                              .....ii
Multiplying (i) by 3 and (ii) by 4, we get
15x-16x=16800-17360-x=-560x=560
Substituting x = 560 in (i), we have
5×560+4y=56004y=5600-2800y=28004=700
Hence, the cost of a chair and that of a table are respectively ₹ 560 and ₹ 700.
 

Page No 145:

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.

Answer:

Let the cost of a spoon be ₹and that of a fork be ₹y. Then
23x+17y=1770                                          .....i17x+23y=1830                                          .....ii
Adding (i) and (ii), we get
40x+40y=3600x+y=90                                            .....iii
Now, subtracting (ii) from (i), we get
6x-6y=-60x-y=-10                                               .....iv
Adding (iii) and (iv), we get
2x=80x=40
Substituting x = 40 in (iii), we get
40+y=90y=50
Hence, the cost of a spoon that of a fork are ₹40 and ₹50 respectively.



Page No 146:

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?

Answer:

Let the x and y be the number of 50-paisa and 25-paisa conis respectively. Then
x+y=50                                                  .....i0.5x+0.25y=19.50                               .....ii
Multiplying (ii) by 2 and subtracting it from (i), we get
0.5y=50-39y=110.5=22 
Subtracting y = 22 in (i), we get
x+22=50x=50-22=28
Hence, the number of 25-paisa and 50-paisa conis are 22 and 28 respectively.

Page No 146:

Question 4:

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

Answer:

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.

Page No 146:

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.

Answer:

Let the first number be x and the second number be y.
Then, we have:
2x + 3y = 92                       ....(i)
4x − 7y = 2                         ....(ii)
On multiplying (i) by 7 and (ii) by 3, we get:
14x + 21y = 644                 ....(iii)
12x − 21y = 6                     ....(iv)
On adding (iii) and (iv), we get:
26x = 650
x = 25
On substituting x = 25 in (i), we get:
2 × 25 + 3y = 92
⇒ 50 + 3y = 92
⇒ 3y = (92 − 50) = 42
y = 14
Hence, the first number is 25 and the second number is 14.

Page No 146:

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.

Answer:

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.

Page No 146:

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.

Answer:

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.

Page No 146:

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.

Answer:

We know:
Dividend = Divisor × Quotient + Remainder

Let the larger number be x and the smaller be y.
Then, we have:
3x = y × 4 + 8 or 3x − 4y = 8              ....(i)
5y = x × 3 + 5 or −3x + 5y = 5             ....(ii)
On adding (i) and (ii), we get:
y = (8 + 5) = 13
On substituting y = 13 in (i) we get:
3x − 4 × 13 = 8
⇒ 3x = (8 + 52) = 60
x = 20
Hence, the larger number is 20 and the smaller number is 13.

Page No 146:

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.

Answer:

Let the required numbers be x and y.
Now, we have:
x+2y+2=12
By cross multiplication, we get:
2x + 4 = y + 2
⇒ 2xy  = −2                  ....(i)
Again, we have:
x-4y-4=511
By cross multiplication, we get:
11x − 44 = 5y − 20
⇒ 11x − 5y = 24                ....(ii)
On multiplying (i) by 5, we get:
10x − 5y = −10                 ....(iii)
On subtracting (iii) from (ii), we get:
x = (24 + 10) = 34
On substituting x = 34 in (i), we get:
2 × 34 − y = −2
⇒ 68 − y = −2
y = (68 + 2) = 70
Hence, the required numbers are 34 and 70.

Page No 146:

Question 10:

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

Answer:

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=25228=9
On substituting y = 9 in (i), we get:
x = 14 + 9 = 23
Hence, the required numbers are 23 and 9.

Page No 146:

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.

Answer:

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.

Page No 146:

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.

Answer:

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.

Page No 146:

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.

Answer:

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.

Page No 146:

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.

Answer:

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.

Page No 146:

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.

Answer:

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.



Page No 147:

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.

Answer:

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
y+x=±y-x2+4xy
y+x=±4+4×35 =±144 =±12
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.

Page No 147:

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.

Answer:

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
x+y=±x-y2+4xy
x+y=±49+4×18             =±49+72             =±121=±11
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.

Page No 147:

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.

Answer:

Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per the question
10y+x+10x+y=12111x+11y=121x+y=11                                                  .....i
Since the digits differ by 3, so
x-y=3                                                   ......ii
Adding (i) and (ii), we get
2x=14x=7
Putting x = 7 in (i), we get
7+y=11y=4
Changing the role of x and y, x = 4 and y = 7
Hence, the two-digit number is 74 or 47.

Page No 147:

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 34. Find the fraction.

Answer:

Let the required fraction be xy.
Then, we have:
x + y = 8                        ....(i)
And, x+3y+3=34
⇒ 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 35.

Page No 147:

Question 20:

If 2 is added to the numerator of a fraction, it reduces to 12 and if 1 is subtracted from the denominator, it reduces to 13. Find the fraction.

Answer:

Let the required fraction be xy.
Then, we have:
x+2y=12
⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2xy = −4                      .....(i)

Again, xy-1=13
⇒ 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 310.

Page No 147:

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 34. Find the fraction.

Answer:

Let the required fraction be xy.
Then, we have:
y = x + 11
yx = 11                        ....(i)
Again, x+8y+8=34
⇒ 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 2536.

Page No 147:

Question 22:

Find a fraction which becomes 12 when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes 13 when 7 is subtracted from the numerator and 2 is subtracted from the denominator.

Answer:

Let the required fraction be xy.
Then, we have:
x-1y+2=12
⇒ 2(x − 1) = 1(y + 2)
⇒ 2x − 2 = y + 2
⇒ 2xy = 4                    ....(i)

Again, x-7y-2=13
⇒ 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 1526.

Page No 147:

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]

Answer:

Let the fraction be xy
As per the question
x+y=4+2xy-x=4                                                    .....i
After changing the numerator and denominator
New numerator = x + 3
New denominator = + 3
Therefore
x+3y+3=233x+3=2y+33x+9=2y+62y-3x=3                                                .....ii
Multiplying (i) by 3 and subtracting (ii), we get
3y-2y=12-3y=9
Now, putting y = 9 in (i), we get
9-x=4x=9-4=5
Hence, the fraction is 59.

Page No 147:

Question 24:

The sum of two numbers is 16 and the sum of their reciprocals is 13. Find the numbers.

Answer:

Let the larger number be x and the smaller number be y.
Then, we have:
(x + y) = 16                      ....(i)
And, 1x+1y=13                  ....(ii)
x+yxy=13
⇒ 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) = ±64 = ±8
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.

Page No 147:

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.

Answer:

Let the number of students in classroom A be x
Let the number of students in classroom B be y.
If 10 students are transferred from A to B, then we have:
x − 10 = y + 10
x y = 20                 ...(i)
If 20 students are transferred from B to A, then we have:
2(y − 20) = x + 20
⇒ 2y − 40 = x + 20
⇒ −x + 2y = 60            ...(ii)
On adding (i) and (ii), we get:
y = (20 + 60) = 80
On substituting y = 80 in (i), we get:
x − 80 = 20
x = (20 + 80) = 100
Hence, the number of students in classroom A is 100 and the number of students in classroom B is 80.

