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#### Question 1:

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

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

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

 x 1 −2 4 y 0 2 −2

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

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

#### Question 2:

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

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

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

 x 0 2 − 2 y 2 − 1 5

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

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

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

#### Question 3:

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

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 xy + 1 = 0

xy = −1
y = x + 1
Putting x = 0, we get y = 1
Putting x =  −1, we get y = 0
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 0 3

Now, plot the points A(0, 1), B( −1, 0) and C(2, 3) 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  xy + 1 = 0.

Graph of 3x + 2y − 12 = 0
3x + 2y = 12
⇒ 2y = (12 − 3x)

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 QC and extend it on both ways.
Thus, PQ is the graph of  3x + 2y − 12 = 0. The two graph lines intersect at C(2, 3).
x = 2 and y = 3 are the solutions of the given system of equations.

#### Question 4:

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

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

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

 x −1 2 5 y 2 0 −2

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

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

#### Question 5:

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

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 = 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, plot 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 ways.
Thus, AC is the graph of  2x − 3y = 1.

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

#### Question 6:

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

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 = 5

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

 x −1 2 5 y 3 −1 −5

Now, plot the points A(−1 , 3) , B(2 , −1) and C(5, −5) 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 = 5.

Graph of 2y x = 7
2y x = 7
⇒ 2y = (7 + x)
$y=\frac{7+x}{2}$.............(ii)
Putting x = −1, we get y = 3.
Putting x = 3, we get y = 5.
Putting x = −3, we get y = 2.
Thus, we have the following table for the equation 2yx = 7.
 x −1 3 −3 y 3 5 2
Now, plot the points P(3, 5), Q(−3, 2). The point A(−1, 3) has already been plotted. Join PA and QA and extend it on both ways.
Thus, PQ is the graph of  2yx = 7. The two graph lines intersect at A(−1 , 3).
∴ The solution of the given system of equations is x = −1 and y = 3.

#### Question 7:

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

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

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

 x −2 0 2 y 0 −1 −2

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

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

#### Question 8:

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

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

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

 x 1 −5 7 y 2 6 −2

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

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

#### Question 9:

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

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

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

 x −2 3 8 y 0 2 4

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

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

#### Question 10:

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

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

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

 x 0 −1 1 y −1 2 −4

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

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

#### Question 11:

Check graphically whether the pair of equations  is consistent. Also, find the coordinates of the points, where the graphs of the equations meet the y-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 3x − 2y + 2 = 0

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

 x 0 2 −2 y 1 4 −2

Now, plot the points A(0, 1), B(2 , 4) and C(−2, −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 3x − 2y + 2 = 0 and it meets the y-axis at A(0, 1).

Graph of $\frac{3}{2}x-y+3=0$
$\frac{3}{2}x-y+3=0$
$y=\frac{3}{2}x+3$...............(ii)
Putting x = 0, we get y = 3.
Putting x = 2, we get y = 6.
Putting x = −2, we get y = 0.
Thus, we have the following table for the equation $\frac{3}{2}x-y+3=0$.
 x 0 2 −2 y 3 6 0
Now, plot the points P(0, 3), Q(2, 6) and R (−2, 0) on the same graph paper as above. Join PQ and PR to get the graph line QR. Extend it on both ways.
Then, QR is the graph of the equation $\frac{3}{2}x-y+3=0$ and it meets the y-axis at P(0, 3). It is clear from the graph that the two graph lines are parallel and do not intersect even when produced.
Hence, the given system of equations has no solutions.

#### Question 12:

Represent the following system of linear equations graphically. From the graph, find the points, where the lines intersect y-axis:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 3x + y − 5 = 0

3x + y − 5 = 0
y = (−3x + 5) ...........(i)
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 3x + y − 5 = 0.

 x 0 1 2 y 5 2 −1

Now, plot the points A(0, 5), B(1 , 2) and C(2, −1) 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  3x + y − 5 = 0.

Graph of 2xy − 5 = 0
2xy − 5 = 0
y = (2x − 5) ............(ii)
Putting x = 0, we get y = −5.
Putting x = 1, we get y = −3.
Putting x = 3, we get y = 1.
Thus, we have the following table for the equation 2xy − 5 = 0.
 x 0 1 3 y −5 −3 1
Now, plot the points P(0, −5), Q(1, −3) and R( 3, 1) on the same graph paper. Join PQ and QR to get the graph line PR. Extend it on both ways.
Then, PR is the graph of the equation 2xy − 5 = 0. Hence, the lines (i) and (ii) intersect y-axis at (0, 5) and (0, −5), respectively.

#### Question 13:

Check graphically whether the pair of equations  is consistent. Also, find the coordinates of the points, where the graphs of the equations meet the y-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 3x + 5y = 15

3x + 5y = 15
⇒ 5y = (15 − 3x)
$y=\frac{15-3x}{5}$ ...........(i)
Putting x = 0, we get y = 3.
Putting x = 5, we get y = 0.
Putting x = −5, we get y = 6.
Thus, we have the following table for the equation 3x + 5y = 15.

 x 0 5 −5 y 3 0 6

Now, plot the points A(0, 3), B(5, 0) and C(−5, 6) 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 + 5y = 15.

Graph of xy = 5
xy = 5
y = (x − 5) ............(ii)
Putting x = 0, we get y = −5.
Putting x = 5, we get y = 0.
Putting x = 2, we get y = −3.
Thus, we have the following table for the equation xy  = 5.
 x 0 5 2 y −5 0 −3
Now, plot the points P(0, −5) and Q(2, −3). The point B(0, 5) has already been plotted. Join PQ and QB to get the graph line PB. Extend it on both ways.
Then, PB is the graph of the equation xy = 5. It is clear from the graph that the given system of equation is consistent.
Hence, the lines (i) and (ii) intersect y-axis at A(0, 3) and P(0, −5), respectively.

#### Question 14:

Solve the following system of equations graphically:

Also, find the points, where these lines meet the x-axis.

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

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

 x 1 3 5 y 2 1 0

Now, plot the points A(1, 2), B(3, 1) and C(5, 0) 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 = 5.

Graph of 2x − 3y = −4
2x − 3y = −4
⇒ 3y = (2x + 4)
$y=\frac{2x+4}{3}$............(ii)
Putting x = 1, we get y = 2.
Putting x = −2, we get y = 0.
Putting x = 4, we get y = 4.
Thus, we have the following table for the equation 2x − 3y = −4.
 x 1 −2 4 y 2 0 4
Now, plot the points P(4, 4) and Q(−2, 0). The point A(1, 2) has already been plotted. Join PA and QA to get the graph line PQ. Extend it on both ways.
Then, PQ is the graph of the equation 2x − 3y = −4. The two graph lines intersect at A(1, 2).
∴ The solution of the given system of equations x = 1 and y = 2.
Hence, the lines (i) and (ii) intersect x-axis at Q(−2, 0) and C(5, 0), respectively.

