Rs Aggarwal 2020 2021 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 2020 2021 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 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 93:

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

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

 x 1 −2 4 y 0 2 −2

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

Graph of x − 2y = 8
x − 2y = 8
⇒ 2y = ( x − 8)
...(ii)
Putting x = 2, we get y = −3
Putting x = 4, we get y = −2
Putting x = 0, we get y = −4
Thus, we have the following table for the equation x - 2y = 8.
 x 2 4 0 y − 3 − 2 − 4
Now, plot the points P(0, − 4) and Q(2, − 3). The point C(4, −2) has already been plotted. Join PQ and QC and extend it on both ways.
Thus, line PC is the graph of x − 2y = 8.

The two graph lines intersect at C(4, −2).
x = 4 and y = −2 are the solutions of the given system of equations.

#### Page No 93:

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

2x + 3y = 8
⇒ 3y = (8 − 2x)
$y=\frac{8-2x}{3}$ ...........(i)
Putting x = 1, we get y = 2.
Putting x = −5, we get y = 6.
Putting x = 7, we get y = −2.
Thus, we have the following table for the equation 2x + 3y = 8.

 x 1 −5 7 y 2 6 −2

Now, plot the points A(1, 2), B(5, −6) and C(7, −2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.

Graph of x − 2y + 3 = 0
x − 2y + 3 = 0
⇒ 2y = (x + 3)
$y=\frac{x+3}{2}$ ..........(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = −3, we get y = 0.
Thus, we have the following table for the equation x − 2y + 3 = 0.
 x 1 3 −3 y 2 3 0
Now, plot the points P(3, 3) and Q(−3, 0). The point A(1, 2) has already been plotted. Join AP and QA and extend it on both ways.
Thus, PQ is the graph of x − 2y + 3 = 0.

The two graph lines intersect at A(1, 2).
x = 1 and y = 2

#### Page No 93:

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

2x − 5y + 4 = 0
⇒ 5y = (2x + 4)
$y=\frac{2x+4}{5}$ ...........(i)
Putting x = −2, we get y = 0.
Putting x = 3, we get y = 2.
Putting x = 8, we get y = 4.
Thus, we have the following table for the equation 2x − 5y + 4 = 0.

 x −2 3 8 y 0 2 4

Now, plot the points A(−2 , 0), B(3, 2) and C(8, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x − 5y + 4 = 0.

Graph of 2x + y − 8 = 0
2x + y − 8 = 0
y = (8 − 2x) ..........(ii)
Putting x = 1, we get y = 6.
Putting x = 3, we get y = 2.
Putting x = 2, we get y = 4.
Thus, we have the following table for the equation 2x + y − 8 = 0.
 x 1 3 2 y 6 2 4
Now, plot the points P(1, 6) and Q(2, 4). The point B(3, 2) has already been plotted. Join PQ and QB and extend it on both ways.
Thus, PB is the graph of 2x + y − 8 = 0.

The two graph lines intersect at B(3, 2).
x = 3 and y = 2

#### Page No 93:

The given equations are:

From (i), write y in terms of x

Now, substitute different values of x in (iii) to get different values of y
For = 0, $y=\frac{12-3x}{2}=\frac{12-0}{2}=6$
For = 2, $y=\frac{12-3x}{2}=\frac{12-6}{2}=3$
For = 4, $y=\frac{12-3x}{2}=\frac{12-12}{2}=0$
Thus, the table for the first equation (3x + 2y = 12) is

 x 0 2 4 y 6 3 0

Now, plot the points A(0, 6), B(2, 3) and C(4, 0) on a graph paper and join
A, B and C to get the graph of
3x + 2y = 12.
From (ii), write y in terms of x

Now, substitute different values of x in (iv) to get different values of y
For = 0, $y=\frac{5x-4}{2}=\frac{0-4}{2}=-2$
For = 2, $y=\frac{5x-4}{2}=\frac{10-4}{2}=3$
For = 4, $y=\frac{5x-4}{2}=\frac{20-4}{2}=8$
Thus, the table for the first equation (5x − 2y = 4) is

 x 0 2 4 y −2 3 8

Now, plot the points D(0, −2), E(2, 3) and F(4, 8) on the same graph paper and join
D, E and F to get the graph of
5x − 2y = 4.

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

#### Page No 93:

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

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

 x 0 −1 1 y −1 2 −4

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

Graph of 2x − 3y + 8 = 0
2x − 3y + 8 = 0
⇒ 3y = (2x + 8)
$y=\frac{2x+8}{3}$
Putting x = −1, we get y = 2.
Putting x = 2, we get y = 4.
Putting x = −4, we get y = 0.
Thus, we have the following table for the equation 2x − 3y + 8 = 0.
 x −1 2 −4 y 2 4 0
Now, plot the points P(2, 4), Q(−4, 0). The point B(−1 , 2) has already been plotted. Join PB and BQ and extend it on both ways.
Thus, PQ is the graph of 2x − 3y + 8 = 0.

The two graph lines intersect at B(−1, 2).
x = −1 and y = 2

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation ( 2x + 3y + 5 = 0 ) is

 x −1 2 5 y −1 −3 −5

Now, plot the points A(−1,−1), B(2,−3) and C(5,−5) on a graph paper and join
them to get the graph of 2x + 3y + 5 = 0.

From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation ( 3x  2y  12 = 0 ) is

 x 0 2 4 y −6 −3 0

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

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

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation ( 2x  3y + 13 = 0 ) is

 x −5 1 4 y 1 5 7

Now, plot the points A(−5,1), B(1,5) and C(4,7) on a graph paper and join
A, B and C to get the graph of
2x  3y + 13 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation ( 3x  2y + 12 = 0 ) is

 x −4 −2 0 y 0 3 6

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

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

#### Page No 93:

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

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

 x −1 2 5 y 2 0 −2

Now, plot the points A(−1 , 2), B(2 , 0) and C(5, −2) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of  2x + 3y = 4.

Graph of 3x  y = −5
3x y  = −5
y = (3x + 5)              ...(ii)
Putting x = −1, we get y = 2
Putting x = 0, we get y = 5
Putting x = −2, we get y = −1
Thus, we have the following table for the equation 3xy = − 5 = 0.
 x −1 0 −2 y 2 5 −1
Now, plot the points P(0, 5), Q(−2 , −1). The point A(−1 , 2) has already been plotted. Join PA and QA and extend it on both ways.
Thus, PQ is the graph of  3xy = −5.

The two graph lines intersect at A(−1 , 2).
x = −1 and y = 2 are the solutions of the given system of equations.

#### Page No 93:

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

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

 x −2 0 2 y 0 −1 −2

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

Graph of  3x + 2y − 2 = 0
3x + 2y − 2 = 0
⇒ 2y = (2 − 3x)
$y=\frac{2-3x}{2}$...............(ii)
Putting x = 0, we get y = 1.
Putting x = 2, we get y = −2.
Putting x = 4, we get y = −5.
Thus, we have the following table for the equation 3x + 2y − 2 = 0.
 x 0 2 4 y 1 −2 −5
Now, plot the points P(0, 1) and Q(4, −5). The point C(2, −2) has already been plotted. Join PC and QC and extend it on both ways.
Thus, PQ is the graph of 3x + 2y − 2 = 0.

The two graph lines intersect at A(2, −2).
x = 2 and y = −2

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (x  y + 3 = 0) is

 x −3 −1 1 y 0 2 4

Now, plot the points A(−3,0), B(1,2) and C(1,4) on a graph paper and join
A, B and C to get the graph of
x  y + 3 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation ( 2x + 3y  4 = 0 ) is

 x −4 −1 2 y 4 2 0

Now, plot the points D(−4,4), E(−1,2) and F(2,0) on the same graph paper and join
D, E and F to get the graph of 2x − 3y − 4 = 0.

From the graph it is clear that, the given lines intersect at (−1,2).
So, the solution of the given system of equation is (−1,2).
The vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2) and (2,0).
Now, draw a perpendicular from the intersection point E on the x-axis. So,

Hence, the vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2)
and
(2,0) and its area is 5 sq. units.

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (2x − 3y + 4 = 0) is

 x −2 1 4 y 0 2 4

Now, plot the points A(−2,0), B(1,2) and C(4,4) on a graph paper and join
A, B and C to get the graph of 2
x  3y + 4 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation ( x + 2y  5 = 0 ) is

 x −3 1 5 y 4 2 0

Now, plot the points D(−3,4), B(1,2) and F(5,0) on the same graph paper and join
D, E and F to get the graph of x + 2y − 5 = 0.

From the graph it is clear that, the given lines intersect at (1,2).
So, the solution of the given system of equations is (1,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (1,2) and (5,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,

Hence, the vertices of the triangle formed by the given lines and the x-axis are (−2,0), (1,2) and (5,0)
and the area of the triangle is
7 sq. units.

#### Page No 93:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 4x − 3y + 4 = 0
4x − 3y + 4 = 0
⇒ 3y = (4x + 4)
$y=\frac{4x+4}{3}$............(i)
Putting x = −1, we get y = 0.
Putting x = 2, we get y = 4.
Putting x = 5, we get y = 8.
Thus, we have the following table for the equation 4x − 3y + 4 = 0

 x −1 2 5 y 0 4 8

Now, plot the points A(−1, 0), B( 2, 4) and C(5, 8) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of  4x − 3y + 4 = 0.