Page No 147:

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.

Answer:

Let fixed charges be ₹x and rate per km be ₹y.
Then as per the question
x+80y=1330                                              .....ix+90y=1490                                              .....ii
Subtracting (i) from (ii), we get
10y=160y=16010=16
Now, putting y = 16, we have
x+80×16=1330x=1330-1280=50
Hence, the fixed charges be ₹50 and the rate per km is ₹16.



Page No 148:

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.

Answer:

Let the fixed charges be ₹x and the cost of food per day be ₹y.
Then as per the question
x+25y=4500                                              .....ix+30y=5200                                              .....ii
Subtracting (i) from (ii), we get
5y=700y=7005=140
Now, putting = 140, we have
x+25×140=4500x=4500-3500=1000
Hence, the fixed charges is ₹1000 and the cost of the food per day is ₹140.

Page No 148:

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?

Answer:

Let the the amounts invested at 10% and 8% be ₹x and ₹y respectively.
Then as per the question
x×10×1100+y×8×1100=135010x+8y=135000                                        .....i
After the amounts interchanged but the rate being the same, we have
x×8×1100+y×10×1100=1350-458x+10y=130500                                        .....ii
Adding (i) and (ii) and dividing by 9, we get
2x+2y=29500                                            .....iii           
Subtracting (ii) from (i), we get
2x-2y=4500                                              .....iv               
Now, adding (iii) and (iv), we have
4x=34000x=340004=8500
Putting x = 8500 in (iii), we get
2×8500+2y=295002y=29500-17000=12500y=125002=6250
Hence, the amounts invested are ₹8,500 at 10% and ₹6,250 at 8%.

Page No 148:

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.

Answer:

Let the monthly income of A and B are ₹x and ₹y respectively.
Then as per the question
xy=54y=4x5                                                     .....i
Since each save ₹9,000, so
Expenditure of A = ​₹x-9000
Expenditure of B = ​₹y-9000
The ratio of expenditures of A and B are in the ratio 7 : 5.
x-9000y-9000=757y-63000=5x-450007y-5x=18000                                     .....ii
From (i), substitute y=4x5 in (ii) to get
7×4x5-5x=1800028x-25x=900003x=90000x=30000
Now, putting x = 30000, we get
y=4×300005=4×6000=24000
Hence, the monthly incomes of A and B are ​₹30,000 and ​₹24,000.

Page No 148:

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.

Answer:

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
100+25100×x+100+10100×y=1520125100x+110100y=152025x+22y-30400=0                              .....i
When the profit on chair and table are 10% and 25% respectively, then

100+10100×x+100+25100×y=1535110100x+125100y=153522x+25y-30700=0                               .....ii
Solving (i) and (ii) by cross multiplication, we get
x22-30700-25-30400=y-3040022--3070025=12525-2222x7600-6754=y7675-6688=1003×47x846=y987=1003×47x=100×8463×47, y=100×9873×47
x=600, y=700
Hence, the cost of chair and table are ​₹600 and ​₹700 respectively.

Page No 148:

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.

Answer:

Let X and Y be the cars starting from points A and B, respectively and let their speeds be x km/h and y km/h, respectively.
Then, we have the following cases:
Case I: When the two cars move in the same direction
In this case, let the two cars meet at point M.

Distance covered by car X in 7 hours = 7x km
Distance covered by car Y in 7 hours = 7y km
∴ AM = (7x) km and BM = (7y) km
⇒ (AM − BM) = AB
⇒ (7x − 7y) = 70
⇒ 7(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.

Page No 148:

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.

Answer:

Let the original speed be x kmph and let the time taken to complete the journey be y hours. 
∴ Length of the whole journey = (xy) km
Case I:
When the speed is (x + 5) kmph and the time taken is (y − 3) hrs:
Total journey = (x + 5) (y − 3) km
⇒ (x + 5) (y − 3) = xy
xy + 5y − 3x − 15 = xy
⇒ 5y − 3x = 15                ...(i)
Case II:
When the speed is (x − 4) kmph and the time taken is (y + 3) hrs:
Total journey = (x − 4) (y + 3) km
⇒ (x − 4) (y + 3) = xy
xy − 4y + 3x − 12 = xy
⇒ 3x − 4y = 12                ...(ii)
On adding (i) and (ii), we get:
y = 27
On substituting y = 27 in (i), we get:
5 × 27 − 3x = 15
⇒ 135 − 3x = 15
⇒ 3x = 120
x = 40

∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km

Page No 148:

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.

Answer:

Let the speed of the train and taxi be x km/h and y km/h respectively.
Then as per the question
3x+2y=11200                                            .....i
When the speeds of the train and taxi are 260 km and 240 km respectively, then
260x+240y=112+66013x+12y=28100                                             .....ii
Multiplying (i) by 6 and subtracting (ii) from it, we get
18x-13x=66200-281005x=10200x=100
Putting x = 100 in (i), we have
3100+2y=112002y=11200-3100=140y=80
Hence, the speed of the train and that of the taxi are 100 km/h and 80 km/h respectively.

Page No 148:

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.

Answer:

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
AC-BC=160x×8-y×8=160x-y=20                                                  .....i
Case-2: When they travel in opposite direction

From the figure
AC+BC=160x×2+y×2=160x+y=80                                                  .....ii
Adding (i) and (ii), we get
2x=100x=50 km/h
Putting = 50 in (ii), we have
50+y=80y=80-50=30 km/h
Hence, the speeds of the cars are 50 km/h and 30 km/h.

Page No 148:

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.

Answer:

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
x+y×4060=8x+y=12 .....i
When the sailor goes upstream, then
x-y×1=8x-y=8 .....ii
Adding (i) and (ii), we get
2x=20x=10
Putting x = 10 in (i), we have
10+y=12y=2
Hence, the speeds of the sailor in staill water and the current are 10 km/h and 2 km/h respectively.



Page No 149:

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.

Answer:

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 = 12x-y hrs
Time taken to cover 40 km downstream = 40x+y hrs
Total time taken = 8 hrs
12x-y + 40x+y = 8        ....(i)

Again, we have:
Time taken to cover 16 km upstream = 16x-y hrs
Time taken to cover 32 km downstream = 32x+y hrs
Total time taken = 8 hrs

16x-y + 32x+y = 8        ....(ii)
Putting 1x-y=u and 1x+y=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=28=14
On substituting u=14 in (iii), we get:
40v = 5
v=540=18
Now, we have:
u=14
1x-y=14x-y=4             ....(v)
v=18
1x+y=18x+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

Page No 149:

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.