#### Question 15:

Solve the following system of equations graphically:

Determine the vertices of the triangle formed by these lines and the x-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 4x − 5y + 16 = 0

4x − 5y + 16 = 0
⇒ 5y = (4x +16)
$y=\frac{4x+16}{5}$ ...........(i)
Putting x = 1, we get y = 4.
Putting x = −4, we get y = 0.
Putting x = 6, we get y = 8.
Thus, we have the following table for the equation 4x − 5y + 16 = 0.

 x 1 −4 6 y 4 0 8

Now, plot the points A(1, 4), B(−4, 0) and C(6, 8) 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 4x − 5y + 16 = 0.

Graph of 2x + y − 6 = 0
2x + y − 6 = 0
y = (−2x + 6) ...........(ii)
Putting x = 1, we get y = 4.
Putting x = 3, we get y = 0.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 2x + y − 6 = 0.
 x 1 3 2 y 4 0 2
Now, plot the points P(3, 0) and Q(2, 2). The point A(1, 4) has already been plotted. Join PQ and QA to get the graph line PA. Extend it on both ways.
Then, PA is the graph of the equation 2x + y − 6 = 0. The two graph lines intersect at A(1, 4).
∴ The solution of the given system of equations is x = 1 and y = 4.
Hence, these lines form ΔBAP with the x-axis, whose vertices are B(−4, 0), A(1, 4) and P(3, 0).

#### Question 16:

Solve the following system of linear equations graphically:

Shade the region between the lines and the x-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x − 3y − 17 = 0

2x − 3y − 17 = 0
⇒ 3y = (2x − 17)
$y=\frac{2x-17}{3}$ ...........(i)
Putting x = 1, we get y = −5.
Putting x = 4, we get y = −3.
Putting x = 7, we get y = −1.
Thus, we have the following table for the equation 2x − 3y − 17 = 0.

 x 1 4 7 y −5 −3 −1

Now, plot the points A(1, −5), B( 4, −3) and C(7, −1) 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 − 17 = 0.

Graph of 4x + y − 13 = 0
4x + y − 13 = 0
y = (−4x + 13) ...........(ii)
Putting x = 4, we get y = −3.
Putting x = 2, we get y = 5.
Putting x = 3, we get y = 1.
Thus, we have the following table for the equation 4x + y − 13 = 0.
 x 4 2 3 y −3 5 1
Now, plot the points P(2, 5) and Q(3, 1). The point B(4, −3) has already been plotted. Join PQ and QB to get the graph line PB. Extend it on both ways.
Then, PB is the graph of the equation 4x + y − 13 = 0. The two graph lines intersect at B(4, −3).
∴ The solution of the given system of equations is x = 4 and y = −3.
These graph lines intersect the x-axis at R and S.
Hence, the region bounded by these lines and the x-axis has been shaded.

#### Question 17:

Solve the following system of equations graphically:

Shade the region bounded by these lines and the y-axis.

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

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

 x 0 1 2 y −4 0 4

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

Graph of 3x + 2y = 14
3x + 2y = 14
⇒ 2y = (14 − 3x)
$y=\frac{14-3x}{2}$  ...........(ii)
Putting x = 0, we get y = 7.
Putting x = 2, we get y = 4.
Putting x = 4 we get y = 1.
Thus, we have the following table for the equation 3x + 2y = 14.
 x 0 2 4 y 7 4 1
Now, plots the points P(0, 7) and Q(4, 1). The point C(2, 4) 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 = 14. The two graph lines intersect at C(2, 4).
∴ The solution of the given system of equations is x = 2 and y = 4.

These graph lines intersect the y-axis at P and A.
Hence, the region bounded by these lines and the y-axis has been shaded.

#### Question 18:

Solve the following system of linear equations graphically:

Shade the region bounded by these lines and the y-axis.

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

 x 1 2 0 y 1 3 −1

Now, plot the points A(1, 1), B( 2, 3) and C(0, −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 2xy = 1.

Graph of xy = −1
xy = −1
y = (x + 1) ............(ii)
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Putting x = 0, we get y = 1.
Thus, we have the following table for the equation xy = −1.
 x 1 2 0 y 2 3 1
Now, plot the points P(1, 2) and Q(0, 1). The point B(2, 3) has already been plotted. Join PB and PQ to get the graph line BQ. Extend it on both ways.
Then, BQ is the graph of the equation xy = −1. The two graph lines intersect at B(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
These graph lines intersect the y-axis at C and Q.
Hence, the region bounded by these lines and the y-axis has been shaded.

#### Question 19:

Solve the following system of linear equations graphically:

Find the points, were these lines meet the y-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x − 5y + 4 = 0
2x − 5y + 4 = 0
⇒ 5y = (2x + 4)
$y=\frac{2x+4}{5}$   ............(i)
Putting x = 3, we get y = 2.
Putting x = −2, we get y = 0.
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 = (−2x + 8) ............(ii)
Putting x = 3, we get y = 2.
Putting x = 1, we get y = 6.
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), Q(2, 4). The point B(3, 2) has already been plotted. Join PQ and QB to get the graph line PB. Extend it on both ways.
Then, PB is the graph of the equation 2x + y − 8 = 0. The two graph lines intersect at B(3, 2).
∴ The solution of the given system of equations is x = 3 and y = 2.
These graph lines intersect the y-axis at R(0,0.8) and S(0, 8).

#### Question 20:

Solve graphically the system of equations

Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 4x − 5y − 20 = 0
4x − 5y − 20 = 0
⇒ 5y = (4x − 20)
$y=\frac{4x-20}{5}$............(i)
Putting x = 0, we get y = −4.
Putting x = 5, we get y = 0.
Putting x = 10, we get y = 4.
Thus, we have the following table for the equation 4x − 5y − 20 = 0.

 x 0 5 10 y −4 0 4

Now, plot the points A(0, −4), B( 5, 0) and C(10, 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  4x − 5y − 20 = 0.