Graph of 4x + 3y − 20 = 0
4x + 3y − 20 = 0
⇒ 3y = (−4x + 20)
$y=\frac{-4x+20}{3}$ ............(ii)
Putting x = 2, we get y = 4.
Putting x = −1, we get y = 8.
Putting x = 5, we get y = 0.
Thus, we have the following table for the equation 4x + 3y − 20 = 0.
 x 2 −1 5 y 4 8 0
Now, plot the points P(1, −8) and Q(5 , 0). The point B(2, 4) has already been plotted. Join PB and QB to get the graph line PQ. Extend it on both ways.
Then, line PQ is the graph of the equation 4x + 3y − 20 = 0.

The two graph lines intersect at B(2, 4).
∴ The solution of the given system of equations is x = 2 and y = 4.
Clearly, the vertices of ΔABQ formed by these two lines and x-axis are Q(5, 0), B(2, 4) and A(−1, 0).
Now, consider ΔABQ.
Here, height = 4 units and base (AQ) = 6 units
∴ Area of ΔABQ = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×6×4\right)=12$ sq. units.

#### Page No 93:

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

 x −1 1 2 y 0 2 3

Now, plot the points A(-1, 0), B( 1, 2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of  xy + 1 = 0.

Graph of 3x + 2y − 12 = 0
3x + 2y − 12 = 0
⇒ 2y = (−3x + 12)
$y=\frac{-3x+12}{2}$  ............(ii)
Putting x = 0, we get y = 6.
Putting x = 2, we get y = 3.
Putting x = 4, we get y = 0.
Thus, we have the following table for the equation 3x + 2y − 12 = 0.
 x 0 2 4 y 6 3 0
Now, plot the points P(0, 6) and Q(4 , 0). The point C(2, 3) has already been plotted. Join PC and CQ to get the graph line PQ. Extend it on both ways.
Then, PQ is the graph of the equation 3x + 2y − 12 = 0.

The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔACQ formed by these two lines and the x-axis are Q(4, 0), C(2, 3) and A(−1, 0).
Now, consider ΔACQ.
Here, height = 3 units and base (AQ) = 5 units
Area of ΔACQ = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×5×3\right)=7.5$ sq. units.

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (x − 2y + 2 = 0) is

 x −2 2 4 y 0 2 3

Now, plot the points A(−2,0), B(2,2) and C(4,3) on a graph paper and join
A, B and C to get the graph of
x  2y + 2 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (2x + y  6 = 0 ) is

 x 1 3 4 y 4 0 −2

Now, plot the points D(1,4), E(3,0) and F(4,−2) on the same graph paper and join
D, E and F to get the graph of 2x + y − 6 = 0.

From the graph it is clear that, the given lines intersect at (2,2).
So, the solution of the given system of equations is (2,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,

Hence, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0) and the area of the triangle is 5 sq. units.

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (2x − 3y + 6 = 0) is

 x −3 0 3 y 0 2 4

Now, plot the points A(−3,0), B(0,2) and C(3,4) on a graph paper and join
A, B and C to get the graph of 2
x  3y + 6 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (2x + 3y  18 = 0 ) is

 x 0 3 9 y 6 4 0

Now, plot the points D(0,6), E(3,4) and F(9,0) on the same graph paper and join
D, E and F to get the graph of 2x + 3y − 18 = 0.

From the graph it is clear that, the given lines intersect at (3,4).
So, the solution of the given system of equations is (3,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4).
Now, draw a perpendicular from the intersection point (or C) on the y-axis. So,

Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4) and its area is 6 sq. units.

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (4x y  4 = 0) is

 x 0 1 2 y −4 0 4

Now, plot the points A(0,−4), B(1,0) and C(2,4) on a graph paper and join
A, B and C to get the graph of 4
x  y  4 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (3x + 2y  14 = 0 ) is

 x 0 4 $\frac{14}{3}$ y 7 1 0

Now, plot the points D(0,7), E(4,1) and $F\left(\frac{14}{3},0\right)$ on the same graph paper and join
D, E and F to get the graph of 3x + 2y − 14 = 0.

From the graph it is clear that, the given lines intersect at (,4).
So, the solution of the given system of equations is (2,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4).
Now, draw a perpendicular from the intersection point  on the y-axis. So,

Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4) and its area is 11 sq. units.

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (x − y  5 = 0) is

 x 0 2 5 y −5 −3 0

Now, plot the points A(0,−5), B(2,−3) and C(5,0) on a graph paper and join
A, B and C to get the graph of
x  y  5 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (3x + 5y  15 = 0 ) is

 x −5 0 5 y 6 3 0

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

From the graph it is clear that, the given lines intersect at (5,0).
So, the solution of the given system of equations is (5,0).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0).
Now,

Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0) and its area is 20 sq. units.

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (2x − 5y + 4 = 0) is

 x −2 0 3 y 0 $\frac{4}{5}$ 2

Now, plot the points A(−2,0), $B\left(0,\frac{4}{5}\right)$ and C(3,2) on a graph paper and join
A, B and C to get the graph of 2
x  5y + 4 = 0.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (2x + y  8 = 0 ) is

 x 0 2 4 y 8 4 0

Now, plot the points D(0,8), E(2,4) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 2x + y − 8 = 0.

From the graph it is clear that, the given lines intersect at (3,2).
So, the solution of the given system of equations is (3,2).
The vertices of the triangle formed by the system of equations and y-axis are (0,8), $\left(0,\frac{4}{5}\right)$ and (3,2).
Draw a perpendicular from point C to the y-axis. So,

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

#### Page No 93:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 5xy = 7
5xy = 7
y = (5x − 7) .........(i)
Putting x = 0, we get y = −7.
Putting x = 1, we get y = −2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation 5xy = 7.

 x 0 1 2 y −7 −2 3

Now, plot the points A(0, −7) , B( 1, −2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of  5xy = 7.

Graph of xy + 1 = 0
xy + 1 = 0
y = (x + 1) ...........(ii)
Putting x = 0, we get y = 1.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation xy + 1 = 0.
 x 0 1 2 y 1 2 3
Now, plot the points P(0, 1) and Q(1, 2). The point C(2, 3) has already been plotted. Join PQ and QC to get the graph line PC. Extend it on both ways.
Then, PC is the graph of the equation xy + 1 = 0.

The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔAPC formed by these two lines and the y-axis are P(0, 1), C(2, 3) and A(0, −7).
Now, consider ΔAPC.
Here, height = 2 units and base (AP) = 8 units
∴ Area of ΔAPC = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×8×2\right)=8$ sq. units.

#### Page No 93:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (2x − 3y = 12) is

 x 0 3 6 y −4 −2 0

Now, plot the points A(0,−4), B(3,−2) and C(6,0) on a graph paper and join
A, B and C to get the graph of 2
x  3y = 12.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (x + 3y = 6 ) is

 x 0 3 6 y 2 1 0

Now, plot the points D(0,2), E(3,1) and F(6,0) on the same graph paper and join
D, E and F to get the graph of x + 3y = 6.

From the graph it is clear that, the given lines intersect at (6,0).
So, the solution of the given system of equations is (6,0).
The vertices of the triangle formed by the system of equations and y-axis are (0,2), (6,0) and (0,−4).
Now,

Hence, the veritices of the triangle are (0,2), (6,0) and (0,−4) and its area is 18 sq. units.

#### Page No 94:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (2x + 3y = 6) is

 x −3 3 6 y 4 0 −2

Now, plot the points A(−3,4), B(3,0) and C(6,−2) on a graph paper and join
A, B and C to get the graph of 2
x + 3y = 6.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (4x + 6y = 12 ) is

 x −6 0 9 y 6 2 −4

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

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

#### Page No 94:

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

 x 1 0 2 y −2 −5 1

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

Graph of 6x − 2y = 10
6x − 2y = 10
⇒ 2y = (6x − 10)
$y=\frac{6x-10}{2}$      ...........(ii)
Putting x = 0, we get y = −5.
Putting x = 1, we get y = −2.
Putting x = 2, we get y =  1.

Thus, we have the following table for the equation 6x − 2y = 10.
 x 0 1 2 y −5 −2 1
These are the same points as obtained for the graph line of equation (i).

It is clear from the graph that these two lines coincide.
Hence, the given system of equations has infinitely many solutions.

#### Page No 94:

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

 x 3 1 2 y 0 4 2

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

Graph of 6x + 3y = 18
6x + 3y = 18 ⇒ 3y = (18 − 6x)
$y=\frac{18-6x}{3}$...........(ii)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 6x + 3y = 18.
 x 3 1 2 y 0 4 2
These are the same points as obtained for the graph line of equation (i).

It is clear from the graph that these two lines coincide.
Hence, the given system of equations has an infinite number of solutions.

#### Page No 94:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (x − 2y = 5) is

 x −5 1 3 y −5 −2 −1

Now, plot the points A(−5,−5), B(1,−2) and C(3,−1) on a graph paper and join
A, B and C to get the graph of
x  2y = 5.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (3x  6y = 15 ) is

 x −3 −1 5 y −4 −3 0

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

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

#### Page No 94:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (x − 2y = 6) is

 x −2 0 2 y −4 −3 −2

Now, plot the points A(−2,−4), B(0,−3) and C(2,−2) on a graph paper and join
A, B and C to get the graph of
x  2y = 6.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (3x  6y = 0 ) is

 x −4 0 4 y −2 0 2

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

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

#### Page No 94:

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

 x 2 −1 −4 y 0 2 4

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

Graph of 4x + 6y = 12
4x + 6y = 12
⇒ 6y = (−4x + 12)
$y=\frac{-4x+12}{6}$      ...........(ii)
Putting x = 3, we get y = 0.
Putting x = 0, we get y = 2.
Putting x = 6, we get y = −2.
Thus, we have the following table for the equation 4x + 6y = 12.
 x 3 0 6 y 0 2 −2

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

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

#### Page No 94:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (2x y = 6) is

 x 0 2 4 y 6 2 −2

Now, plot the points A(0,6), B(2,2) and C(4,−2) on a graph paper and join
A, B and C to get the graph of 2
x + y = 6.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (6x + 3y = 20 ) is

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

Now, plot the points  on the same graph paper and join
D, E and F to get the graph of 6x + 3y = 20.