Answer:

Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days.
∴ One man's one day's work = 1x
And, one boy's one day's work = 1y
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) = 14
2x+5y=14
2u+5v=14              ...(i)           Here, 1x=u and 1y=v
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) = 13
3x+6y=13
3u+6v=13             ....(ii)           Here, 1x=u and 1y=v
On multiplying (i) by 6 and (ii) by 5, we get:
12u+30v=64             ....(iii)
15u+30v=53             ....(iv)
On subtracting (iii) from (iv), we get:
3u=53-64=212=16
u=16×3=1181x=118x=18
On substituting u=118 in (i), we get:
2×118+5v=145v=14-19=536
v=536×15=1361y=136y=36
Hence, one man alone can finish the work in 18 days and one boy alone can finish the work in 36 days.

Page No 149:

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.

Answer:

Let the length of the room be x metres and the breadth of the room be y metres.
Then, we have:
Area of the room = xy
According to the question, we have:
x = y + 3
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.

Page No 149:

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.

Answer:

Let the length and the breadth of the rectangle be x m and y m, respectively.
∴ Area of the rectangle = (xy) sq. m

Case 1: When the length is reduced by 5 m and the breadth is increased by 3 m:
New length = (x − 5) m
New breadth = (y + 3) m
∴ New area = (x − 5) (y + 3) sq. m
xy − (x − 5) (y + 3) = 8
xy − [xy − 5y + 3x − 15] = 8
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.

Page No 149:

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.

Answer:

Let the length and breadth of the rectangle be x m and y m respectively.
Case1: When length is increased by 3 m and breadth is decreased by 4 m
xy-x+3y-4=67xy-xy+4x-3y+12=674x-3y=55                                              .....i
Case2: When length is reduced by 1 m and breadth is increased by 4 m
x-1y+4-xy=89xy+4x-y-4-xy=894x-y=93                                                .....ii
Subtracting (i) from (ii), we get
2y=38y=19
Now, putting y = 19 in (ii), we have
4x-19=934x=93+19=112x=28
Hence, length = 28 m and breadth = 19 m.

Page No 149:

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?

Answer:

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
x+y=4150                                                   .....i
Case 2: One full and one half reserved first class tickets cost ₹6,255
x+y+12x+y=62553x+4y=12510                                         .....ii
Substituting y=4150-x from (i) in (ii), we get
3x+44150-x=125103x-4x+16600=12510x=16600-12510=4090
Now, putting x = 4090 in (i), we have
4090+y=4150y=4150-4090=60
Hence, cost of basic first class full fare = ₹4,090 and reservation charge = ₹60.

Page No 149:

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.

Answer:

Let the the present age of the man be x years and that of his son be y years.
After 5 years man's age = x + 5
After 5 years ago son's age = y + 5
As per the question
x+5=3y+5x-3y=10                                           .....i
5 years ago man's age = x − 5
5 years ago son's age = y − 5
As per the question
x-5=7y-5x-7y=-30                                             .....ii
Subtracting (ii) from (i), we have
4y=40y=10
Putting y = 10 in (i), we get
x-3×10=10x=10+30=40
Hence, man's present age = 40 years and son's present age = 10 years.

Page No 149:

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.

Answer:

Let the man's present age be x years.
Let his son's present age be y years.
According to question, we have:
Two years ago:
Age of the man = Five times the age of the son
⇒ (x − 2) = 5(y − 2)
x − 2 = 5y − 10
x − 5y = −8                .....(i)
Two years later:
Age of the man = Three times the age of the son + 8
⇒  (x + 2) = 3(y + 2) + 8
⇒  x + 2 = 3y + 6 + 8
x − 3y = 12                 .....(ii)
Subtracting (i) from (ii), we get:
2y = 20
y = 10
On substituting y = 10 in (i), we get:
x − 5 × 10 = −8
x − 50 = −8
x = (−8 + 50) = 42
Hence, the present age of the man is 42 years and the present age of the son is 10 years.

Page No 149:

Question 44:

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

Answer:

Let the father's present age be x years.
Let her son's present age be y years.
Then, we have:
x + 2y = 70                      ....(i)
And, 2x + y = 95               ....(ii)
On multiplying (ii) by 2, we get:
4x + 2y = 190                 ....(iii)
On subtracting (i) from (iii), we get:
3x = 120
x = 40
On substituting x = 40 in (i), we get:
40 + 2y = 70
⇒ 2y = (70 − 40) = 30
y = 15
Hence, the father's present age is 40 years and her son's present age is 15 years.

Page No 149:

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.

Answer:

Let the woman's present age be x years.
Let her daughter's present age be y years.
Then, we have:
x = 3y + 3
x − 3y = 3               ....(i)
After three years, we have:
(x + 3) = 2(y + 3) + 10
x + 3 = 2y + 6 + 10
x − 2y = 13             ....(ii)
Subtracting (ii) from (i), we get:
y = (3 − 13) = −10
y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 3
x − 30 = 3
x = (3 + 30) = 33 
Hence, the woman's present age is 33 years and her daughter's present age is 10 years.

Page No 149:

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.

Answer:

Let the actual price of the tea and lemon set be ₹x and ₹y respectively.
When gain is ₹7, then
y100×15-x100×5=73y-x=140                                        .....i
When gain is ₹14, then
y100×5+x100×10=14y+2x=280                                         .....ii
Multiplying (i) by 2 and adding with (ii), we have
7y=280+280y=5607=80
Putting y = 80 in (ii), we get
80+2x=280x=2002=100
Hence, actual price of the tea set and lemon set are ₹100 and ₹80 respectively.



Page No 150:

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.

Answer:

Let the fixed charge be ₹x and the charge for each extra day be ₹y.
In case of  Mona, as per the question
x+4y=27                                               .....i
In case of Tanvy, as per the question
x+2y=21                                               .....ii
Subtracting (ii) from (i), we get
2y=6y=3
Now, putting y = 3 in (ii), we have
x+2×3=21x=21-6=15
Hence, the fixed charge be ₹15 and the charge for each extra day is ₹3.

Page No 150:

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?

Answer:

Let x litres and y litres be the amount of acids from 50% and 25% acid solutions respectively.
As per the question
50% of x+25% of y=40% of 100.50x+0.25y=4             2x+y=16                                          .....i
Since, the total volume is 10 litres, so
x+y=10                                                .....ii
Subtracting (ii) from (i), we get
x=6
Now, putting x = 6 in (ii), we have
6+y=10y=4
Hence, volume of 50% acid solution = 6 litres and volume of 25% acid solution = 4 litres.

Page No 150:

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?

Answer:

Let x g and y g be the weight of 18-carat and 12-carat gold respectively.
As per the given condition
18x24+12y24=120×16243x+2y=320                                            .....i
And
x+y=120                                                    .....ii
Multiplying (ii) by 2 and subtracting from (i), we get
x=320-240=80
Now, putting x = 80 in (ii), we have
80+y=120y=40
Hence, the required weight of 18-carat and 12-carat gold bars are 80 g and 40 g respectively.

Page No 150:

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.

Answer:

Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions.
As per the given condition
0.90x+0.97y=21×0.950.90x+0.97y=21×0.95                            .....i
And
x+y=21                                                       .....ii
From (ii), subtitute y=21-x in (i) to get
0.90x+0.9721-x=21×0.950.90x+0.97×21-0.97x=21×0.950.07x=0.97×21-21×0.95x=21×0.020.07=6
Now, putting x = 6 in (ii), we have
6+y=21y=15
Hence, the required quantities are 6 litres and 15 litres.