Graph of 3x + 5y − 15 = 0
3x + 5y − 15 = 0
⇒ 5y = (−3x + 15)
$y=\frac{-3x+15}{5}$  ............(ii)
Putting x = 5, we get y = 0.
Putting x = 0, we get y = 3
Putting x = −5, we get y = 6.
Thus, we have the following table for the equation 3x + 5y − 15 = 0.
 x 5 0 −5 y 0 3 6
Now, plot the points P(0, 3) and Q(−5 , 6). The point B(5, 0) has already been plotted. Join PQ and PB to get the graph line QB. Extend it on both ways.
Then, QB is the graph of the equation 3x + 5y − 15 = 0. The two graph lines intersect at B(5, 0).
∴ The solution of the given system of equations is x = 5 and y = 0.
Clearly, the vertices of ΔPBA formed by these two lines and y-axis are P(0, 3), B(5, 0) and A(0, −4).

#### Question 21:

Solve the following system of linear equations graphically:

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

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

 x −1 2 5 y 0 4 8

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

Graph of 4x + 3y − 20 = 0
4x + 3y − 20 = 0
⇒ 3y = (−4x + 20)
$y=\frac{-4x+20}{3}$ ............(ii)
Putting x = 2, we get y = 4.
Putting x = −1, we get y = 8.
Putting x = 5, we get y = 0.
Thus, we have the following table for the equation 4x + 3y − 20 = 0.
 x 2 −1 5 y 4 8 0
Now, plot the points P(1, −8) and Q(5 , 0). The point B(2, 4) has already been plotted. Join PB and QB to get the graph line PQ. Extend it on both ways.
Then, line PQ is the graph of the equation 4x + 3y − 20 = 0. The two graph lines intersect at B(2, 4).
∴ The solution of the given system of equations is x = 2 and y = 4.
Clearly, the vertices of ΔABQ formed by these two lines and x-axis are Q(5, 0), B(2, 4) and A(−1, 0).
Now, consider ΔABQ.
Here, height = 4 units and base (AQ) = 6 units
∴ Area of ΔABQ = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×6×4\right)=12$ sq. units.

#### Question 22:

Solve the following system of linear equations graphically:

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

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

 x −1 1 2 y 0 2 3

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

Graph of 3x + 2y − 12 = 0
3x + 2y − 12 = 0
⇒ 2y = (−3x + 12)
$y=\frac{-3x+12}{2}$  ............(ii)
Putting x = 0, we get y = 6.
Putting x = 2, we get y = 3.
Putting x = 4, we get y = 0.
Thus, we have the following table for the equation 3x + 2y − 12 = 0.
 x 0 2 4 y 6 3 0
Now, plot the points P(0, 6) and Q(4 , 0). The point C(2, 3) has already been plotted. Join PC and CQ to get the graph line PQ. Extend it on both ways.
Then, PQ is the graph of the equation 3x + 2y − 12 = 0. The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔACQ formed by these two lines and the x-axis are Q(4, 0), C(2, 3) and A(−1, 0).
Now, consider ΔACQ.
Here, height = 3 units and base (AQ) = 5 units
Area of ΔACQ = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×5×3\right)=7.5$ sq. units.

#### Question 23:

Solve graphically the system of equations:

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

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

 x 0 1 2 y −7 −2 3

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

Graph of xy + 1 = 0
xy + 1 = 0
y = (x + 1) ...........(ii)
Putting x = 0, we get y = 1.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation xy + 1 = 0.
 x 0 1 2 y 1 2 3
Now, plot the points P(0, 1) and Q(1, 2). The point C(2, 3) has already been plotted. Join PQ and QC to get the graph line PC. Extend it on both ways.
Then, PC is the graph of the equation xy + 1 = 0. The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔAPC formed by these two lines and the y-axis are P(0, 1), C(2, 3) and A(0, −7).
Now, consider ΔAPC.
Here, height = 2 units and base (AP) = 8 units
∴ Area of ΔAPC = $\frac{1}{2}×\mathrm{base}×\mathrm{height}$ sq. units
= $\left(\frac{1}{2}×8×2\right)=8$ sq. units.

#### Question 24:

Show graphically that the system of equations  is consistent.

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
x − 2y = 2
⇒ 2y = (x − 2)
$y=\frac{x-2}{2}$.........(i)
Putting x = 0, we get y =  −1.
Putting x = 2, we get y = 0.
Putting x = 1, we get y = −0.5.
Thus, we have the following table for the equation x − 2y = 2.

 x 0 2 1 y −1 0 −0.5

Now, plot the points A(0, −1), B( 2, 0) and C(1, −0.5) on the graph paper.
Join AC and BC to get the graph line AB. Extend it on both ways.
Thus, AB is the graph of  x − 2y = 2.

Graph of 4x − 2y = 5
4x − 2y = 5
⇒ 2y = (4x − 5)
$y=\frac{4x-5}{2}$...........(ii)
Putting x = 0, we get y = −2.5.
Putting x = 1, we get y = −0.5.
Putting x = 2, we get y = 1.5.
Thus, we have the following table for the equation 4x − 2y = 5.
 x 0 1 2 y −2.5 −0.5 1.5
Now, plot the points P(0, −2.5) and Q(2, 1.5). The point C(1, −0.5) 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 4x − 2y = 5. The two graph lines intersect at C(1, −0.5).
Hence, the given system of equations is consistent.

#### Question 25:

Show graphically that the system of equations is inconsistent.

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

 x 2 −1 −4 y 0 2 4

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

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

Now, on the same graph, plot the points P(3, 0), Q(0, 2) and R (6, −2).
Join PQ and PR to get the graph line QR. Extend it on both ways.
Then, QR is the graph of the equation 4x + 6y = 12. It is clear from the graph that these two lines are parallel and do not intersect when produced.
Hence, the given system of equations is inconsistent.

#### Question 26:

Show graphically that the system of equations is inconsistent.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2y x = 9
2y x = 9
⇒ 2y = (x + 9)
$y=\frac{x+9}{2}$.........(i)
Putting x = 1, we get y = 5.
Putting x = −1, we get y = 4.
Putting x = −3, we get y = 3.
Thus, we have the following table for the equation 2y x = 9.

 x 1 −1 −3 y 5 4 3

Now, plot the points A(1, 5), B( −1, 4) and C(−3, 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  2y x = 9.

Graph of 4y − 2x = 20
4y − 2x = 20
⇒ 4y = (2x + 20)
$y=\frac{2x+20}{4}$...........(ii)
Putting x = 0, we get y = 5.
Putting x = 2, we get y = 6.
Putting x = −2, we get y = 4.
Thus, we have the following table for the equation 4y − 2x = 20.
 x 0 2 −2 y 5 6 4

Now, on the same graph, plots the points P(0, 5), Q(2, 6) and R(−2, 4).
Join PQ and PR to get the graph line QR. Extend it on both ways.
Then, QR is the graph of the equation 4y − 2x = 20. It is clear from the graph that these two lines are parallel and do not intersect when produced.
Hence, the given system of equations is inconsistent.