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

#### Page No 94:

From the first equation, write y in terms of x

Substitute different values of x in (i) to get different values of y

Thus, the table for the first equation (2x y = 2) is

 x 0 1 2 y 2 0 −2

Now, plot the points A(0,2), B(1,0) and C(2,−2) on a graph paper and join
A, B and C to get the graph of 2
x + y = 2.
From the second equation, write y in terms of x

Now, substitute different values of x in (ii) to get different values of y

So, the table for the second equation (2x + y = 6 ) is

 x 0 1 3 y 6 4 0

Now, plot the points D(0,6), E(1,4) and F(3,0)on the same graph paper and join
D, E and F to get the graph of 2x + y = 6.

From the graph it is clear that, the given lines do not intersect at all when produced. So, these lines are
parallel to each other and therefore, the quadrilateral DABF is a trapezium. The vertices of the
required trapezium are D(0,6), A(0,2), B(1,0) and F(3,0).
Now,

Hence, the area of the rquired trapezium is 8 sq. units.

#### Page No 109:

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:

The given system of equations is

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

#### Page No 110:

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:

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:

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:

The given system of equations can be written as

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

#### Page No 110:

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

and $\frac{5x}{6}-\frac{y}{3}=-7$
⇒ 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:

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

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

#### Page No 110:

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

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

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

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

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

#### Page No 110:

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

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

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

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

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

#### Page No 110:

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

and 2x + 3y + 1 = 0
⇒ 2x + 3y = −1 .............(ii)

On subtracting (ii) from (i), we get:
2x = 8
x = 4

On substituting x = 4 in (i), we get:
16 + 3y = 7
⇒ 3y = (7 − 16) = −9
y = −3

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

#### Page No 110:

The given system of equations is

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

#### Page No 110:

The given system of equations is

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

#### Page No 110:

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:

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:

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

and $\frac{x+2y-14}{3}=\frac{3x+y-12}{11}$
⇒ 11x + 22y − 154 = 9x + 3y − 36
⇒ 11x − 9x + 22y − 3y = −36 + 154
⇒ 2x + 19y = 118 .........(ii)

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

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

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

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

#### Page No 110:

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

Putting $\frac{1}{x}=u$, we get:
5u + 6y = 13 .............(iii)
3u + 4y = 7 ...........(iv)

On multiplying (iii) by 4 and (iv) by 6, we get:
20u + 24y = 52 ...........(v)
18u + 24y = 42 ............(vi)

On subtracting (vi) from (v), we get:
2u = 10 ⇒ u = 5
$⇒\frac{1}{x}=5⇒x=\frac{1}{5}$
On substituting $x=\frac{1}{5}$ in (i), we get:
$\frac{5}{1}{5}}+6y=13$
⇒ 25 + 6y = 13
⇒ 6y = (13 − 25) = −12
y =  −2

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

#### Page No 110:

The given equations are:
$x+\frac{6}{y}=6$ ............(i)
$3x-\frac{8}{y}=5$ .............(ii)
Putting $\frac{1}{y}=v$, we get:
x + 6v = 6 .............(iii)
3x − 8v = 5 ...........(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:
4x + 24v = 24 ...........(v)
9x − 24v = 15 ............(vi)

On adding (v) and (vi), we get:
13x = 39 ⇒ x = 3

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

#### Page No 110:

The given equations are:
$2x-\frac{3}{y}=9$ ............(i)
$3x+\frac{7}{y}=2$ .............(ii)
Putting $\frac{1}{y}=v$, we get:
2x − 3v = 9 .............(iii)
3x + 7v = 2 ...........(iv)

On multiplying (iii) by 7 and (iv) by 3, we get:
14x − 21v = 63 .............(v)
9x + 21v = 6............(vi)

On adding (v) and (vi), we get:
23x = 69 ⇒ x = 3

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

#### Page No 110:

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

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
9u − 4v = 8 .............(iii)
13u + 7v = 101 ...........(iv)

On multiplying (iii) by 7 and (iv) by 4, we get:
63u − 28v = 56 .............(v)
52u + 28v = 404............(vi)

On adding (v) and (vi), we get:
115u = 460 ⇒ u = 4
$⇒\frac{1}{x}=4⇒x=\frac{1}{4}$
On substituting $x=\frac{1}{4}$ in (i), we get:
$\frac{9}{1}{4}}-\frac{4}{y}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒36-\frac{4}{y}=8⇒\frac{4}{y}=\left(36-8\right)=28\phantom{\rule{0ex}{0ex}}$
$⇒y=\frac{4}{28}=\frac{1}{7}$
Hence, the required solution is $x=\frac{1}{4}$ and $y=\frac{1}{7}$.

#### Page No 110:

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

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

On multiplying (iii) by 4 and (iv) by 3, we get:
20u − 12v = 4 .............(v)
27u + 12v = 90 ...........(vi)

On adding (iv) and (v), we get:
47u = 94 ⇒ u = 2
$⇒\frac{1}{x}=2⇒x=\frac{1}{2}$

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

#### Page No 110:

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

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

#### Page No 110:

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

From equation (i), we have:

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

For equation (ii), we have:

$\frac{8x+9y}{xy}=5\phantom{\rule{0ex}{0ex}}$
$⇒\frac{8}{y}+\frac{9}{x}=5$
.............(iv)
On substituting , we get:
4v + 6u = 3 ...........(v)
8v + 9u = 5 ...........(vi)

On multiplying (v) by 9 and (vi) by 6, we get:
36v + 54u = 27 ..............(vii)
48v + 54u = 30 ...............(viii)

On subtracting (vii) from (viii), we get:
12v = 3 $⇒v=\frac{3}{12}=\frac{1}{4}$
$⇒\frac{1}{y}=\frac{1}{4}⇒y=4$

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

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

#### Page No 110:

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

From equation (i), we have:

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

From equation (ii), we have:

$\frac{3x+2y}{xy}=13\phantom{\rule{0ex}{0ex}}$
$⇒\frac{3}{y}+\frac{2}{x}=13$
.............(iv)
On substituting , we get:
v + u = 5 ...........(v)
3v + 2u = 13 ...........(vi)

On multiplying (v) by 2, we get:
2v + 2u = 10 ..............(vii)
On subtracting (vii) from (vi), we get:
v = 3
$⇒\frac{1}{y}=3⇒y=\frac{1}{3}$
On substituting $y=\frac{1}{3}$ in (iii), we get:
$\frac{1}{1}{3}}+\frac{1}{x}=5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒3+\frac{1}{x}=5⇒\frac{1}{x}=2⇒x=\frac{1}{2}$
Hence, the required solution is $x=\frac{1}{2}$ and $y=\frac{1}{3}$ or x = 0 and y = 0.

#### Page No 110:

The given equations are

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

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

Now, substituting v = 1 in (iii), we get

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

#### Page No 111:

The given equations are:
$\frac{3}{x+y}+\frac{2}{x-y}=2$          ...(i)
$\frac{9}{x+y}-\frac{4}{x-y}=1$           ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, we get:
3u + 2v = 2            ...(iii)
9u − 4v = 1              ...(iv)
On multiplying (iii) by 2, we get:
6u + 4v = 4               ...(v)
On adding (iv) and (v), we get:
15u = 5
$⇒u=\frac{5}{15}=\frac{1}{3}$
$⇒\frac{1}{x+y}=\frac{1}{3}⇒x+y=3$             ...(vi)
On substituting $u=\frac{1}{3}$ in (iii), we get:
1 + 2v = 2
⇒ 2v = 1
$⇒v=\frac{1}{2}$
$⇒\frac{1}{x-y}=\frac{1}{2}⇒x-y=2$                ...(vii)
On adding (vi) and (vii), we get:
2x = 5
$x=\frac{5}{2}$
On substituting $x=\frac{5}{2}$ in (vi), we get:

Hence, the required solution is .

#### Page No 111:

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

#### Page No 111:

The given equations are:
$\frac{44}{x+y}+\frac{30}{x-y}=10$        ...(i)
$\frac{55}{x+y}+\frac{40}{x-y}=13$         ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ , we get:
44u + 30v = 10          ...(iii)
55u + 40v = 13          ...(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
176u + 120v = 40           ...(v)
165u + 120v = 39           ...(vi)
On subtracting (vi) from (v), we get:
11u = 1
$⇒u=\frac{1}{11}$
$⇒\frac{1}{x+y}=\frac{1}{11}⇒x+y=11$         ...(vii)
On substituting $u=\frac{1}{11}$ in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
$⇒v=\frac{6}{30}=\frac{1}{5}$
$⇒\frac{1}{x-y}=\frac{1}{5}⇒x-y=5$     ...(viii)
On adding (vii) and (viii), we get:
2x = 16
x = 8
On substituting x = 8 in (vii), we get:
8 + y = 11
y = (11 − 8) = 3
Hence, the required solution is x = 8 and y = 3.