Page No 150:

Question 51:

The larger of the two supplementary angles exceeds the smaller by 18. Find them.

Answer:

Let x and y be the supplementary angles, where x > y.
As per the given condition
x+y=180                                                  .....i
And
x-y=18                                                     .....ii
Adding (i) and (ii), we get
2x=198x=99
Now, substituting x=99 in (ii), we have
99-y=18x=99-18=81
Hence, the required angles are 99 and 81.

Page No 150:

Question 52:

In ABC, A=x, B=3x-2, C=y and C-B=9. Find the three angles.

Answer:

C-B=9y-3x-2=9y-3x+2=9y-3x=7                                          .....i
The sum of all the angles of a triangle is 180, therefore
A+B+C=180x+3x-2+y=1804x+y=182                                     .....ii
Subtracting (i) from (ii), we have
7x=182-7=175x=25
Now, substituting x=25 in (i), we have
y=3x+7=3×25+7=82
Thus
A=x=25B=3x-2=75-2=73C=y=82
Hence, the angles are 25, 73 and 82.

Page No 150:

Question 53:

In a cyclic quadrilateral ABCD, it is given that A=2x+4, B=y+3, C=2y+10 and D=4x-5 .
Find the four angles.

Answer:

The opposite angles of cyclic quadrilateral are supplementary, so
A+C=1802x+4+2y+10=180x+y=83                                                 .....i
And
B+D=180y+3+4x-5=1804x+y=182                                           .....ii
Subtracting (i) from (ii), we have
3x=99x=33
Now, substituting x=33 in (i), we have
33+y=83y=83-33=50
Therefore
A=2x+4=2×33+4=70B=y+3=50+3=53C=2y+10=2×50+10=110D=4x-5=4×33-5=132°-5=127
Hence, A=70, B=53, C=110 and D=127.



Page No 155:

Question 1:

Write the number of solutions of the following pair of linear equations:
x+2y-8=0, 2x+4y=16

Answer:

The given equations are
x+2y-8=0                                                 .....i2x+4y-16=0                                            .....ii
Which is of the form a1x+b1y+c1=0 and a2x+b2y+c2=0, where a1=1, b1=2, c1=-8, a2=2, b2=4 and c2=-18
Now
a1a2=12b1b2=24=12c1c2=-8-16=12
a1a2=b1b2=c1c2=12
Thus, the pair of linear equations are coincident and therefore has infinitely many solutions.

Page No 155:

Question 2:

Find the value of k for which the following pair of linear equations have infinitely many solutions:
2x+3y=7, k-1x+k+2y=3k

Answer:

The given equations are
2x+3y-7=0                                                .....ik-1x+k+2y-3k=0                            .....ii
Which is of the form a1x+b1y+c1=0 and a2x+b2y+c2=0, where a1=2, b1=3, c1=-7, a2=k-1, b2=k+2 and c2=-3k
For the given pair of linear equations to have infinitely many solutions, we must have
a1a2=b1b2=c1c22k-1=3k+2=-7-3k2k-1=3k+2, 3k+2=-7-3k and 2k-1=-7-3k2k+2=3k-1, 9k=7k+14 and 6k=7k-7
k=7, k=7 and k=7
Hence, k = 7.

Page No 155:

Question 3:

For what value of k does the following pair of linear equations have infinitely many solutions?:
10x+5y-k-5=0, 20x+10y-k=0

Answer:

The given pair of linear equations are
10x+5y-k-5=0                                    .....i20x+10y-k=0                                          .....ii
Which is of the form a1x+b1y+c1=0 and a2x+b2y+c2=0, where a1=10, b1=5, c1=-k-5, a2=20, b2=10 and c2=-k
For the given pair of linear equations to have infinitely many solutions, we must have
a1a2=b1b2=c1c21020=510=-k-5-k12=k-5k2k-10=kk=10
Hence, k = 10.

Page No 155:

Question 4:

For what value of k will the following pair of linear equations have no solution?:
2x+3y=9, 6x+k-2y=3k-2

Answer:

The given pair of linear equations are
2x+3y-9=0                                        .....i6x+k-2y-3k-2=0                    .....ii
Which is of the form a1x+b1y+c1=0 and a2x+b2y+c2=0, where a1=2, b1=3, c1=-9, a2=6, b2=k-2 and c2=-3k-2
For the given pair of linear equations to have no solution, we must have
a1a2=b1b2c1c226=3k-2-9-3k-226=3k-2, 3k-2-9-3k-2k=11, 3k-293k-2
k=11, 33k-29k-2k=11, 13 true
Hence, k = 11.

Page No 155:

Question 5:

Write the number of solutions of the following pair of linear equations:
x+3y-4=0, 2x+6y-7=0

Answer:

The given pair of linear equations are
x+3y-4=0                                                 .....i2x+6y-7=0                                               .....ii
Which is of the form a1x+b1y+c1=0 and a2x+b2y+c2=0, where a1=1, b1=3, c1=-4, a2=2, b2=6 and c2=-7
Now
a1a2=12b1b2=36=12c1c2=-4-7=47a1a2=b1b2c1c2
Thus, the pair of the given linear equations has no solution.

Page No 155:

Question 6:

Write the value of k for which the system of equations 3x+ky=0, 2x-y=0 has a unique solution.

Answer:

The given pair of linear equations is
3x+ky=0                                                 .....i2x-y=0                                                   .....ii
Which is of the form a1x+b1y+c1=0 and a2x+b2y+c2=0, where a1=3, b1=k, c1=0, a2=2, b2=-1 and c2=0
For the system to have a unique solution, we must have
a1a2=b1b232k-1k-32
Hence, k-32.

Page No 155:

Question 7:

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

Answer:

Let the numbers be x and y, where x > y.
Then as per the question
x-y=5                                                    .....ix2-y2=65                                               .....ii
Dividing (ii) by (i), we get
x2-y2x-y=655x-yx+yx-y=13x+y=13                                                  .....iii
Now, adding (i) and (ii), we have
2x=18x=9
Substituting x = 9 in (iii), we have
9+y=13y=4
Hence, the numbers are 9 and 4.

Page No 155:

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.

Answer:

Let the cost of 1 pen and 1 pencil are ₹x and ₹y respectively.
Then as per the question
5x+8y=120                                                  .....i8x+5y=153                                                  .....ii
Adding (i) and (ii), we get
13x+13y=273x+y=21                                                  .....iii
Subtracting (i) from (ii), we get
3x-3y=33x-y=11                                                  .....iv
Now, adding (iii) and (iv), we get
2x=32x=16
Substituting x = 16 in (iii), we have
16+y=21y=5
Hence, the cost of 1 pen and 1 pencil are respectively ₹16 and ₹5.