#### Question 27:

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

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

 x 1 0 2 y −2 −5 1

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

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

Thus, we have the following table for the equation 6x − 2y = 10.
 x 0 1 2 y −5 −2 1
These are the same points as obtained for the graph line of equation (i). It is clear from the graph that these two lines coincide.
Hence, the given system of equations has infinitely many solutions.

#### Question 28:

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

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

 x 3 1 2 y 0 4 2

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

Graph of 6x + 3y = 18
6x + 3y = 18 ⇒ 3y = (18 − 6x)
$y=\frac{18-6x}{3}$...........(ii)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 6x + 3y = 18.
 x 3 1 2 y 0 4 2
These are the same points as obtained for the graph line of equation (i). It is clear from the graph that these two lines coincide.
Hence, the given system of equations has an infinite number of solutions.

#### Question 1:

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

The given system of equation is:
x + y = 8 .........(i)
2x − 3y = 1 ........(ii)

On multiplying (i) by 3, we get:
3x + 3y = 24 .......(iii)

On adding (ii) and (iii), we get:
5x = 25
x = 5

On substituting the value of x = 5 in (i), we get:
5 + y = 8
y = (8 − 5) = 3

Hence, the solution is x = 5 and y = 3.

#### Question 2:

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

The given system of equation is:
xy = 3 .........(i)
3x − 2y = 10 ........(ii)

On multiplying (i) by 2, we get:
2x − 2y = 6 .......(iii)

On subtracting (iii) and (ii), we get:
x = 4

On substituting the value of x = 4 in (i), we get:
4 − y = 3
y = (4 − 3) = 1

Hence, the solution is x = 4 and y = 1.

#### Question 3:

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

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

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

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

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

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

#### Question 4:

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

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

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

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

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

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

#### Question 5:

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

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

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

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

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

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

#### Question 6:

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

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

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

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

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

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

#### Question 7:

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

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

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

#### Question 8:

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

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

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

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

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

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

#### Question 9:

$11x+15y+23=0,\phantom{\rule{0ex}{0ex}}7x-2y-20=0.$

The given equations are:
11x + 15y + 23 = 0 ........(i)
7x − 2y − 20 = 0 .............(ii)

On multiplying (i) by 2 and (ii) by 15, we get:
22x + 30y + 46 = 0 .......(iii)
105x − 30y − 300 = 0 .......(iv)

On adding (iii) and (iv), we get:
127x = 254
x = 2

On substituting x = 2 in (i), we get:
22 + 15y + 23 = 0
⇒ 15y = (−23 − 22) = −45
y = −3

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

#### Question 10:

$2x-5y+8=0,\phantom{\rule{0ex}{0ex}}x-4y+7=0.$

The given equations are:
2x − 5y + 8 = 0   ........(i)
x − 4y + 7 = 0  .............(ii)

On multiplying (i) by 4 and (ii) by 5, we get:
8x − 20y + 32 = 0 ........(iii)
5x − 20y + 35 = 0  .............(iv)

On subtracting (iii) from (iv), we get:
−3x = −3
x = 1

On substituting x = 1 in (i), we get:
2 − 5y + 8 = 0
⇒ 5y = 10
y = 2

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

#### Question 11:

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

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

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

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

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

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

#### Question 12:

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

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

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

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

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

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

#### Question 13:

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

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

and $\frac{5x}{6}-\frac{y}{3}+7=0$
⇒ 5x − 2y = −42..........(ii)

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

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

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

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

#### Question 14:

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

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

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

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

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

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

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

#### Question 15:

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

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

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

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

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

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

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

#### Question 16:

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

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

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

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

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

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

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

#### Question 17:

$.8x+.3y=3.8,\phantom{\rule{0ex}{0ex}}.4x-.5y=0.6.$

The given equations are:
.8x + .3y = 3.8 ........(i)
.4x − .5y = 0.6 ..........(ii)

On multiplying (i) by 10 and (ii) by 10, we get:
8x + 3y = 38 ..........(iii)
4x − 5y = 6 .........(iv)

On multiplying (iii) by 5 and (iv) by 3, we get:
40x + 15y = 190 .......(v)
12x − 15y = 18 ..........(vi)

On adding (v) from (vi), we get:
52x = 208
x = 4

On substituting x = 4 in (iii), we get:
32 + 3y = 38
⇒ 3y = (38 − 32) = 6
y = 2

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

#### Question 18:

$.05x+.2y=.07,\phantom{\rule{0ex}{0ex}}.3x-.1y=.03.$

The given equations are:
.05x + .2y = .07 ........(i)
.3x − .1y = .03 ..........(ii)

On multiplying (i) by 100 and (ii) by 100, we get:
5x + 20y = 7..........(iii)
30x − 10y = 3.........(iv)

On multiplying (iv) by 2, we get:
60x − 20y = 6 ..........(v)

On adding (v) from (vi), we get:
65x = 13
$x=\frac{13}{65}=\frac{1}{5}=0.2$

On substituting x = 0.2 in (iii), we get:
1 + 20y = 7
⇒ 20y = (7 − 1) = 6
$y=\frac{6}{20}=\frac{3}{10}=0.3$

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

#### Question 19:

$mx-ny={m}^{2}+{n}^{2},\phantom{\rule{0ex}{0ex}}x+y=2m.$

The given equations are:
mxny = m2 + n2........(i)
x + y = 2m ........(ii)

On multiplying (ii) by n, we get:
nx + ny = 2mn .......(iii)

On adding (i) and (iii), we get:
mx + nx = m2 + n2 + 2mn
⇒ (m + n)x = (m + n)2
$x=\frac{{\left(m+n\right)}^{2}}{\left(m+n\right)}=m+n$

On substituting x = m + n in (i), we get:
m(m + n) − ny = m2 + n2
m2 + mnny = m2 + n2
⇒ −ny = n2mn
$y=\frac{n\left(n-m\right)}{-n}=m-n$

Hence, the solution is x = (m + n) and y = (mn).