#### Page No 111:

The given equations are

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

Multiplying  (iii) by 9 and (iv) by 2 and adding, we get

Now, substituting $u=\frac{4}{15}$ in (iii), we get

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

#### Page No 111:

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:

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:

The given equations are

Adding (i) and (ii), we get

Subtracting (i) from (ii), we get

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

#### Page No 111:

The given equations can be written as

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

#### Page No 111:

The given equations are

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

Adding (iii) and (iv), we get

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

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

#### Page No 111:

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

#### Page No 111:

The given equations are

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

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

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

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

#### Page No 111:

The given equations can be written as

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

#### Page No 111:

The given equations are

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

#### Page No 111:

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

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

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

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

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

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

#### Page No 111:

The given equations are

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

#### Page No 111:

The given equations are

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

#### Page No 111:

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

and 6(bxay) = 3b − 2a
⇒ 6bx − 6ay = 3b − 2a ...................(ii)

On multiplying (i) by a and (ii) by b, we get:
6a2x + 6aby = 3a2 + 2ab ................(iii)
6b2x − 6aby = 3b2 − 2ab ....................(iv)

On adding (iii) and (iv), we get:
6(a2 + b2)x = 3(a2 + b2)
$⇒x=\frac{3\left({a}^{2}+{b}^{2}\right)}{6\left({a}^{2}+{b}^{2}\right)}=\frac{1}{2}$

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

#### Page No 111:

The given equations are

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

#### Page No 111:

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

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

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

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

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

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

#### Page No 111:

The given equations are:

$\frac{bx}{a}+\frac{ay}{b}={a}^{2}+{b}^{2}$
By taking LCM, we get:
$\frac{{b}^{2}x+{a}^{2}y}{ab}={a}^{2}+{b}^{2}$
b2x + a2y = (ab)a2 + b2
b2x + a2y = a3b + ab3 .......(i)

Also, x + y =  2ab........(ii)

On multiplying (ii) by a2,  we get:
a2x + a2y = 2a3b.........(iii)

On subtracting (iii) from (i), we get:
(b2a2)x = a3b + ab3  − 2a3b
⇒ (b2a2)x = −a3b + ab3
⇒ (b2a2)x = ab(b2 a2)
⇒ (b2a2)x = ab(b2a2)
$x=\frac{ab\left({b}^{2}-{a}^{2}\right)}{\left({b}^{2}-{a}^{2}\right)}=ab$

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

#### Page No 111:

The given equations are

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

#### Page No 111:

The given equations are

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

Hence, .

#### Page No 111:

The given equations are

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

Hence, .

#### Page No 117:

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

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

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

#### Page No 117:

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

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

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

#### Page No 117:

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

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

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

#### Page No 117:

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

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

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

#### Page No 117:

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

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

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

#### Page No 117:

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

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

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

#### Page No 117:

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:

$\frac{x}{\left[\left(-2\right)×\left(-16\right)-\left(-3\right)×\left(-3\right)\right]}=\frac{y}{\left[\left(-3\right)×22-\left(-16\right)×7\right]}=\frac{1}{\left[7×\left(-3\right)-22×\left(-2\right)\right]}$
$\frac{x}{\left(32-9\right)}=\frac{y}{\left(-66+112\right)}=\frac{1}{\left(-21+44\right)}$
⇒ $\frac{x}{\left(23\right)}=\frac{y}{46}=\frac{1}{\left(23\right)}$

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

#### Page No 117:

The given equations may be written as:
$\frac{x}{6}+\frac{y}{15}-4=0$           ...(i)
$\frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0$       ...(ii)
Here,
By cross multiplication, we have:

$\frac{x}{\left[\frac{1}{15}×\left(-\frac{19}{4}\right)-\left(-\frac{1}{12}\right)×\left(-4\right)\right]}=\frac{y}{\left[\left(-4\right)×\frac{1}{3}-\left(\frac{1}{6}\right)×\left(-\frac{19}{4}\right)\right]}=\frac{1}{\left[\frac{1}{6}×\left(-\frac{1}{12}\right)-\frac{1}{3}×\frac{1}{15}\right]}$
$\frac{x}{\left(-\frac{19}{60}-\frac{1}{3}\right)}=\frac{y}{\left(-\frac{4}{3}+\frac{19}{24}\right)}=\frac{1}{\left(-\frac{1}{72}-\frac{1}{45}\right)}$
$\frac{x}{\left(-\frac{39}{60}\right)}=\frac{y}{\left(-\frac{13}{24}\right)}=\frac{1}{\left(-\frac{13}{360}\right)}$

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

#### Page No 117:

Taking $\frac{1}{x}=u$  and $\frac{1}{y}=v$, the given equations become:
u + v = 7
2u + 3v = 17

The given equations may be written as:
u + v − 7 = 0            ...(i)
2u + 3v − 17 = 0      ...(ii)

Here, a1 = 1, b1 = 1, c1 = −7, a2 = 2, b2 = 3 and c2 = −17
By cross multiplication, we have:

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

#### Page No 117:

Taking $\frac{1}{x+y}=u$  and $\frac{1}{x-y}=v$, the given equations become:
5u − 2v + 1 = 0      ...(i)
15u + 7v − 10 = 0    ...(ii)
Here, a1 = 5, b1 = −2, c1 = 1, a2 = 15, b2 = −7 and c2 = −10
By cross multiplication, we have:

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

Again, the above equations (iii) and (iv) may be written as:
x + y − 5 = 0           ...(v)
xy − 1 = 0          ...(vi)
Here, a1 = 1, b1 = 1, c1 = −5, a2 = 1, b2 = −1 and c2 = −1
By cross multiplication, we have:

∴ $\frac{x}{\left[1×\left(-1\right)-\left(-5\right)×\left(-1\right)\right]}=\frac{y}{\left[\left(-5\right)×1-\left(-1\right)×1\right]}=\frac{1}{\left[1×\left(-1\right)-1×1\right]}$

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

#### Page No 117:

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

$\frac{x}{\left(-\frac{b}{a}\right)×\left(-2ab\right)-\left(-b\right)×\left(-\left(a+b\right)\right)}=\frac{y}{-\left(a+b\right)×a-\left(-2ab\right)×\frac{a}{b}}=\frac{1}{\frac{a}{b}×\left(-b\right)-a×\left(-\frac{b}{a}\right)}$
$\frac{x}{2{b}^{2}-b\left(a+b\right)}=\frac{y}{-a\left(a+b\right)+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{2{b}^{2}-ab-{b}^{2}}=\frac{y}{-{a}^{2}-ab+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{{b}^{2}-ab}=\frac{y}{{a}^{2}-ab}=\frac{1}{-\left(a-b\right)}$
$\frac{x}{-b\left(a-b\right)}=\frac{y}{a\left(a-b\right)}=\frac{1}{-\left(a-b\right)}$
$x=\frac{-b\left(a-b\right)}{-\left(a-b\right)}=b,y=\frac{a\left(a-b\right)}{-\left(a-b\right)}=-a$
Hence, x = b and y = −a is the required solution.

#### Page No 117:

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

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

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

#### Page No 117:

Substituting  in the given equations, we get

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

Hence, x = a and y = b.

#### Page No 128:

The given system of equations is:
3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 3, b1= 5, c1 = −12 and a2 = 5, b2 = 3, c2 = −4
For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{3}{5}\ne \frac{5}{3}$
Hence, the given system of equations has a unique solution.

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

#### Page No 128:

The system of equations can be written as

The given equations are of the form

where
Now,

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

Hence, .

#### Page No 128:

The given system of equations are:
$\frac{x}{3}+\frac{y}{2}=3$
$\frac{2x+3y}{6}=3$
2x + 3y = 18
⇒ 2x + 3y − 18 = 0                 ...(i)
and
x − 2y = 2
x − 2y − 2 = 0                    ...(ii)
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 2, b1= 3, c1 = −18 and a2 = 1, b2 = −2, c2 = −2
For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{2}{1}\ne \frac{3}{-2}$
Hence, the given system of equations has a unique solution.

Again, the given equations are:
2x + 3y − 18 = 0           ...(iii)
x − 2y − 2 = 0               ...(iv)
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y − 36 = 0           ...(v)
3x − 6y − 6 = 0             ...(vi)
On adding (v) and (vi), we get:
7x = 42
x = 6
On substituting x = 6 in (iii), we get:
2(6) + 3y = 18
⇒ 3y = (18 − 12) = 6
y = 2
Hence, x = 6 and y = 2 is the required solution.

#### Page No 128:

The given system of equations are

This system is of the form

where
Now, for the given system of equations to have a unique solution, we must have
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{k}\ne \frac{3}{-6}\phantom{\rule{0ex}{0ex}}⇒k\ne -4$
Hence, $k\ne -4$.

#### Page No 128:

The given system of equations are

This system of equations is of the form

where
Now, for the given system of equations to have a unique solution, we must have
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{3}\ne \frac{-k}{2}\phantom{\rule{0ex}{0ex}}⇒k\ne -\frac{2}{3}$
Hence, $k\ne -\frac{2}{3}$.

#### Page No 128:

The given system of equations is

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

#### Page No 128:

The given system of equations is

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

#### Page No 129:

The given system of equations:
4x − 5y = k
⇒ 4x − 5yk = 0          ...(i)
And, 2x − 3y = 12
⇒ 2x − 3y − 12 = 0        ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 4, b1= −5, c1 = −k and a2 = 2, b2 = −3, c2 = −12
For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
i.e.  $\frac{4}{2}\ne \frac{-5}{-3}$
$⇒2\ne \frac{5}{3}⇒6\ne 5$
Thus, for all real values of k, the given system of equations will have a unique solution.

#### Page No 129:

The given system of equations:
kx + 3y = (k − 3)
⇒ kx + 3y − (k − 3) = 0            ....(i)
And, 12x + ky = k
⇒ 12x + kyk = 0                   ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −(k − 3) and a2 = 12, b2 = k, c2 = −k
For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
i.e.  $\frac{k}{12}\ne \frac{3}{k}$
$⇒{k}^{2}\ne 36⇒k\ne ±6$
Thus, for all real values of k other than $±6$, the given system of equations will have a unique solution.