Page No 155:

Question 9:

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

Answer:

Let the larger number be x and the smaller number be y.
Then as per the question
x+y=80                                                       .....ix=4y+5x-4y=5                                                       .....ii
Subtracting (ii) from (i), we get
5y=75y=15
Now, putting y = 15 in (i), we have
x+15=80x=65
Hence, the numbers are 65 and 15.

Page No 155:

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.

Answer:

Let the ones digit and tens digit be x and y respectively.
Then as per the question
x+y=10                                                       .....i10y+x-18=10x+yx-y=-2                                                      .....ii
Adding (i) and (ii), we get
2x=8x=4
Now, putting x = 4 in (i), we have
4+y=10y=6
Hence, the number is 64.

Page No 155:

Question 11:

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

Answer:

Let the number of stamps of 20 p and 25 p be x and y respectively.
As per the question
x+y=47                                                       .....i0.20x+0.25y=104x+5y=200                                                 .....ii
From (i), we get
y=47-x
Now, substituting y=47-x in (ii), we have
4x+547-x=2004x-5x+235=200x=235-200=35
Putting x = 35 in (i), we get
35+y=47y=47-35=12
Hence, the number of 20 p stamps and 25 p stamps are 35 and 12 respectively.

Page No 155:

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?

Answer:

Let the number of hens and cow be x and y respectively.
As per the question
x+y=48                                                       .....i2x+4y=140x+2y=70                                                     .....ii
Subtracting (i) from (ii), we have
y=22
Hence, the number of cows is 22.

Page No 155:

Question 13:

If 2x+3y=9xy and 4x+9y=21xy, find the values of x and y.

Answer:

The given pair of equation is
2x+3y=9xy                                               .....i4x+9y=21xy                                               .....ii
Multiplying (i) and (ii) by xy, we have
3x+2y=9                                                    .....iii9x+4y=21                                                  .....iv
Now, multiplying (iii) by 2 and subtracting from (iv), we get
9x-6x=21-18x=33=1
Putting x = 1 in (iii), we have
3×1+2y=9y=9-32=3
Hence, x = 1 and y = 3.

Page No 155:

Question 14:

If x4+y3=512 and x2+y=1, then find the value of (x + y).

Answer:

The given pair of equations is
x4+y3=512                                               .....ix2+y=1                                                       .....ii
Multiplying (i) by 12 and (ii) by 4, we get
3x+4y=5                                                    .....iii2x+4y=4                                                       .....iv
Now, subtracting (iv) from (iii), we get
x=1
Putting x = 1 in (iv), we have
2+4y=44y=2y=12
x+y=1+12=32
Hence, the value of x + y is 32.

Page No 155:

Question 15:

If 12x+17y=53 and 17x+12y=63, then find the value of (x + y).

Answer:

The given pair of equations is
12x+17y=53                                               .....i17x+12y=63                                               .....ii
Adding (i) and (ii), we get
29x+29y=116x+y=4         Dividing by 4
Hence, the value of x + y is 4.



Page No 156:

Question 16:

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

Answer:

The given system is
3x+5y=0                                                     .....ikx+10y=0                                                   .....ii
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
a1a2=b1b23k=510=12k=6
Hence, k = 6.

Page No 156:

Question 17:

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

Answer:

The given system is
kx-y-2=0                                                  .....i6x-2y-3=0                                                .....ii
Here, a1=k, b1=-1, c1=-2. a2=6, b2=-2 and c2=-3.
For the system, to have a unique solution, we must have
a1a2b1b2k6-1-2=12k3
Hence, k3.

Page No 156:

Question 18:

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

Answer:

The given system is
2x+3y-5=0                                                .....i4x+ky-10=0                                              .....ii
Here, a1=2, b1=3, c1=-5. a2=4, b2=k and c2=-10.
For the system, to have an infinite number of solutions, we must have
a1a2=b1b2=c1c224=3k=-5-1012=3k=12k=6
Hence, k = 6.

Page No 156:

Question 19:

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

Answer:

The given system is
2x+3y-1=0                                                .....i4x+6y-4=0                                                .....ii
Here, a1=2, b1=3, c1=-1. a2=4, b2=6 and c2=-4.
Now,
a1a2=24=12b1b2=36=12c1c2=-1-4=14
Thus, a1a2=b1b2c1c2 and therefor the given system has no solution.

Page No 156:

Question 20:

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

Answer:

The given system is
x+2y-3=0                                                 .....i5x+ky+7=0                                               .....ii
Here, a1=1, b1=2, c1=-3, a2=5, b2=k and c2=7.
For the system to be inconsistent, we must have
a1a2=b1b2c1c215=2k-3715=2k k=10 
Hence, k = 10.

Page No 156:

Question 21:

Solve: 3x+y+2x-y=2 and 9x+y-4x-y=1.

Answer:

The given system of equations is
3x+y+2x-y=2                                       .....i9x+y-4x-y=1                                       .....ii
Substituting 1x+y=u and 1x-y=v in (i) and (ii), the given equations are changed to
3u+2v=2                                              .....iii9u-4v=1                                              .....iv
Multiplying (i) by 2 and adding it with (ii), we get
15u=4+1u=13
Multiplying (i) by 3 and subtracting (ii) from it, we get
6v+4v=6-1u=510=12
Therefore
x+y=3                                                          .....vx-y=2                                                          .....vi
Now, adding (v) and (vi) we have
2x=5x=52
Substituting x=52 in (v), we have
52+y=3y=3-52=12
Hence, x=52 and y=12.



Page No 158:

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

Answer:

The given system of equations is
2x+3y=12                                                  .....i3x-2y=5                                                    .....ii
Multiplying (i) by 2 and (ii) by 3 and then adding, we get
4x+9x=24+15x=3913=3
Now, putting x = 3 in (i), we have
2×3+3y=12y=12-63=2
Thus, x = 3 and y = 2.
Hence, the correct answer is option (c).

Page No 158:

Question 2:

If x-y=2 and 2x+y=15 then
(a) x = 4, y = 2          (b) x = 5, y = 3          (c) x = 6, y = 4          (d) x = 7, y = 5

Answer:

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

Page No 158:

Question 3:

If 2x3-y2+16=0 and x2+2y3=3 then
(a) x = 2, y = 3          (b) x = − 2, y = 3          (c) x = 2, y = − 3          (d) x = − 2, y = − 3

Answer:

The given system of equations is
2x3-y2=-16                                          .....ix2+2y3=3                                                 .....ii
Multiplying (i) and (ii) by 6, we get
4x-3y=-1                                                 .....iii3x+4y=18                                                   .....iv
Multiplying (iii) by 4 and (iv) by 3 and adding, we get
16x+9x=-4+54x=5025=2
Now, putting x = 2 in (iv), we have
3×2+4y=18y=18-64=3
Thus, x = 2 and y = 3.
Hence, the correct answer is option (a).

Page No 158:

Question 4:

If 1x+2y=4 and 3y-1x=11 then
(a) x = 2, y = 3          (b) x = − 2, y = 3          (c) x=-12,y=3          (d)  x=-12,y=13

Answer:

The given system of equations is
1x+2y=4                                                   .....i3y-1x=11                                                 .....ii
Adding (i) and (ii), we get
2y+3y=155y=15y=515=13
Now, putting y=13 in (i), we have
1x+2×3=41x=4-6x=-12
Thus, x=-12,y=13.
Hence, the correct answer is option (d).