#### Question 20:

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

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

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

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

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

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

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

#### Question 21:

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

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

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

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

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

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

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

#### Question 22:

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

The given equations are:

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

Also, x + y =  2ab........(ii)

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

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

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

#### Question 23:

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

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

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

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

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

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

#### Question 24:

$2\left(ax-by\right)+\left(a+4b\right)=0,\phantom{\rule{0ex}{0ex}}2\left(bx+ay\right)+\left(b-4a\right)=0.$

The given equations are:
2(axby) + (a + 4b) = 0
⇒ 2ax − 2by + a + 4b = 0 ...............(i)

and 2(bx + ay) + b − 4a = 0
⇒ 2bx + 2ay + b − 4a = 0 ...................(ii)

On multiplying (i) by a and (ii) by b, we get:
2a2x − 2aby + a2 + 4ab = 0  ................(iii)
2b2x + 2aby + b2 − 4ab  = 0  ....................(iv)

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

On substituting $x=\frac{-1}{2}$ in (i), we get:
$2a×\frac{-1}{2}-2by+a+4b=0$
⇒ −2by + 4b  = 0
⇒ 2by = 4b
y = $\frac{4b}{2b}=2$

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

#### Question 25:

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

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

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

On subtracting (ii) from (i), we get:
34x − 34y = −34
⇒ 34(xy) = −34
⇒ (xy) = −1...........(iv)

On adding (iii) and (iv), we get:
2x = 5 − 1= 4
x = 2

On subtracting (iv) from (iii), we get:
2y = 5 + 1 = 6
y = 3

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

#### Question 26:

$37x+43y=123,\phantom{\rule{0ex}{0ex}}43x+37y=117.$

The given equations are:

37x + 43y = 123 ...........(i)
43x + 37y = 117 .............(ii)

On adding (i) and (ii), we get:
80x + 80y = 240
⇒80(x + y) = 240
⇒ (x + y) = 3 .............(iii)

On subtracting (i) from (ii), we get:
6x − 6y = −6
⇒ 6(xy) = −6
⇒ (xy)  = −1 ............(iv)

On adding (iii) and (iv), we get:
2x = 3 − 1 = 2 ⇒ x = 1

On subtracting (iv) from (iii), we get:
2y = 3 + 1 = 4
y = 2

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

#### Question 27:

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

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

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

On subtracting (ii) from (i), we get:
86x − 86y = 86
⇒ 86(xy) = 86
xy = 1 ...............(iv)

On adding (iii) and (iv), we get :
2x = 6 ⇒ x = 3

On substituting x = 3 in (iii), we get:
3 + y = 5
y = 5 − 3 = 2

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

#### Question 28:

$41x-17y=99,\phantom{\rule{0ex}{0ex}}17x-41y=75.$

The given equations are:
41x − 17y = 99 .............(i)
17x − 41y = 75 .............(ii)
On adding (i) and (ii), we get:
58x − 58y = 174
⇒ 58(x y) = 174
xy = 3 ............(iii)

On subtracting (ii) from (i), we get:
24x + 24y = 24
⇒ 24(x + y) = 24
x + y = 1 ...............(iv)

On adding (iii) and (iv), we get :
2x = 4 ⇒ x = 2

On substituting x = 2 in (iii), we get:
2 − y = 3
y = 2 − 3 = −1

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

#### Question 29:

$\frac{4}{x}+3y=8,\phantom{\rule{0ex}{0ex}}\frac{6}{x}-4y=-5.$

The given equations are:
$\frac{4}{x}+3y=8$ ............(i)
$\frac{6}{x}-4y=-5$ .............(ii)
Putting $\frac{1}{x}=u$, we get:
4u + 3y = 8 .............(iii)
6u − 4y = −5 ...........(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:
16u + 12y = 32 ...........(v)
18u − 12y = −15 ............(vi)

On adding (v) and (vi), we get:
34u = 17
$⇒u=\frac{17}{34}=\frac{1}{2}$
$⇒\frac{1}{x}=\frac{1}{2}⇒x=2$

On substituting x = 2 in (i), we get:
$\frac{4}{2}+3y=8$
⇒ 2 + 3y = 8
⇒ 3y = (8 − 2) = 6
y = 2
Hence, the required solution is x = 2 and y = 2.

#### Question 30:

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

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

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

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

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

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

#### Question 31:

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

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

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

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

#### Question 32:

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

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

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

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

#### Question 33:

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

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

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

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

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

#### Question 34:

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

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

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

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

#### Question 35:

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

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

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

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

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

#### Question 36:

The given equations are:
$\frac{1}{7x}+\frac{1}{6y}=3\phantom{\rule{0ex}{0ex}}$............(i)
$\frac{1}{2x}-\frac{1}{3y}=5$ .............(ii)

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
$\frac{u}{7}+\frac{v}{6}=3$.............(iii)
On multiplying (iii) by 42, we get:
6u + 7v = 126 .........(iv)

Again, $\frac{u}{2}-\frac{v}{3}=5$ .............(v)

On multiplying (v) by 6, we get:
3u − 2v = 30 .............(vi)

On multiplying (iv) by 2 and (vi) by 7, we get:
12u + 14v = 252 ..............(vii)
21u − 14v = 210 ................(viii)

On adding (vii) and (viii), we get:
33u = 462 ⇒ u = 14
$⇒\frac{1}{x}=14⇒x=\frac{1}{14}$

On substituting $x=\frac{1}{14}$ in (i), we get:
$\frac{1}{7×\frac{1}{14}}+\frac{1}{6y}=3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$2+\frac{1}{6y}=3⇒\frac{1}{6y}=\left(3-2\right)=1\phantom{\rule{0ex}{0ex}}$
$⇒y=\frac{1}{6}$

Hence, the required solution is $x=\frac{1}{14}$ and $y=\frac{1}{6}$.

#### Question 37:

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

From equation (i), we have:

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

For equation (ii), we have:

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

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

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

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

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

#### Question 38:

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

From equation (i), we have:
$\frac{3x+9y}{xy}=11\phantom{\rule{0ex}{0ex}}$
$⇒\frac{3x}{xy}+\frac{9y}{xy}=11⇒\frac{3}{y}+\frac{9}{x}=11$ .........(iii)

From equation (ii), we have:

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

On substituting $\frac{1}{y}=v$ and $\frac{1}{x}=u$ , we get:
3v + 9u = 11 ...........(v)
6v + 3u = 7 ...........(vi)

On multiplying (vi) by 3, we get:
18v + 9u = 21 ..............(vii)

On subtracting (v) 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 (iv), we get:
$\frac{6}{\left(3}{2}\right)}+\frac{3}{x}=7\phantom{\rule{0ex}{0ex}}$
$⇒4+\frac{3}{x}=7⇒\frac{3}{x}=3⇒3x=3\phantom{\rule{0ex}{0ex}}$
$⇒x=1$
Hence, the required solution is x = 1 and $y=\frac{3}{2}$.