#### Page No 129:

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

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

#### Page No 129:

The given system of equations can be written as

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

#### Page No 129:

The given system of equations is:
kx + 2y = 5
kx + 2y − 5= 0                ...(i)
3x − 4y = 10
⇒ 3x − 4y − 10 = 0             ...(ii)
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = k, b1= 2, c1 = −5 and a2 = 3, b2 = −4, c2 = −10
(i) For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{k}{3}\ne \frac{2}{-4}⇒k\ne \frac{-3}{2}$
Thus for all real values of k other than $\frac{-3}{2}$, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$⇒\frac{k}{3}=\frac{2}{-4}\ne \frac{-5}{-10}$

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

#### Page No 129:

The given system of equations is:
x + 2y = 5
x + 2y − 5= 0                      ...(i)
3x + ky + 15 = 0          ...(ii)
These equations are of the form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 1, b1= 2, c1 = −5 and a2 = 3, b2 = k, c2 = 15
(i) For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e., $\frac{1}{3}\ne \frac{2}{k}⇒k\ne 6$
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$⇒\frac{1}{3}=\frac{2}{k}\ne \frac{-5}{15}$

Hence, the required value of k is 6.

#### Page No 129:

The given system of equations is:
x + 2y = 3
x + 2y − 3= 0                    ....(i)
And, 5xky + 7 = 0          ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 1, b1= 2, c1 = −3 and a2 = 5, b2 = k, c2 = 7
(i) For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, i.e. $\frac{1}{5}\ne \frac{2}{k}⇒k\ne 10$
Thus, for all real values of k​, other than 10, the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{1}{5}=\frac{2}{k}\ne \frac{-3}{7}$

$⇒k=10,k\ne \frac{14}{-3}$
Hence, the required value of k is 10.
There is no value of k for which the given system of equations has an infinite number of solutions.

#### Page No 129:

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0                            ....(i)
And, (k − 1)x + (k + 2)y = 3k
⇒ (k − 1)x + (k + 2)y − 3k = 0         ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= 3, c1 = −7 and a2 = (k − 1), b2 = (k + 2), c2 = −3k
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(k-1\right)}=\frac{3}{\left(k+2\right)}=\frac{-7}{-3k}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{2}{\left(k-1\right)}=\frac{3}{\left(k+2\right)}=\frac{7}{3k}$

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

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

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

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

#### Page No 129:

The given system of equations:
2x + (k − 2)y = k
⇒ 2x + (k − 2)yk = 0                  ....(i)
And, 6x + (2k − 1)y = (2k + 5)
⇒ 6x + (2k − 1)y − (2k + 5) = 0      ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= (k − 2), c1 = −k and a2 = 6, b2 = (2k − 1), c2 = −(2k + 5)
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{6}=\frac{\left(k-2\right)}{\left(2k-1\right)}=\frac{-k}{-\left(2k+5\right)}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{1}{3}=\frac{\left(k-2\right)}{\left(2k-1\right)}=\frac{k}{\left(2k+5\right)}$

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

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

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

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

#### Page No 129:

The given system of equations:
kx + 3y = (2k + 1)
kx + 3y − (2k + 1) = 0                  ...(i)
And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y − (7k + 1) = 0        ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −(2k + 1) and a2 = 2(k + 1), b2 = 9, c2 = −(7k + 1)
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$

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

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

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

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

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

#### Page No 129:

The given system of equations:
5x + 2y = 2k
⇒ 5x + 2y − 2k= 0                          ...(i)
And, 2(k + 1)x + ky = (3k + 4)
⇒ 2(k + 1)x + ky − (3k + 4) = 0       ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= 2, c1 = −2k and a2 = 2(k + 1), b2 = k, c2 = −(3k + 4)
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{-2k}{-\left(3k+4\right)}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$

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

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

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

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

#### Page No 129:

The given system of equations:
(k − 1)xy = 5
⇒ (k − 1)xy − 5 = 0                           ...(i)
And, (k + 1)x + (1 − k)y = (3k + 1)
⇒ (k + 1)x + (1 − k)y − (3k + 1) = 0      ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = (k − 1), b1= −1, c1 = −5 and a2 = (k + 1), b2 = (1 − k), c2 = −(3k + 1)
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$

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

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

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

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

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

#### Page No 129:

The given system of equations can be written as

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

Hence, k = 6.

#### Page No 129:

The given system of equations:
(a − 1)x + 3y = 2
⇒ (a − 1)x + 3y − 2 = 0        ...(i)
and 6x + (1 − 2b)y = 6
⇒ 6x + (1 − 2b)y − 6= 0       ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
where, a1 = (a − 1), b1= 3, c1 = −2 and a2 = 6, b2 = (1 − 2b), c2 = −6
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{\left(a-1\right)}{6}=\frac{3}{\left(1-2b\right)}=\frac{-2}{-6}$
$⇒\frac{a-1}{6}=\frac{3}{\left(1-2b\right)}=\frac{1}{3}$
$⇒\frac{a-1}{6}=\frac{1}{3}\mathrm{and}\frac{3}{\left(1-2\mathrm{b}\right)}=\frac{1}{3}$
⇒ 3a − 3 = 6 and 9 = 1 − 2b
⇒ 3a = 9 and 2b = −8
a = 3 and b = −4
∴​ a = 3 and b = −4

#### Page No 130:

The given system of equations:
(2a − 1)x + 3y = 5
⇒ (2a − 1)x + 3y − 5 = 0          ....(i)
And, 3x + (b − 1)y = 2
⇒ 3x + (b − 1)y − 2 = 0            ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = (2a − 1), b1= 3, c1 = −5 and a2 = 3, b2 = (b − 1), c2 = −2
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{\left(2a-1\right)}{3}=\frac{3}{\left(b-1\right)}=\frac{-5}{-2}$
$⇒\frac{\left(2a-1\right)}{3}=\frac{3}{\left(b-1\right)}=\frac{5}{2}$

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

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

#### Page No 130:

The given system of equations:
2x − 3y = 7
⇒ 2x − 3y − 7 = 0                                ....(i)
And, (a + b)x − (a + b − 3)y = 4a + b
⇒ (a + b)x − (a + b − 3)y − (4a + b) = 0 ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1= −3, c1 = −7 and a2 = (a + b), b2 = −(a + b − 3), c2 = −(4a + b)
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{a+b}=\frac{-3}{-\left(a+b-3\right)}=\frac{-7}{-\left(4a+b\right)}$
$⇒\frac{2}{a+b}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}$

⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b − 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b − 21
a = 5b                  ....(iii)
And, 5a = 4b − 21     ....(iv)
On substituting a = 5b in (iv), we get:
25b = 4b − 21
⇒ 21b = −21
b = −1
On substituting b = −1 in (iii), we get:
a = 5(−1) = −5
a = −5 and b = −1

#### Page No 130:

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0                                   ....(i)
And, (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
⇒ (a + b + 1)x + (a + 2b + 2)y − [4(a + b) + 1] = 0    ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 2, b1=  3, c1 = −7 and a2 = (a + b + 1), b2 = (a + 2b + 2), c2 = −[4(a + b) + 1]
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{-7}{-\left[4\left(a+b\right)+1\right]}$
$⇒\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{7}{4\left(a+b\right)+1}$
$⇒\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}\mathrm{and}\frac{3}{\left(\mathrm{a}+2\mathrm{b}+2\right)}=\frac{7}{4\left(a+b\right)+1}$

⇒ 2(a + 2b + 2) = 3(a + b+ 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)
⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14
ab − 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
ab = 1 and 5a − 2b = 11
a = (b + 1)     ....(iii)
5a − 2b = 11     ....(iv)
On substituting a = (b + 1) in (iv), we get:
5(b + 1) − 2b = 11
⇒ 5b + 5 − 2b = 11
⇒ 3b = 6
b = 2
On substituting b = 2 in (iii), we get:
a = 3
∴​ a = 3 and b = 2

#### Page No 130:

The given system of equations can be written as

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

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

Hence, a = 5 and b = 1.

#### Page No 130:

The given system of equations can be written as

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

Substituting a = 4 in a + b = 12, we get
$4+b=12⇒b=12-4=8$
Hence, a = 4 and b = 8.

#### Page No 130:

The given system of equations:
8x + 5y = 9
8x + 5y − 9 = 0                    ....(i)
kx + 10y = 15
kx + 10y − 15= 0                 ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 8, b1= 5, c1 = −9 and a2 = k, b2 = 10, c2 = −15
In order that the given system has no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$

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

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

#### Page No 130:

The given system of equations:
kx + 3y = 3
kx + 3y − 3 = 0                  ....(i)
12x + ky = 6
12x + ky − 6= 0                 ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = k, b1= 3, c1 = −3 and a2 = 12, b2 = k, c2 = −6
In order that the given system of equations has no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
i.e.  $\frac{k}{12}=\frac{3}{k}\ne \frac{-3}{-6}$
$\frac{k}{12}=\frac{3}{k}$ and $\frac{3}{k}\ne \frac{1}{2}$

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

#### Page No 130:

The given system of equations:
3xy − 5 = 0                    ...(i)
And, 6x − 2y + k = 0            ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 3, b1= −1, c1 = −5 and a2 = 6, b2 = −2, c2 = k
In order that the given system of equations has no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
i.e. $\frac{3}{6}=\frac{-1}{-2}\ne \frac{-5}{k}$
$\frac{-1}{-2}\ne \frac{-5}{k}⇒k\ne -10$
Hence, equations (i) and (ii) will have no solution if $k\ne -10$.

#### Page No 130:

The given system of equations can be written as

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

Hence, $k=-6$.