Page No 158:

Question 5:

If 2x+y+25=3x-y+13=3x+2y+16 then
(a) x = 1, y = 1          (b) x = − 1, y = − 1          (c) x = 1, y = 2          (d) x = 2, y = 1

Answer:

Consider 2x+y+25=3x-y+13 and 3x-y+13=3x+2y+16. Now, simplifying these equations, we get
32x+y+2=53x-y+16x+3y+6=15x-5y+59x-8y=1                                                .....i
And
63x-y+1=33x+2y+118x-6y+6=9x+6y+33x-4y=-1                                              .....ii
Multiplying (ii) by 2 and subtracting it from (i)
9x-6x=1+2x=1
Now, putting x = 1 in (ii), we have
3×1-4y=-1y=3+14=1
Thus, x = 1, y = 1.
Hence, the correct answer is option (a).

Page No 158:

Question 6:

If 3x+y+2x-y=2 and 9x+y-4x-y=1 then
(a) x=12, y=32          (b) x=52, y=12          (c) x=32, y=12          (d) x=12, y=52

Answer:

The given equations are
3x+y+2x-y=2                                        .....i9x+y-4x-y=1                                        .....ii
Substituting 1x+y=u and 1x-y=v in (i) and (ii), the new system becomes
3u+2v=2                                                    .....iii9u-4v=1                                                    .....iv
Now, multiplying (iii) by 2 and adding it with (iv), we get
6u+9u=4+1u=515=13
Again, multiplying (iii) by 3 and subtracting (iv) from it, we get
6v+4v=6-1v=510=12 
Therefore
x+y=3                                                          .....vx-y=2                                                          .....vi
Adding (v) and (vi), we get
2x=3+2x=52
Substituting x=52, in (v), we have
52+y=3y=3-52=12
Thus, x=52 and y=12.
Hence, the correct answer is option (b).

Page No 158:

Question 7:

If 4x+6y=3xy and 8x+9y=5xy then
(a) x = 2, y = 3          (b) x = 1, y = 2          (c) x = 3, y = 4          (d) x = 1, y = − 1

Answer:

The given equations are
4x+6y=3xy                                                 .....i8x+9y=5xy                                                 .....ii
Dividing (i) and (ii) by xy, we get
6x+4y=3                                                   .....iii9x+8y=5                                                   .....iv
Multiplying (iii) by 2 and subtracting (iv) from it, we get
12x-9x=6-53x=1x=3
Substituting x = 3 in (iii), we get
63+4y=34y=1y=4 
Thus, x = 3 and y = 4.
Hence, the correct answer is option (c).

Page No 158:

Question 8:

If 29x+37y=103 and 37x+29y=95 then
(a) x = 1, y = 2          (b) x = 2, y = 1          (c) x = 3, y = 2          (d) x = 2, y = 3

Answer:

The given equations are
29x+37y=103                                                 .....i37x+29y=95                                                   .....ii
Adding (i) and (ii), we get
66x+66y=198x+y=3                                                    .....iii
Subtracting (i) from (ii), we get
8x-8y=-8x-y=-1                                                 .....iv
Adding (iii) and (iv), we get
2x=2x=1 
Substituting x = 1 in (iii), we have
1+y=3y=2
Thus, x = 1 and y = 2.
Hence, the correct answer is option (a).

Page No 158:

Question 9:

If 2x+y=2x-y=8 then the value of y is
(a) 12            (b) 32            (c) 0            (d) none of these

Answer:

2x+y=2x-y=8x+y=x-yy=0
Hence, the correct answer is option (c).



Page No 159:

Question 10:

If 2x+3y=6 and 1x+12y=2 then
(a) x=1, y=23            (b) x=23, y=1            (c) x=1, y=32            (d) x=32, y=1

Answer:

The given equations are
2x+3y=6                                                   .....i1x+12y=2                                                 .....ii
Multiplying (ii) by 2 and subtracting it from (ii), we get
3y-1y=6-42y=2y=1
Substituting y = 1 in (ii), we get
1x+12=21x=2-12=32x=23
Hence, the correct answer is option (b).

Page No 159:

Question 11:

The system kx-y=2 and 6x-2y=3 has a unique solution only when
(a) k = 0            (b) k0            (c) k = 3            (d) k3

Answer:

The given equations are
kx-y-2=0                                                 .....i6x-2y-3=0                                               .....ii
Here, a1=k, b1=-1, c1=-2, a2=6, b2=-2 and c2=-3.
For the given system to have a unique solution, we must have
a1a2b1b2k6-1-2k3
Hence, the correct answer is option (d).

Page No 159:

Question 12:

The system x-2y=3 and 3x+ky=1 has a unique solution only when

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

Answer:

The correct option is (b).

The given system of equations can be written as follows:
x − 2y − 3 = 0 and 3x + ky − 1 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −2, c1 = −3 and a2 = 3, b2 = k and c2 = −1
a1a2=13,b1b2=-2k and c1c2=-3-1=3
These graph lines will intersect at a unique point when we have:
 a1a2b1b213-2kk-6
Hence, k has all real values other than −6.

Page No 159:

Question 13:

The system x+2y=3 and 5x+ky+7=0 has no solution, when

(a) k = 10
(b) k10
(c) k=-73
(d) k = −21

Answer:

The correct option is (a).

The given system of equations can be written as follows:
x + 2y − 3 = 0 and 5x + ky + 7 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 5, b2 = k and c2 = 7
a1a2=15,b1b2=2kandc1c2=-37
For the system of equations to have no solution, we must have:
a1a2=b1b2c1c2
15=2k-37k=10

Page No 159:

Question 14:

If the lines given by 3x+2ky=2 and 2x+5y+1=0 are parallel, then the value of k is

(a) -54
(b) 25
(c) 32
(d) 154

Answer:

The correct option is (d).

The given system of equations can be written as follows:
3x + 2ky − 2 = 0 and 2x + 5y + 1 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 3, b1 = 2k, c1 = −2 and a2 = 2, b2 = 5 and c2 = 1
a1a2=32,b1b2=2k5 and c1c2=-21
For parallel lines, we have:
a1a2=b1b2c1c2
32=2k5-21
k=154

Page No 159:

Question 15:

For what value of k do the equations kx-2y=3 and 3x+y=5 represent two lines intersecting at a unique point?

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

Answer:

The correct option is (d).

The given system of equations can be written as follows:
kx − 2y − 3 = 0 and 3x + y − 5 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = −2, c1 = −3 and a2 = 3, b2 = 1 and c2 = −5
a1a2=k3,b1b2=-21andc1c2=-3-5=35
Thus, for these graph lines to intersect at a unique point, we must have:
a1a2b1b2
k3-21k-6
Hence, the graph lines will intersect at all real values of k except −6.