#### Question 39:

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

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

From equation (i), we have:

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

From equation (ii), we have:

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

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

#### Question 40:

The given equations are:
$\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ ............(i)

$\frac{4}{x}+\frac{9}{y}=\frac{21}{xy}$............(ii)
On multiplying (i) and (ii) by xy, we get:
$\frac{2xy}{x}+\frac{3xy}{y}=9\phantom{\rule{0ex}{0ex}}$
⇒ 2y + 3x = 9 .............(iii)
and $\frac{4xy}{x}+\frac{9xy}{y}=21\phantom{\rule{0ex}{0ex}}$
⇒ 4y + 9x = 21 ...........(iv)
On multiplying (iii) by 3, we get:
6y + 9x = 27 .............(v)
On subtracting (iv) from (v), we get:
2y = 6 ⇒ y = 3
On substituting y = 3 in (iii), we get:
$2×3+3x=9\phantom{\rule{0ex}{0ex}}$
$6+3x=9⇒3x=\left(9-6\right)=3\phantom{\rule{0ex}{0ex}}$
$⇒x=\frac{3}{3}=1$
Hence, the required solution is x = 1 and  y = 3.

#### Question 41:

The given equations are:
$\frac{x+y}{xy}=2\phantom{\rule{0ex}{0ex}}$
$⇒\frac{x}{xy}+\frac{y}{xy}=2\phantom{\rule{0ex}{0ex}}$
$⇒\frac{1}{y}+\frac{1}{x}=2$ ...........(i)

and $\frac{x-y}{xy}=6\phantom{\rule{0ex}{0ex}}$
$⇒\frac{x}{xy}-\frac{y}{xy}=6\phantom{\rule{0ex}{0ex}}$
$⇒\frac{1}{y}-\frac{1}{x}=6$ ...........(ii)
Putting $\frac{1}{y}=v$ and $\frac{1}{x}=u$, we get:
v + u = 2 ..........(iii)
vu = 6 .............(iv)
On adding (iii) and (iv), we get:
2v = 8 ⇒ v = 4
$⇒\frac{1}{y}=4⇒y=\frac{1}{4}$
On substituting $y=\frac{1}{4}$ in (i), we get:
$\frac{1}{1}{4}}+\frac{1}{x}=2\phantom{\rule{0ex}{0ex}}$
$⇒4+\frac{1}{x}=2⇒\frac{1}{x}=\left(2-4\right)=-2\phantom{\rule{0ex}{0ex}}$
$⇒x=\frac{-1}{2}$
Hence, the required solution is $x=\frac{-1}{2}$ and $y=\frac{1}{4}$.

#### Question 42:

The given equations are:
$\frac{xy}{x+y}=\frac{6}{5}⇒6x+6y=5xy\phantom{\rule{0ex}{0ex}}$
$⇒\frac{6x+6y}{xy}=5⇒\frac{6x}{xy}+\frac{6y}{xy}=5$
$⇒\frac{6}{y}+\frac{6}{x}=5$
$⇒\frac{6}{x}+\frac{6}{y}=5$...........(i)
and $\frac{xy}{y-x}=6⇒6y-6x=xy\phantom{\rule{0ex}{0ex}}$
$⇒\frac{6y-6x}{xy}=1⇒\frac{6y}{xy}-\frac{6x}{xy}=1$
$⇒\frac{6}{x}-\frac{6}{y}=1$ ...........(ii)
Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
6u + 6v = 5 ..........(iii)
6u − 6v = 1 .............(iv)
On adding (iii) and (iv), we get:
12u = 6 ⇒ $u=\frac{1}{2}$
$⇒\frac{1}{x}=\frac{1}{2}⇒x=2$
On substituting x = 2 in (i), we get:
$\frac{6}{2}+\frac{6}{y}=5⇒3+\frac{6}{y}=5\phantom{\rule{0ex}{0ex}}$
$⇒\frac{6}{y}=\left(5-3\right)=2\phantom{\rule{0ex}{0ex}}$
$⇒2y=6⇒y=3$
Hence, the required solution is x = 2 and y = 3.

#### Question 43:

$\frac{x+1}{2}+\frac{y-1}{3}=8,\frac{x-1}{3}+\frac{y+1}{2}=9.$

The given equations are:
$\frac{x+1}{2}+\frac{y-1}{3}=8\phantom{\rule{0ex}{0ex}}$ ..............(i)
$\frac{x-1}{3}+\frac{y+1}{2}=9$ ...............(ii)
On multiplying each equation by 6, we get:
3(x + 1) + 2(y − 1) = 48
⇒ 3x + 3 + 2y − 2 = 48
⇒ 3x + 2y = 47 .............(iii)
Also, 2(x − 1) + 3(y + 1) = 54
⇒ 2x − 2 + 3y + 3 = 54
⇒ 2x + 3y = 53 ..........(iv)
On multiplying (iii) by 3 and (iv) by 2, we get:
9x + 6y = 141 .........(v)
4x + 6y = 106 ..........(vi)
On subtracting (vi) from (v), we get:
5x = (141 − 106) = 35
x = 7
On substituting x = 7 in (iii), we get:
3 × 7 + 2y = 47
⇒ 2y = (47 − 21) = 26
y = 13
Hence, required solution is x = 7 and y = 13.

#### Question 44:

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

#### Question 45:

The given equations are:
$\frac{40}{x+y}+\frac{2}{x-y}=5$             ...(i)
$\frac{25}{x+y}-\frac{3}{x-y}=1$             ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, we get:
40u + 2v = 5               ...(iii)
25u − 3v = 1               ...(iv)
On multiplying (iii) by 3 and (iv) by 2, we get:
120u + 6v = 15    ...(v)
50u − 6v = 2             ...(vi)
On adding (v) and (vi), we get:
170u = 17
$⇒u=\frac{17}{170}=\frac{1}{10}$
$⇒\frac{1}{x+y}=\frac{1}{10}⇒x+y=10$        ...(vii)
On substituting $u=\frac{1}{10}$ in (vi), we get:
5 − 6v = 2 ⇒ 6v = 3
$⇒v=\frac{3}{6}=\frac{1}{2}$
$⇒\frac{1}{x-y}=\frac{1}{2}⇒x-y=2$               ...(viii)
On adding (vii) and (viii), we get:
2x = 12
x = 6
On substituting x = 6 in (vii), we get:
6 + y = 10
y = (10 − 6) = 4
Hence, the required solution is x = 6 and y = 4.