#### Page No 130:

The given system of equations:
5x − 3y = 0         ....(i)
2x + ky = 0         ....(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 5, b1= −3, c1 = 0 and a2 = 2, b2 = k, c2 = 0
For a non-zero solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}$
$⇒\frac{5}{2}=\frac{-3}{k}\phantom{\rule{0ex}{0ex}}$
$⇒5k=-6⇒k=\frac{-6}{5}$
Hence, the required value of k is $\frac{-6}{5}$.

#### Page No 151:

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

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

#### Page No 151:

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

Adding (i) and (ii), we get

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

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

#### Page No 152:

Let the x and y be the number of 50-paisa and 25-paisa conis respectively. Then

Multiplying (ii) by 2 and subtracting it from (i), we get

Subtracting y = 22 in (i), we get
$x+22=50\phantom{\rule{0ex}{0ex}}⇒x=50-22=28$
Hence, the number of 25-paisa and 50-paisa conis are 22 and 28 respectively.

#### Page No 152:

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:

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:

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:

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:

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:

Let the required numbers be x and y.
Now, we have:
$\frac{x+2}{y+2}=\frac{1}{2}$
By cross multiplication, we get:
2x + 4 = y + 2
⇒ 2xy  = −2                  ....(i)
Again, we have:
$\frac{x-4}{y-4}=\frac{5}{11}$
By cross multiplication, we get:
11x − 44 = 5y − 20
⇒ 11x − 5y = 24                ....(ii)
On multiplying (i) by 5, we get:
10x − 5y = −10                 ....(iii)
On subtracting (iii) from (ii), we get:
x = (24 + 10) = 34
On substituting x = 34 in (i), we get:
2 × 34 − y = −2
⇒ 68 − y = −2
y = (68 + 2) = 70
Hence, the required numbers are 34 and 70.

#### Page No 152:

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

#### Page No 152:

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:

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:

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:

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:

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:

Let the tens and the units digits of the required number be x and y​, respectively.
Then, we have:
xy = 35                      ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ 9(yx) = 18
yx = 2                 ....(ii)

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

y + x = 12              .....(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2y = 2 + 12 = 14
y = 7
On substituting y = 7 in (ii), we get:
7 − x = 2
x = (7 − 2) = 5
∴ Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.

#### Page No 153:

Let the tens and the units digits of the required number be x and y​, respectively.
Then, we have:
xy = 18                       ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) − 63 = 10y + x
⇒ 9x − 9y = 63
⇒ 9(xy) = 63
xy = 7                ....(ii)

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

x + y = 11              ....(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2x = 7 + 11 = 18
x = 9
On substituting x = 9 in (ii), we get:
9 − y = 7
y = (9 − 7) = 2
∴ Number = (10x + y) = 10 × 9 + 2 = 90 + 2 = 92
Hence, the required number is 92.

#### Page No 153:

Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per the question

Since the digits differ by 3, so

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

#### Page No 153:

Let the required fraction be $\frac{x}{y}$.
Then, we have:
x + y = 8                        ....(i)
And, $\frac{x+3}{y+3}=\frac{3}{4}$
⇒ 4(x + 3) = 3(y + 3)
⇒ 4x + 12 = 3y + 9
⇒ 4x − 3y = −3              ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 24                  ....(iii)
On adding (ii) and (iii), we get:
7x = 21
x = 3
On substituting x = 3 in (i), we get:
3 + y = 8
y = (8 − 3) = 5
∴​ ​x = 3 and y = 5
Hence, the required fraction is $\frac{3}{5}$.

#### Page No 153:

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

Again, $\frac{x}{y-1}=\frac{1}{3}$
⇒ 3x = 1(y − 1)
⇒ 3x y = −1                     .....(ii)
On subtracting (i) from (ii), we get:
x = (−1 + 4) = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = −4
⇒ 6 − y = −4
y = (6 + 4) = 10
x = 3 and y = 10
Hence, the required fraction is $\frac{3}{10}$.

#### Page No 153:

Let the required fraction be $\frac{x}{y}$.
Then, we have:
y = x + 11
yx = 11                        ....(i)
Again, $\frac{x+8}{y+8}=\frac{3}{4}$
⇒ 4(x + 8) =3(y + 8)
⇒ 4x + 32 = 3y + 24
⇒ 4x − 3y = −8                    ....(ii)
On multiplying (i) by 4, we get:
4y − 4x = 44                        ....(iii)
On adding (ii) and (iii), we get:
y = (−8 + 44) = 36
On substituting y = 36 in (i), we get:
36 − x = 11
x = (36 − 11) = 25
∴​ ​x = 25 and  y = 36
Hence, the required fraction is $\frac{25}{36}$.

#### Page No 153:

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

Again, $\frac{x-7}{y-2}=\frac{1}{3}$
⇒ 3(x − 7) = 1(y − 2)
⇒ 3x − 21 = y − 2
⇒ 3xy = 19                  ....(ii)
On subtracting (i) from (ii), we get:
x = (19 − 4) = 15
On substituting x = 15 in (i), we get:
2 × 15 − y = 4
⇒ 30 − y = 4
⇒  y = 26
∴​ ​x = 15 and y = 26
Hence, the given fraction is $\frac{15}{26}$.

#### Page No 153:

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

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

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

#### Page No 153:

Let the larger number be x and the smaller number be y.
Then, we have:
(x + y) = 16                      ....(i)
And, $\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$                  ....(ii)
$\frac{x+y}{xy}=\frac{1}{3}$
⇒ 3(x + y) = xy
⇒ 3 × 16 = xy      [Since from (i), we have: x + y = 16]
xy = 48                         ....(iii)
We know:
(xy)2 = (x + y)2 − 4xy
(xy)2 = (16)2 4 × 48 = 256 − 192 = 64
∴ (xy) =
Since x is larger and y is smaller, we have:
xy = 8                       .....(iv)
On adding (i) and (iv), we get:
2x = 24
x = 12
On substituting x = 12 in (i), we get:
12 + y = 16 ⇒ y = (16 − 12) = 4
Hence, the required numbers are 12 and 4.

#### Page No 153:

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:

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

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

#### Page No 154:

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

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

#### Page No 154:

Let the the amounts invested at 10% and 8% be ₹x and ₹y respectively.
Then as per the question

After the amounts interchanged but the rate being the same, we have

Adding (i) and (ii) and dividing by 9, we get

Subtracting (ii) from (i), we get

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

#### Page No 154:

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

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

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

#### Page No 154:

Let the cost price of the chair and table be ₹x and ₹y respectively.
Then as per the question
Selling price of chair + Selling price of table = 1520

When the profit on chair and table are 10% and 25% respectively, then

Solving (i) and (ii) by cross multiplication, we get

Hence, the cost of chair and table are ​₹600 and ​₹700 respectively.

#### Page No 154:

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:

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:

Let the speed of the train and taxi be x km/h and y km/h respectively.
Then as per the question

When the speeds of the train and taxi are 260 km and 240 km respectively, then

Multiplying (i) by 6 and subtracting (ii) from it, we get
$\frac{18}{x}-\frac{13}{x}=\frac{66}{200}-\frac{28}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{x}=\frac{10}{200}⇒x=100\phantom{\rule{0ex}{0ex}}$
Putting x = 100 in (i), we have
$\frac{3}{100}+\frac{2}{y}=\frac{11}{200}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{y}=\frac{11}{200}-\frac{3}{100}=\frac{1}{40}\phantom{\rule{0ex}{0ex}}⇒y=80$
Hence, the speed of the train and that of the taxi are 100 km/h and 80 km/h respectively.

#### Page No 154:

Let the speed of the car A and B be x km/h and km/h respectively . Let x > y.
Case-1: When they travel in the same direction

From the figure

Case-2: When they travel in opposite direction

From the figure

Adding (i) and (ii), we get

Putting = 50 in (ii), we have

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

#### Page No 154:

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

#### Page No 155:

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,

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:

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

#### Page No 155:

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:

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:

Let the length and breadth of the rectangle be x m and y m respectively.
Case1: When length is increased by 3 m and breadth is decreased by 4 m

Case2: When length is reduced by 1 m and breadth is increased by 4 m

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

#### Page No 155:

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

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

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

#### Page No 155:

Let the the present age of the man be x years and that of his son be y years.
After 5 years man's age = x + 5
After 5 years ago son's age = y + 5
As per the question

5 years ago man's age = x − 5
5 years ago son's age = y − 5
As per the question

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

#### Page No 155:

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:

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:

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:

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

When gain is ₹14, then

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

#### Page No 156:

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

In case of Tanvy, as per the question

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

#### Page No 156:

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

According to the question,

#### Page No 156:

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

According to the question,

#### Page No 156:

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.

$3\left(x+y\right)+5=2\left(x+5+y+5\right)\phantom{\rule{0ex}{0ex}}⇒3x+3y+5=2x+10+2y+10\phantom{\rule{0ex}{0ex}}⇒3x+3y-2x-2y=20-5\phantom{\rule{0ex}{0ex}}⇒x+y=15\phantom{\rule{0ex}{0ex}}⇒3\left(x+y\right)=45$

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

#### Page No 156:

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,

#### Page No 156:

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,

#### Page No 156:

Let x litres and y litres be the amount of acids from 50% and 25% acid solutions respectively.
As per the question

Since, the total volume is 10 litres, so

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

#### Page No 156:

Let x g and y g be the weight of 18-carat and 12-carat gold respectively.
As per the given condition

And

Multiplying (ii) by 2 and subtracting from (i), we get
$x=320-240=80$
Now, putting x = 80 in (ii), we have
$80+y=120⇒y=40$
Hence, the required weight of 18-carat and 12-carat gold bars are 80 g and 40 g respectively.