Page No 159:

Question 16:

The pair of equations x+2y+5=0 and -3x-6y+1=0 has

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

Answer:

The correct option is (d).

The given system of equations can be written as:
x + 2y + 5 = 0 and −3x − 6y + 1 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = 5 and a2 = −3, b2 = −6 and c2 = 1
a1a2=1-3,b1b2=2-6=1-3andc1c2=51
∴ a1a2=b1b2c1c2 
Hence, the given system has no solution.

Page No 159:

Question 17:

The pair of equations 2x+3y=5 and 4x+6y=15 has

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

Answer:

The correct option is (d).

The given system of equations can be written as follows:
2x + 3y − 5 = 0 and 4x + 6y − 15 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = −5 and a2 = 4, b2 = 6 and c2 = −15
a1a2=24=12,b1b2=36=12andc1c2=-5-15=13
∴ ​a1a2=b1b2c1c2 
Hence, the given system has no solution.

Page No 159:

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

Answer:

The correct option is (d).

If a pair of linear equations is consistent, then the two graph lines either intersect at a point or coincide.

Page No 159:

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

Answer:

The correct option is (a).

If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since there is no common solution, their graph lines do not intersect. Hence, they are parallel.

Page No 159:

Question 20:

In a ABC, C=3B=2(A+B), then B=?
(a) 20°
(b) 40°
(c) 60°
(d) 80°

Answer:

The correct option is (b).

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

Page No 159:

Question 21:

In a cyclic quadrilateral ABCD, it is being given that A=(x+y+10)°,B=(y+20)°,C=(x+y-30)° and D=(x+y)°. Then,B=?
(a) 70°
(b) 80°
(c) 100°
(d) 110°

Answer:

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



Page No 160:

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

Answer:

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.

Page No 160:

Question 23:

In a given fraction, if 1 is subtracted from the numerator and 2 is added to the denominator, it becomes 12. If 7 is subtracted from the numerator and 2 is subtracted from the denominator, it becomes 13. The fraction is
(a) 1324

(b) 1526

(c) 1627

(d) 1621

Answer:

Let the fraction be xy
It is given that x-1y+2=12
2x-2=y+22x-y=4          .....(i)
Also, x-7y-2=13
3x-21=y-23x-y=19          .....(ii)
Subtract (ii) from (i), we get
x=15
Put the value of x in equation (i), we get;
2(15)-y=430-y=4y=30-4=26
Therefore, the fraction is xy=1526
Hence, the correct answer is option (b)

Page No 160:

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

Answer:

The correct option is (d).
Let the man's present age be x years.
Let his son's present age be y years.
Five years later:
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x − 3y = 10            ....(i)
Five years ago:
(x − 5) = 7(y − 5)
x − 5 = 7y − 35
x − 7y = −30           ....(ii)
On subtracting (i) from (ii), we get:
−4y = −40 ⇒ y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 10 ⇒ x − 30 = 10 ⇒ x = (10 + 30) = 40 years
Hence, the man's present age is 40 years.

Page No 160:

Question 25:

The graphs of the equations 6x-2y+9=0 and 3x-y+12=0 are two lines which are

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

Answer:

The correct option is (b).

The given equations are as follows:
 6x-2y+9=0 and 3x-y+12=0
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
a1a2=63=21,b1b2=-2-1=21 and c1c2=912=34
a1a2=b1b2c1c2
The given system has no solution.
Hence, the lines are parallel.

Page No 160:

Question 26:

The graphs of the equations 2x+3y-2=0 and x-2y-8=0 are two which are

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

Answer:

The correct option is (c).

The given equations are as follows:
 2x+3y-2=0 and x-2y-8=0
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
a1a2=21,b1b2=3-2andc1c2=-2-8=14
a1a2b1b2
The given system has a unique solution.
Hence, the lines intersect exactly at one point.

Page No 160:

Question 27:

The graphs of the equations 5x-15y=8 and 3x-9y=245 are two which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

Answer:

The correct option is (a).

The given system of equations can be written as follows:
 5x-15y-8=0 and 3x-9y-245=0
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 = -245
a1a2=53,b1b2=-15-9=53 and c1c2=-8×5-24=53
a1a2=b1b2=c1c2
The given system of equations will have an infinite number of solutions.
Hence, the lines are coincident.



Page No 163:

Question 1:

The graphic representation of the equations x+2y=3 and 2x+4y+7=0 gives a pair of
(a) parallel lines
(b) intersecting lines
(c) coincident lines
(d) none of these

Answer:

The correct option is (a).

The given system of equations can be written as follows:
x + 2y − 3 = 0 and 2x + 4y + 7 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 2, b2 = 4 and c2 = 7
a1a2=12,b1b2=24=12andc1c2=-37
a1a2=b1b2c1c2 
So, the given system has no solution.
Hence, the lines are parallel.

Page No 163:

Question 2:

If 2x-3y=7 and (a+b)x-(a+b-3)y=4a+b 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

Answer:

The correct option is (d).

The given system of equations can be written as follows:
2x − 3y − 7 = 0 and (a + b)x − (a + b − 3)y − (4a + b) = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = −3, c1 = −7 and a2 = (a + b), b2 =  −(a + b − 3) and c2 = −(4a + b)
a1a2=2a+b,b1b2=-3-a+b-3=3a+b-3 and c1c2=-7-4a+b=74a+b
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
2a+b=3a+b-3=74a+b
Now, we have:
2a+b=3a+b-32a+2b-6=3a+3b
a + b + 6 = 0                                  ...(i)
Again, we have:
3a+b-3=74a+b12a+3b=7a+7b-21
⇒ 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

Page No 163:

Question 3:

The pair of equations 2x+y=5, 3x+2y=8 has
(a) a unique solution
(b) two solutions
(c) no solution
(d) infinitely many solutions

Answer:

The correct option is (a).

The given system of equations can be written as follows:
2x + y − 5 = 0 and 3x + 2y − 8 = 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 1, c1 = −5 and a2 = 3, b2 = 2 and c2 = −8
a1a2=23,b1b2=12 and c1c2=-5-8=58
a1a2b1b2
The given system has a unique solution.
Hence, the lines intersect at one point.

Page No 163:

Question 4:

If x = −y and y > 0, which of the following is wrong?
(a) x2y>0
(b) x+y=0
(c) xy<0
(d) 1x-1y=0

Answer:

The correct option is (d).

Given:
x = −y and y > 0
Now, we have:
(i) x2y 
On substituting x = −y, we get:
(−y)2y = y3 > 0 (∵ 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) 1x-1y=0
y-xxy=0
On substituting x = −y, we get:
y--y-yy=02y-y2=02y=0y=0
Hence, from the above equation, we get y = 0, which is wrong.

Page No 163:

Question 5:

Show that the system of equations -x+2y+2=0 and 12x-14y-1=0 has a unique solution.

Answer:

The given system of equations:
 -x+2y+2=0 and 12x-14y-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 = 2 and a2 = 12, b2-14 and c2 = −1
a1a2=-112=-2,b1b2=2-14=-8 and c1c2=2-1=-2
a1a2b1b2
The given system has a unique solution.
Hence, the lines intersect at one point.