#### Question 46:

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

Hence, the required solution is .

#### Question 47:

The given equations are:
$\frac{10}{x+y}+\frac{2}{x-y}=4$         ...(i)
$\frac{15}{x+y}-\frac{5}{x-y}=-2$        ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ , we get:
10u + 2v = 4             ...(iii)
15u − 5v = −2            ...(iv)
On multiplying (iii) by 5 and (iv) by 2, we get:
50u + 10v = 20          ...(v)
30u − 10v = −4           ...(vi)
On adding (v) and (vi), we get:
80u = 16
$⇒u=\frac{16}{80}=\frac{1}{5}$
$⇒\frac{1}{x+y}=\frac{1}{5}⇒x+y=5$        ...(vii)
On substituting $u=\frac{1}{5}$ in (iii), we get:
2 + 2v = 4
⇒ 2v = 2
$⇒v=1$
$⇒\frac{1}{x-y}=1⇒x-y=1$            ...(viii)
On adding (vii) and (viii), we get:
$2x=6\phantom{\rule{0ex}{0ex}}⇒x=3$
On substituting $x=3$ in (vii), we get:
$3+y=5⇒y=2$
Hence, the required solution is .

Note : There is an error in the question, the correct equations are:
$\frac{10}{x+y}+\frac{2}{x-y}=4$         ...(i)
$\frac{15}{x+y}-\frac{5}{x-y}=-2$        ...(ii)

#### Question 48:

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

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

#### Question 49:

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

The given equations are:
(ab)x + (a + b)y = a2 − 2abb2     ...(i)
(a + b)(x + y) = a2 + b2                  ...(ii)
From (ii), we have:
(a + b)x + (a + b)y = a2 + b2            ...(iii)
On subtracting (iii) from (i), we get:
(abab)x = (a2 − 2abb2 a2b2 )
⇒ −2bx = −2ab − 2b2
⇒ −2bx = −2b(a + b)
x = (a + b)
On substituting x = (a + b) in (i), we get:
(ab)(a + b) + (a + b)y = a2 − 2abb2
a2b2 + (a + b)y = a2 − 2abb2
⇒ (a + b)y = −2ab
$y=\frac{-2ab}{\left(a+b\right)}$
Hence, the required solution is $x=\left(a+b\right)$ and $y=\frac{-2ab}{\left(a+b\right)}$.

#### Question 50:

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

#### Question 1:

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

The given equations are:
x + 2y + 1 = 0       ...(i)
2x − 3y − 12 = 0      ...(ii)
Here, a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = −3 and c2 = −12
By cross multiplication, we have: $\frac{x}{\left[2×\left(-12\right)-1×\left(-3\right)\right]}=\frac{y}{\left[1×2-1×\left(-12\right)\right]}=\frac{1}{\left[1×\left(-3\right)-2×2\right]}$
$\frac{x}{\left(-24+3\right)}=\frac{y}{\left(2+12\right)}=\frac{1}{\left(-3-4\right)}$
$\frac{x}{\left(-21\right)}=\frac{y}{\left(14\right)}=\frac{1}{\left(-7\right)}$

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

#### Question 2:

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

The given equations may be written as:
2x + 5y − 1 = 0         ...(i)
2x + 3y − 3 = 0         ...(ii)
Here, a1 = 2, b1 = 5, c1 = −1, a2 = 2, b2 = 3 and c2 = −3
By cross multiplication, we have: $\frac{x}{\left[5×\left(-3\right)-3×\left(-1\right)\right]}=\frac{y}{\left[\left(-1\right)×2-\left(-3\right)×2\right]}=\frac{1}{\left[2×3-2×5\right]}$
$\frac{x}{\left(-15+3\right)}=\frac{y}{\left(-2+6\right)}=\frac{z}{\left(6-10\right)}$
$\frac{x}{-12}=\frac{y}{4}=\frac{1}{-4}$

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

#### Question 3:

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

The given equations are:
3x − 2y + 3 = 0       ...(i)
4x + 3y − 47 = 0     ...(ii)
Here, a1 = 3, b1 = −2 , c1 = 3, a2 = 4, b2 =  3 and c2 = −47
By cross multiplication, we have: $\frac{x}{\left[\left(-2\right)×\left(-47\right)-3×3\right]}=\frac{y}{\left[3×4-\left(-47\right)×3\right]}=\frac{1}{\left[3×3-\left(-2\right)×4\right]}$
$\frac{x}{\left(94-9\right)}=\frac{y}{\left(12+141\right)}=\frac{z}{\left(9+8\right)}$
$\frac{x}{85}=\frac{y}{153}=\frac{1}{17}$

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

#### Question 4:

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

The given equations are:
6x − 5y − 16 = 0        ...(i)
7x − 13y + 10 = 0      ...(ii)
Here, a1 = 6, b1 = −5 , c1 = −16, a2 = 7, b2 = −13 and c2 = 10
By cross multiplication, we have: $\frac{x}{\left[\left(-5\right)×10-\left(-16\right)×\left(-13\right)\right]}=\frac{y}{\left[\left(-16\right)×7-10×6\right]}=\frac{1}{\left[6×\left(-13\right)-\left(-5\right)×7\right]}$
$\frac{x}{\left(-50-208\right)}=\frac{y}{\left(-112-60\right)}=\frac{z}{\left(-78+35\right)}$
$\frac{x}{\left(-258\right)}=\frac{y}{\left(-172\right)}=\frac{1}{\left(-43\right)}$

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

#### Question 5:

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

The given equations are:
3x + 2y + 25 = 0        ...(i)
2x + y + 10 = 0          ...(ii)
Here, a1 = 3, b1 = 2 , c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have: $\frac{x}{\left[2×10-25×1\right]}=\frac{y}{\left[25×2-10×3\right]}=\frac{1}{\left[3×1-2×2\right]}$
$\frac{x}{\left(20-25\right)}=\frac{y}{\left(50-30\right)}=\frac{1}{\left(3-4\right)}$
$\frac{x}{\left(-5\right)}=\frac{y}{20}=\frac{1}{\left(-1\right)}$