#### Page No 156:

Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions.
As per the given condition

And

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

#### Page No 156:

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

And

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

#### Page No 156:

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

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

#### Page No 156:

The opposite angles of cyclic quadrilateral are supplementary, so

And

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

#### Page No 161:

The given equations are

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

#### Page No 161:

The given equations are

Which is of the form , where
For the given pair of linear equations to have infinitely many solutions, we must have

Hence, k = 7.

#### Page No 162:

The given pair of linear equations are

Which is of the form , where
For the given pair of linear equations to have infinitely many solutions, we must have
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{10}{20}=\frac{5}{10}=\frac{-\left(k-5\right)}{-k}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{k-5}{k}\phantom{\rule{0ex}{0ex}}⇒2k-10=k⇒k=10$
Hence, k = 10.

#### Page No 162:

The given pair of linear equations are

Which is of the form , where
For the given pair of linear equations to have no solution, we must have

Hence, k = 11.

#### Page No 162:

The given pair of linear equations are

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

#### Page No 162:

The given pair of linear equations is

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

#### Page No 162:

Let the numbers be x and y, where x > y.
Then as per the question

Dividing (ii) by (i), we get

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

#### Page No 162:

Let the cost of 1 pen and 1 pencil are ₹x and ₹y respectively.
Then as per the question

Adding (i) and (ii), we get

Subtracting (i) from (ii), we get

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

#### Page No 162:

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

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

#### Page No 162:

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

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

#### Page No 162:

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

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

#### Page No 162:

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

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

#### Page No 162:

The given pair of equation is

Multiplying (i) and (ii) by xy, we have

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

#### Page No 162:

The given pair of equations is

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

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

#### Page No 162:

The given pair of equations is

Adding (i) and (ii), we get

Hence, the value of x + y is 4.

#### Page No 162:

The given system is

This is a homogeneous system of linear differential equation, so it always has a zero
solution i.e.,  x = y = 0.
But to have a nonzero solution, it must have infinitely many solutions.
For this, we have
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{k}=\frac{5}{10}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k=6$
Hence, k = 6.

#### Page No 162:

The given system is

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

#### Page No 162:

The given system is

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

#### Page No 162:

The given system is

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

#### Page No 162:

The given system is

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

Hence, k = 10.

#### Page No 162:

The given system of equations is

Substituting in (i) and (ii), the given equations are changed to

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

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

#### Page No 164:

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

Again, we have:
$\frac{x-5}{y-5}=\frac{1}{2}$
⇒ 2(x − 5) = 1(y − 5)
⇒ 2x − 10 = y  − 5
⇒ 2xy = 5                             ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 20                             ....(iii)
On subtracting (i) from (iii), we get:
3x = (20 − (−1)) = 20 + 1 = 21
⇒ 3x = 21
x = 7
On substituting x = 7 in (i), we get:
5 × 7 − 4y = −1
⇒ 35 − 4y =  −1
⇒ 4y = 36
y = 9
∴ ​x = 7 and  y = 9
Hence, the required fraction is $\frac{7}{9}$.

#### Page No 164:

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

$\frac{x}{\left(-\frac{b}{a}\right)×\left(-2ab\right)-\left(-b\right)×\left(-\left(a+b\right)\right)}=\frac{y}{-\left(a+b\right)×a-\left(-2ab\right)×\frac{a}{b}}=\frac{1}{\frac{a}{b}×\left(-b\right)-a×\left(-\frac{b}{a}\right)}$
$\frac{x}{2{b}^{2}-b\left(a+b\right)}=\frac{y}{-a\left(a+b\right)+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{2{b}^{2}-ab-{b}^{2}}=\frac{y}{-{a}^{2}-ab+2{a}^{2}}=\frac{1}{-a+b}$
$\frac{x}{{b}^{2}-ab}=\frac{y}{{a}^{2}-ab}=\frac{1}{-\left(a-b\right)}$
$\frac{x}{-b\left(a-b\right)}=\frac{y}{a\left(a-b\right)}=\frac{1}{-\left(a-b\right)}$

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

#### Page No 165:

The given system of equations is

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

#### Page No 165:

The given system of equations is

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

#### Page No 165:

The given system of equations is

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

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

#### Page No 165:

The given system of equations is

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

#### Page No 165:

Consider . Now, simplifying these equations, we get

And

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

#### Page No 165:

The given equations are

Substituting in (i) and (ii), the new system becomes

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

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

#### Page No 165:

The given equations are

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

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

#### Page No 165:

The given equations are

Adding (i) and (ii), we get

Subtracting (i) from (ii), we get

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

#### Page No 165:

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

#### Page No 165:

The given equations are

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

#### Page No 165:

The given equations are

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

#### Page No 165:

The correct option is (b).

The given system of equations can be written as follows:
x − 2y − 3 = 0 and 3x + ky − 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −2, c1 = −3 and a2 = 3, b2 = k and c2 = −1

These graph lines will intersect at a unique point when we have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$$\frac{1}{3}\ne \frac{-2}{k}⇒k\ne -6$
Hence, k has all real values other than −6.

#### Page No 165:

The correct option is (a).

The given system of equations can be written as follows:
x + 2y − 3 = 0 and 5x + ky + 7 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 5, b2 = k and c2 = 7
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{5},\frac{{b}_{1}}{{b}_{2}}=\frac{2}{k}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-3}{7}$
For the system of equations to have no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{1}{5}=\frac{2}{k}\ne \frac{-3}{7}⇒k=10$

#### Page No 166:

The correct option is (d).

The given system of equations can be written as follows:
3x + 2ky − 2 = 0 and 2x + 5y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 3, b1 = 2k, c1 = −2 and a2 = 2, b2 = 5 and c2 = 1

For parallel lines, we have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{3}{2}=\frac{2k}{5}\ne \frac{-2}{1}$
$k=\frac{15}{4}$

#### Page No 166:

The correct option is (d).

The given system of equations can be written as follows:
kx − 2y − 3 = 0 and 3x + y − 5 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = −2, c1 = −3 and a2 = 3, b2 = 1 and c2 = −5
$\frac{{a}_{1}}{{a}_{2}}=\frac{k}{3},\frac{{b}_{1}}{{b}_{2}}=\frac{-2}{1}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-3}{-5}=\frac{3}{5}$
Thus, for these graph lines to intersect at a unique point, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
$\frac{k}{3}\ne \frac{-2}{1}⇒k\ne -6$
Hence, the graph lines will intersect at all real values of k except −6.

#### Page No 166:

The correct option is (d).

The given system of equations can be written as:
x + 2y + 5 = 0 and −3x − 6y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = 5 and a2 = −3, b2 = −6 and c2 = 1
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{-3},\frac{{b}_{1}}{{b}_{2}}=\frac{2}{-6}=\frac{1}{-3}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{5}{1}$
∴ $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Hence, the given system has no solution.

#### Page No 166:

The correct option is (d).

The given system of equations can be written as follows:
2x + 3y − 5 = 0 and 4x + 6y − 15 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = −5 and a2 = 4, b2 = 6 and c2 = −15
$\frac{{a}_{1}}{{a}_{2}}=\frac{2}{4}=\frac{1}{2},\frac{{b}_{1}}{{b}_{2}}=\frac{3}{6}=\frac{1}{2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-5}{-15}=\frac{1}{3}$
∴ ​$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Hence, the given system has no solution.

#### Page No 166:

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:

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:

The correct option is (b).

Let
$\angle C=3\angle B=\left(3y\right)°$
Now, $\angle A+\angle B+\angle C=180°$
x + y + 3y = 180
x + 4y = 180              ...(i)
Also, $\angle C=2\left(\angle A+\angle B\right)$
⇒ 3y = 2(x + y)
⇒ 2x − y = 0                  ...(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0                    ...(iii)
On adding (i) and (iii) we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
x = 20 and  y = 40
∴​ $\angle B=y°=40°$

#### Page No 166:

The correct option is (b).
$\angle A=\left(x+y+10\right)°$
$\angle B=\left(y+20°\right)$
$\angle C=\left(x+y-30\right)°$
$\angle D=\left(x+y\right)°$
We have:
$\angle A+\angle C=180°$ and $\angle B+\angle D=180°$      [Since ABCD is a cyclic quadrilateral]
Now, $\angle A+\angle C=\left(x+y+10\right)°+\left(x+y-30\right)°=180°$
⇒ 2x + 2y − 20 = 180
x + y − 10 = 90
x + y = 100                   ....(i)
Also, $\angle B+\angle D=\left(y+20\right)°+\left(x+y\right)°=180°$
x + 2y + 20 = 180
x + 2y = 160                ....(ii)
On subtracting (i) from (ii), we get:
y = (160 − 100) = 60
On substituting y = 60 in (i), we get:
x + 60 = 100 ⇒ x = (100 − 60) = 40
∴​ $\angle B=\left(y+20°\right)=\left(60+20\right)°=80°$

#### Page No 166:

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:

Let the fraction be $\frac{x}{y}$
It is given that $\frac{x-1}{y+2}=\frac{1}{2}$

Also, $\frac{x-7}{y-2}=\frac{1}{3}$

Subtract (ii) from (i), we get
$x=15$
Put the value of x in equation (i), we get;
$2\left(15\right)-y=4\phantom{\rule{0ex}{0ex}}30-y=4\phantom{\rule{0ex}{0ex}}⇒y=30-4=26$
Therefore, the fraction is $\frac{x}{y}=\frac{15}{26}$
Hence, the correct answer is option (b)

#### Page No 166:

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:

The correct option is (b).

The given equations are as follows:

They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 6, b1 = −2, c1 = 9 and a2 = 3, b2 = −1 and c2 = 12

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

#### Page No 167:

The correct option is (c).

