Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 3 Linear Equations In Two Variables are provided here with simple step-by-step explanations. These solutions for Linear Equations In Two Variables are extremely popular among Class 10 students for Maths Linear Equations In Two Variables Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 93:

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 93:

Answer:

Given:3x+2y=12    ...1x-y+1=0    ...2x-y+1=0y=x+1     ...3Substituting the value of y in 1, we get3x+2x+1=123x+2x+2=125x=12-25x=10x=2        ...4Substituting the value of x in 3, we gety=2+1=3Hence, x=2 and y=3.

Page No 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 93:

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 94:

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 94:

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 94:

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 94:

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 94:

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 94:

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 94:

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 94:

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 109:

Answer:

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

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

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

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

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

Page No 109:

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 110:

Answer:

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

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

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

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

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

Page No 110:

Answer:

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

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

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

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

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

Page No 110:

Answer:

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

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

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

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

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

Page No 110:

Answer:

Given:3x-5y=4    ...12y+7=9x    ...22y+7=9x2y=9x-7y=9x-72     ...3Substituting the value of y in 1, we get3x-59x-72=46x-59x-72=46x-45x+35=8-39x=8-35-39x=-27x=-27-39x=913        ...4Substituting the value of x in 3, we gety=9913-72y=81-9126y=-1026y=-513Hence, x=913 and y=-513.

Page No 110:

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 110:

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 110:

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 110:

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 110:

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 110:

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 110:

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 110:

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 110:

Answer:

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

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

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

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

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

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

Page No 110:

Answer:

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

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

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

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

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

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

Page No 110:

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 110:

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 110:

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 110:

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 110:

Answer:

Given:3x+8y=-11x-2y=2Let 1x=u and 1y=v     ...1Therefore, the equations become3u+8v=-1     ...2u-2v=2           ...3u-2v=2u=2+2v        ...4Substituting the value of u in 2, we get32+2v+8v=-16+6v+8v=-114v=-1-614v=-7v=-714v=-12         ...5y=1v=1-12=-2       1y=vSubstituting the value of v in 4, we getu=2+2-12u=2-1u=1x=1u=11=1       1x=uHence, x=1 and y=-2.

Page No 110:

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 110:

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 110:

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 110:

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 110:

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 110:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 111:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 117:

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 128:

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 128:

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.
 

Page No 128:

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.

Page No 128:

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 128:

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 128:

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 128:

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 129:

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.

Page No 129:

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 129:

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.

Page No 129:

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 129:

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 129:

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.

Page No 129:

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 129:

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.

Page No 129:

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.

Page No 129:

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 129:

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 129:

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 129:

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 129:

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 130:

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 130:

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 130:

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 130:

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 130:

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 130:

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 130:

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 130:

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 130:

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 130:

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 151:

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 151:

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 152:

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 152:

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 152:

Answer:

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

Page No 152:

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 152:

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 152:

Answer:

We know:
Dividend = Divisor × Quotient + Remainder

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

Page No 152:

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 152:

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 152:

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 152:

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 152:

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 152:

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 152:

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 153:

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 153:

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 153:

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 153:

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 153:

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 153:

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 153:

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 153:

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 153:

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 153:

Answer:

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

Page No 153:

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 154:

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 154:

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 154:

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 154:

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 154:

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 154:

Answer:

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

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

Page No 154:

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 154:

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 154:

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 155:

Answer:

Let speed of boat in still water be x km/h and speed of stream be y km/h.
Speed Upstream = (x − y) km/h
Speed downstream = (xy) km/h

According to the question,
30x-y+44x+y=10and40x-y+55x+y=13Let 1x-y=p and 1x+y=q     ...1Therefore, the equation becomes30p+44q=10     ...240p+55q=13     ...340p+55q=1340p=13-55qp=13-55q40     ...4Substituting the value of p in (2), we get3013-55q40+44q=103013-55q40+44q=10313-55q+444q4=1039-165q+176q=4011q=40-3911q=1q=111     ...5Substituting the value of q in (4), we getp=13-5511140p=13-540p=840p=15     ...6From 1, 5 and 6, we getx-y=5 and x+y=11Solving both, we getx=8 and y=3

Hence, the speed of the stream and that of the boat in still water is 3 km/h and 8 km/h, respectively.