Page No 163:

Question 6:

For what values of k is the system of equations kx+3y=k-2, 12x+ky=k inconsistent?

Answer:

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
a1a2=k12,b1b2=3k andc1c2=-k-2-k=k-2k
For inconsistency, we must have:
a1a2=b1b2c1c2
k12=3kk-2kk2=3×12=36
k=36=±6
Hence, the pair of equations is inconsistent if k=±6.

Page No 163:

Question 7:

Show that the equations 9x-10y=21,3x2-5y3=72 have infinitely many solutions.

Answer:

The given system of equations can be written as follows:
 9x-10y-21=0 and 3x2-5y3-72=0
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 = 32, b2-53 and c2 = -72
a1a2=932=6, b1b2=-10-53=6 and c1c2=-21×2-7=6
a1a2=b1b2=c1c2
This shows that the given system equations has an infinite number of solutions.

Page No 163:

Question 8:

Solve the system of equations: x-2y=0, 3x+4y=20

Answer:

The given equations are as follows:
x − 2y = 0                           ....(i)
3x + 4y = 20                       ....(ii)
On multiplying (i) by 2, we get:
2x − 4y = 0                          ....(iii)
On adding (ii) and (iii), we get:
5x = 20 ⇒ x = 4
On substituting x = 4 in (i), we get:
4 − 2y = 0 ⇒ 4 = 2yy = 2
Hence, the required solution is x = 4 and y = 2.

Page No 163:

Question 9:

Show that the paths represented by the equations x-3y=2 and -2x+6y=5 are parallel.

Answer:

The given system of equations can be written as follows:
x − 3y − 2 = 0 and −2x + 6y − 5= 0
The given equations are of the following form:
 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −3, c1 = −2 and a2 = −2, b2 = 6 and c2 = −5
a1a2=1-2=-12,b1b2=-36=-12andc1c2=-2-5=25
∴ ​a1a2=b1b2c1c2 
Thus, the given system of equations has no solution.
Hence, the paths represented by the equations are parallel.

Page No 163:

Question 10:

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

Answer:

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.

Page No 163:

Question 11:

Solve: 23x+29y=98, 29x+23y=110

Answer:

The given equations are as follows:
23x + 29y = 98                              ....(i)
29x + 23y = 110                            ....(ii)
On adding (i) and (ii), we get:
52x + 52y = 208
x + y = 4                                   ....(iii)
On subtracting (i) from (ii), we get:
6x − 6y = 12
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.

Page No 163:

Question 12:

Solve: 6x+3y=7xy and 3x+9y=11xy

Answer:

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

For equation (i), we have:

6x+3yxy=7
6xxy+3yxy=76y+3x=7        ....(iii)

For equation (ii), we have:

3x+9yxy=11
3xxy+9yxy=113y+9x=11      ....(iv)
On substituting 1y=v and 1x=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 = 1015=23
1y=23y=32
On substituting y=32 in (iii), we get:
632+3x=7
4+3x=73x=33x=3
x=1
Hence, the required solution is x = 1 and y=32.

Page No 163:

Question 13:

Find the value of k for which the system of equations 3x+y=1 and kx+2y=5 has (i) a unique solution, (ii) no solution.

Answer:

The given system of equations:
3x + y = 1
⇒ 3x + y − 1= 0                       ....(i)
kx +  2y = 5
kx +  2y − 5 = 0                   ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 3, b1= 1, c1 = −1 and a2 = k, b2 = 2, c2 = −5

(i) For a unique solution, we must have:
a1a2b1b2 i.e. 3k12k6
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:
a1a2=b1b2c1c2
3k=12-1-5
3k=12 and 3k-1-5
k=6, k15
Thus, for k = 6, the given system of equations will have no solution.

Page No 163:

Question 14:

In ABC,C=3B=2(A+B), find the measure of each one of A,B and C.

Answer:

Let A=x° and B=y°
Then, C=3B=3y°
Now, we have:
A+B+C=180°
x + y + 3y = 180
x + 4y = 180              ....(i)
Also, C=2A+B
⇒ 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
∴ ​A=20°, B=40°, C=3×40°=120°



Page No 164:

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.

Answer:

Let the cost of each pencil be Rs. x and that of each pen be Rs. y.
Then, we have:
5x + 7y = 195                           ....(i)
7x + 5y = 153                           ....(ii)
Adding (i) and (ii), we get:
12x + 12y = 348
⇒ 12(x + y) = 348
x + y = 29                             ....(iii)
Subtracting (i) from (ii), we get:
2x − 2y = −42
⇒ 2(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.

Page No 164:

Question 16:

Solve the following system of equations graphically:
2x-3y=1, 4x-3y+1=0

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and the y-axis, respectively.

Graph of 2x − 3y = 1
2x − 3y = 1
⇒ 3y = (2x − 1)
y=2x-13                           ...(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=4x+13                           ...(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.

Page No 164:

Question 17:

Find the angles of a cyclic quadrilateral ABCD in which A=(4x+20)°,B=(3x-5)°,C=(4y)° and D=(7y+5)°.

Answer:

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

Page No 164:

Question 18:

Solve for x and y: 35x+y+14x-y=19, 14x+y+35x-y=37

Answer:

We have:
35x+y+14x-y=19 and 14x+y+35x-y=37
Taking 1x+y=u and 1x-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:


u14×-37- 35×-19=v-19×14--37×35=135×35-14×14
u-518+665=v-266+1295=11225-196
u147=v1029=11029
u=1471029=17,v=10291029=1
1x+y=17,1x-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:

x1×-1--1×-7=y-7×1--1×1=11×-1-1×1

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

Page No 164:

Question 19:

If 1 is added to both of the numerator and denominator of a fraction, it becomes 45. If however, 5 is subtracted from both numerator and denominator, the fraction 12. Find the fraction.

Answer:

Let the required fraction be xy.
Then, we have:
x+1y+1=45
⇒ 5(x + 1) = 4(y + 1)
⇒ 5x + 5 = 4y + 4
⇒ 5x − 4y = −1                       ....(i)

Again, we have:
x-5y-5=12
⇒ 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 79.

Page No 164:

Question 20:

Solve: axb-bya=a+b, ax-by=2ab

Answer:

The given equations may be written as follows:
axb-bya-(a+b)=0                       ....(i)
ax-by-2ab=0                               ....(ii)
Here, a1 = ab, b1 = -ba, c1 = −(a + b), a2 = a, b2 = −b, c2 = −2ab
By cross multiplication, we have:

x-ba×-2ab--b×-a+b=y-a+b×a--2ab×ab=1ab×-b-a×-ba
x2b2-ba+b=y-aa+b+2a2=1-a+b
x2b2-ab-b2=y-a2-ab+2a2=1-a+b
xb2-ab=ya2-ab=1-a-b
x-ba-b=yaa-b=1-a-b
x=-ba-b-a-b=b, y=aa-b-a-b=-a
Hence, x = b and y = −a is the required solution.



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