The given equations are as follows:

They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = −2 and a2 = 1, b2 = −2 and c2 = −8
$\frac{{a}_{1}}{{a}_{2}}=\frac{2}{1},\frac{{b}_{1}}{{b}_{2}}=\frac{3}{-2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-2}{-8}=\frac{1}{4}$
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
The given system has a unique solution.
Hence, the lines intersect exactly at one point.

#### Page No 167:

The correct option is (a).

The given system of equations can be written as follows:

The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 5, b1 = −15, c1 = −8 and a2 = 3, b2 = −9 and c2 = $-\frac{24}{5}$

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

#### Page No 169:

The correct option is (a).

The given system of equations can be written as follows:
x + 2y − 3 = 0 and 2x + 4y + 7 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = −3 and a2 = 2, b2 = 4 and c2 = 7
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{2},\frac{{b}_{1}}{{b}_{2}}=\frac{2}{4}=\frac{1}{2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-3}{7}$
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
So, the given system has no solution.
Hence, the lines are parallel.

#### Page No 169:

The correct option is (d).

The given system of equations can be written as follows:
2x − 3y − 7 = 0 and (a + b)x − (a + b − 3)y − (4a + b) = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = −3, c1 = −7 and a2 = (a + b), b2 =  −(a + b − 3) and c2 = −(4a + b)

For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}$
Now, we have:
$\frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}⇒2a+2b-6=3a+3b$
a + b + 6 = 0                                  ...(i)
Again, we have:
$\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}⇒12a+3b=7a+7b-21\phantom{\rule{0ex}{0ex}}$
⇒ 5a − 4b + 21 = 0                            ...(ii)

On multiplying (i) by 4, we get:
4a + 4b + 24 = 0                              ...(iii)
On adding (ii) and (iii), we get:
9a = −45 ⇒ a = −5
On substituting a = −5 in (i), we get:
−5 + b + 6 = 0 ⇒ b = −1
∴ ​a = −5 and b = −1

#### Page No 170:

The correct option is (a).

The given system of equations can be written as follows:
2x + y − 5 = 0 and 3x + 2y − 8 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 1, c1 = −5 and a2 = 3, b2 = 2 and c2 = −8

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

#### Page No 170:

The correct option is (d).

Given:
x = −y and y > 0
Now, we have:
(i) x2y
On substituting x = −y, we get:
(−y)2y = y3 > 0 (∵ y > 0)
This is true.

(ii) x + y
On substituting x = −y, we get:
(−y) + y = 0
This is also true.

(iii) xy
On substituting x = −y, we get:
(−y) y = −y2 < 0 (∵ y > 0)
This is again true.

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

#### Page No 170:

The given system of equations:

The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = −1, b1 = 2, c1 = 2 and a2 = $\frac{1}{2}$, b2$-\frac{1}{4}$ and c2 = −1

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

#### Page No 170:

The given system of equations can be written as follows:
kx + 3y − (k − 2) = 0 and 12x + kyk = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = 3, c1 = − (k − 2) and a2 = 12, b2 = k and c2 = − k

For inconsistency, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{k}{12}=\frac{3}{k}\ne \frac{\left(k-2\right)}{k}⇒{k}^{2}=\left(3×12\right)=36$
$k=\sqrt{36}=±6$
Hence, the pair of equations is inconsistent if $k=±6$.

#### Page No 170:

The given system of equations can be written as follows:

The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 9, b1 = −10, c1 = −21 and a2 = $\frac{3}{2}$, b2$\frac{-5}{3}$ and c2 = $\frac{-7}{2}$

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

#### Page No 170:

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:

The given system of equations can be written as follows:
x − 3y − 2 = 0 and −2x + 6y − 5= 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = −3, c1 = −2 and a2 = −2, b2 = 6 and c2 = −5
$\frac{{a}_{1}}{{a}_{2}}=\frac{1}{-2}=\frac{-1}{2},\frac{{b}_{1}}{{b}_{2}}=\frac{-3}{6}=\frac{-1}{2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-2}{-5}=\frac{2}{5}$
∴ ​$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Thus, the given system of equations has no solution.
Hence, the paths represented by the equations are parallel.

#### Page No 170:

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:

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:

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

For equation (i), we have:

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

For equation (ii), we have:

$\frac{3x+9y}{xy}=11\phantom{\rule{0ex}{0ex}}$
$⇒\frac{3x}{xy}+\frac{9y}{xy}=11⇒\frac{3}{y}+\frac{9}{x}=11$      ....(iv)
On substituting $\frac{1}{y}=v$ and $\frac{1}{x}=u$ in (iii) and (iv), we get:
6v + 3u = 7                              ....(v)
3v + 9u = 11                            ....(vi)
On multiplying (v) by 3, we get:
18v + 9u = 21                          ....(vii)
On subtracting (vi) from (vii), we get:
15v = 10 ⇒ v = $\frac{10}{15}=\frac{2}{3}$
$\frac{1}{y}=\frac{2}{3}⇒y=\frac{3}{2}$
On substituting $y=\frac{3}{2}$ in (iii), we get:
$\frac{6}{\left(3}{2}\right)}+\frac{3}{x}=7\phantom{\rule{0ex}{0ex}}$
$⇒4+\frac{3}{x}=7⇒\frac{3}{x}=3⇒3x=3\phantom{\rule{0ex}{0ex}}$
$⇒x=1$
Hence, the required solution is x = 1 and $y=\frac{3}{2}$.

#### Page No 170:

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

(i) For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$ i.e. $\frac{3}{k}\ne \frac{1}{2}⇒k\ne 6$
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.

(ii) In order that the given equations have no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{3}{k}=\frac{1}{2}\ne \frac{-1}{-5}$

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

#### Page No 170:

Let
Then, $\angle C=3\angle B=\left(3y\right)°$
Now, we have:
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$
x + y + 3y = 180
x + 4y = 180              ....(i)
Also, $\angle C=2\left(\angle A+\angle B\right)$
⇒ 3y = 2(x + y)
⇒ 2xy = 0                  ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0                    ....(iii)
On adding (i) and (iii), we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
x = 20 and y = 40
∴ ​

#### Page No 170:

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:

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

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

 x −1 2 5 y −1 1 3

Now, plots the points A(−1, −1), B(2, 1) and C(5, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both the sides.
Thus, the line AC is the graph of 2x − 3y = 1.

Graph of 4x − 3y + 1 = 0
4x − 3y + 1 = 0
⇒ 3y = ( 4x + 1)
$y=\frac{4x+1}{3}$                           ...(ii)
Putting x = −1, we get:
y = −1
Putting x = 2, we get:
y = 3
Putting x = 5, we get:
y = 7
Thus, we have the following table for the equation 4x − 3y + 1 = 0.
 x −1 2 5 y −1 3 7
Now, plots the points P(2, 3) and Q(5, 7). The point A(−1, −1) has already been plotted. Join PA and QP to get the graph line AQ. Extend it on both the sides.
Thus, the line AQ is the graph of the equation 4x − 3y + 1 = 0.

The two lines intersect at A(−1, −1).
Thus, x = −1 and y = −1 is the solution of the given system of equations.

#### Page No 170:

Given:
In a cyclic quadrilateral ABCD, we have:
$\angle A=\left(4x+20\right)°$
$\angle B=\left(3x-5\right)°$
$\angle C=\left(4y\right)°$
$\angle D=\left(7y+5\right)°$
$\angle A+\angle C=180°$ and $\angle \mathrm{B}+\angle \mathrm{D}=180°$      [Since ABCD is a cyclic quadrilateral]
Now, $\angle A+\angle C=\left(4x+20\right)°+\left(4y\right)°=180°$
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y  = 180 − 20 = 160
x + y = 40                       ....(i)
Also, $\angle B+\angle D=\left(3x-5\right)°+\left(7y+5\right)°=180°$
⇒ 3x + 7y = 180                 ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 120                     ....(iii)
On subtracting (iii) from (ii), we get:
4y = 60 ⇒ y = 15
On substituting y = 15 in (1), we get:
x + 15 = 40 ⇒ x = (40 − 15) = 25
Therefore, we have:
$\angle A=\left(4x+20\right)°=\left(4×25+20\right)°=120°$
$\angle B=\left(3x-5\right)°=\left(3×25-5\right)°=70°$
$\angle C=\left(4y\right)°=\left(4×15\right)°=60°$
$\angle D=\left(7y+5\right)°=\left(7×15+5\right)°=\left(105+5\right)°=110°$

#### Page No 170:

We have:

Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$:
35u + 14v 19 = 0                    ....(i)
14u + 35v 37 = 0                    ....(ii)
Here, a1 = 35, b1 = 14, c1 = 19, a2 = 14, b2 = 35, c2 = 37
By cross multiplication, we have:

$\frac{u}{-518+665}=\frac{v}{-266+1295}=\frac{1}{1225-196}$
$\frac{u}{147}=\frac{v}{1029}=\frac{1}{1029}$
$u=\frac{147}{1029}=\frac{1}{7},v=\frac{1029}{1029}=1$
$\frac{1}{x+y}=\frac{1}{7},\frac{1}{x-y}=1$
∴ (x + y) = 7                          ....(iii)
And, (x − y) = 1                      ....(iv)

Again, the equations (iii) and (iv) can be written as follows:
x + y 7 = 0                               ....(v)
x y 1 = 0                               ....(vi)
Here, a1 =  1, b1 = 1, c1 = 7 , a2 = 1 , b2 = 1 , c2 = 1
By cross multiplication, we have:

$\frac{x}{\left[1×\left(-1\right)-\left(-1\right)×\left(-7\right)\right]}=\frac{y}{\left[\left(-7\right)×1-\left(-1\right)×1\right]}=\frac{1}{\left[1×\left(-1\right)-1×1\right]}$

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

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