Page No 155:

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 155:

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 155:

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 155:

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 155:

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 155:

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 155:

Answer:

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

Page No 155:

Answer:

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

Page No 155:

Answer:

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

Page No 155:

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 156:

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 156:

Answer:

Let the digit at the tens place be x and digit at the units place be y.
The number is 10x + y.

According to the question,
x+y=9     ...110x+y+27=10y+x     ...210x+y+27=10y+x10x+y-10y-x=-279x-9y=-279x-y=-27x-y=-3x=-3+y     ...3Substituting the value of x in (1), we get-3+y+y=92y=9+32y=12y=6     ...4Substituting the value of y in (3), we getx=-3+6x=3Hence, the number is 36.

 

Page No 156:

Answer:

Let the numerator of the fraction be x and the denominator be y.

According to the question,

x-2y=13   ...(i) and xy-1=12     ...(ii)Solving equation i, we get3x-2=y3x-6=y      ...iiiSubstituting the value of y in equation ii, we getx3x-6-1=122x=3x-6-12x=3x-72x-3x=-7-x=-7x=7      ...(iv)From (iii) and (iv), we getx=7 and y=15Hence, the fraction is 715.

Page No 156:

Answer:

Let the age of first son be x and of second son be y.
Then, the father's present age be 3(x + y).

According to the question,
After 5 years, the father's age will be two times the sum of the ages of his two children.

3x+y+5=2x+5+y+53x+3y+5=2x+10+2y+103x+3y-2x-2y=20-5x+y=153x+y=45

Hence, the present age of the father is 45 years.
 

Page No 156:

Answer:

Let the length of the side of one square be x m and the length of the side of another square be m.

According to the question,
x2+y2=640     ...14x-4y=64      ...24x-4y=644x-y=64x-y=16x=16+y      ...3Substituting the value of x in (1), we get16+y2+y2=640256+y2+32y+y2=6402y2+32y+256-640=02y2+32y-384=02y2+16y-192=0y2+16y-192=0y2+24y-8y-192=0yy+24-8y+24=0y-8y+24=0y=8     ...4      y>0Substituting the value of y in (3), we getx=16+8x=24Hence, the sides of the squares are 24 m and 8 m.

 

Page No 156:

Answer:

Let the length of the side of one square be x m and the length of the side of another square be m.

According to the question,
x2+y2=157     ...14x+4y=68      ...24x+4y=684x+y=68x+y=17x=17-y      ...3Substituting the value of x in (1), we get17-y2+y2=157289+y2-34y+y2=1572y2-34y+289-157=02y2-34y+132=02y2-17y+66=0y2-17y+66=0y2-11y-6y+66=0yy-11-6y-11=0y-6y-11=0y=6, 11     ...4      y>0Substituting the value of y in (3), we getx=17-6 and x=17-11x=11, 6Hence, the sides of the squares are 11 m and 6 m.
 

Page No 156:

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 156:

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 156:

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 156:

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 156:

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 156:

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 161:

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 161:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 162:

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 164:

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:

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.



Page No 165:

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 165:

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 165:

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 165:

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 165:

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 165:

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 165:

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 165:

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 165:

Answer:

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

Page No 165:

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 165:

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 165:

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 165:

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 166:

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 166:

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 166:

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 166:

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 166:

Answer:

The correct option is (d).

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

Page No 166:

Answer:

The correct option is (a).

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

Page No 166:

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 166:

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 166:

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 166:

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 166:

Answer:

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



Page No 167:

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 167:

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 167:

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 169:

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 169:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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 170:

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.



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