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

#### Question 6:

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

The given equations may be written as:
2x + y − 35 = 0          ...(i)
3x + 4y − 65 = 0        ...(ii)
Here, a1 = 2, b1 = 1, c1 = −35, a2 = 3, b2 = 4 and c2 = −65
By cross multiplication, we have: $\frac{x}{\left[1×\left(-65\right)-4×\left(-35\right)\right]}=\frac{y}{\left[\left(-35\right)×3-\left(-65\right)×2\right]}=\frac{1}{\left[2×4-3×1\right]}$
$\frac{x}{\left(-65+140\right)}=\frac{y}{\left(-105+130\right)}=\frac{1}{\left(8-3\right)}$
$\frac{x}{75}=\frac{y}{25}=\frac{1}{5}$

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

#### Question 7:

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

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

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

#### Question 8:

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

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

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

#### Question 9:

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

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

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

#### Question 10:

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

The given equations may be written as:
2ax + 3by − (a + 2b) = 0         ...(i)
3ax + 2by − (2a + b) = 0         ...(ii)
Here, a1 = 2a, b1 = 3b, c1 = −(a + 2b), a2 = 3a, b2 = 2b and c2 = −(2a + b)
By cross multiplication, we have: $\frac{x}{\left[3b×\left(-\left(2a+b\right)\right)-2b×\left(-\left(a+2b\right)\right)\right]}=\frac{y}{\left[-\left(a+2b\right)×3a-2a×\left(-\left(2a+b\right)\right)\right]}=\frac{1}{\left[2a×2b-3a×3b\right]}$
$\frac{x}{\left(-6ab-3{b}^{2}+2ab+4{b}^{2}\right)}=\frac{y}{\left(-3{a}^{2}-6ab+4{a}^{2}+2ab\right)}=\frac{1}{4ab-9ab}$
$\frac{x}{{b}^{2}-4ab}=\frac{y}{{a}^{2}-4ab}=\frac{1}{-5ab}$
$\frac{x}{-b\left(4a-b\right)}=\frac{y}{-a\left(4b-a\right)}=\frac{1}{-5ab}$

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

#### Question 11:

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

The given equations may be written as:
$\frac{x}{a}-\frac{y}{b}=0$                  ...(i)
ax + by − (a2 + b2) = 0       ...(ii)
Here, a1 $\frac{1}{a}$, b1$\frac{-1}{b}$, c1 = 0, a2 = a, b2b and c2 = −(a2 + b2)
By cross multiplication, we have: $\frac{x}{\left[\left(-\frac{1}{b}\right)×\left(-\left({a}^{2}+{b}^{2}\right)\right)-0\right]}=\frac{y}{\left[0-\frac{1}{a}×\left(-\left({a}^{2}+{b}^{2}\right)\right)\right]}=\frac{1}{\left[\frac{b}{a}+\frac{a}{b}\right]}$
$\frac{x}{\left(\frac{{a}^{2}}{b}+\frac{{b}^{2}}{b}\right)}=\frac{y}{\left(\frac{{a}^{2}}{a}+\frac{{b}^{2}}{a}\right)}=\frac{1}{\frac{{b}^{2}+{a}^{2}}{ab}}$
$\frac{x}{\left[\frac{\left({a}^{2}+{b}^{2}\right)}{b}\right]}=\frac{y}{\left[\frac{\left({a}^{2}+{b}^{2}\right)}{a}\right]}=\frac{1}{\left[\frac{\left({a}^{2}+{b}^{2}\right)}{ab}\right]}$

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

#### Question 12:

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

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

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

#### Question 13:

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

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

Hence, x = a2 and y = b2 is the required solution.

#### Question 14:

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

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

Here, a1 = 1, b1 = 1, c1 = −7, a2 = 2, b2 = 3 and c2 = −17
By cross multiplication, we have: $\frac{u}{\left[1×\left(-17\right)-3×\left(-7\right)\right]}=\frac{v}{\left[\left(-7\right)×2-1×\left(-17\right)\right]}=\frac{1}{\left[3-2\right]}$
$\frac{u}{-17+21}=\frac{v}{-14+17}=\frac{1}{1}$
$\frac{u}{4}=\frac{v}{3}=\frac{1}{1}$
$u=\frac{4}{1}=4,v=\frac{3}{1}=3$
$\frac{1}{x}=4,\frac{1}{y}=3$
$x=\frac{1}{4},y=\frac{1}{3}$
Hence, $x=\frac{1}{4}$and $y=\frac{1}{3}$ is the required solution.

#### Question 15:

Taking $\frac{1}{x+y}=u$  and $\frac{1}{x-y}=v$, the given equations become:
5u − 2v + 1 = 0      ...(i)
15u + 7v − 10 = 0    ...(ii)
Here, a1 = 5, b1 = −2, c1 = 1, a2 = 15, b2 = −7 and c2 = −10
By cross multiplication, we have: $\frac{u}{\left[-2×\left(-10\right)-1×7\right]}=\frac{v}{\left[1×15-\left(-10\right)×5\right]}=\frac{1}{\left[35+30\right]}$
$\frac{u}{20-7}=\frac{v}{15+50}=\frac{1}{65}$
$\frac{u}{13}=\frac{v}{65}=\frac{1}{65}$
$u=\frac{13}{65}=\frac{1}{5},v=\frac{65}{65}=1$
$\frac{1}{x+y}=\frac{1}{5},\frac{1}{x-y}=1$
So, (x + y) = 5            ...(iii)
and (xy) = 1           ...(iv)

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

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

#### Question 16:

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

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

#### Question 1:

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

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

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

#### Question 2:

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

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

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

#### Question 3:

Show that the system of equations
$3x-5y=7,6x-10y=3$
is inconsistent.

The given system of equations is:
3x − 5y = 7
⇒ 3x − 5y − 7 = 0           ...(i)
6x − 10y = 3
⇒ 6x − 10y − 3 = 0         ...(ii)
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 3, b1= −5, c1 = −7 and a2 = 6, b2 = −10, c2 = −3
$\frac{{a}_{1}}{{a}_{2}}=\frac{3}{6}=\frac{1}{2},\frac{{b}_{1}}{b2}=\frac{-5}{-10}=\frac{1}{2}$ and $\frac{{c}_{1}}{{c}_{2}}=\frac{-7}{-3}=\frac{7}{3}$
Thus, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
Hence, the given system of equations is inconsistent.

#### Question 4:

Show that the system of equations

has an infinite number of solutions.

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

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

#### Question 5:

For what value of k, the system of equations

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

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

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

#### Question 6:

For what value of k, the system of equations

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

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

Hence, the required value of k is 6.

#### Question 7:

For what value of k does the system of equations

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

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

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

#### Question 8:

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

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

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

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

#### Question 9:

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

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

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

#### Question 10:

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

The given system of equations:
3x + y = 1
⇒ 3x + y − 1 = 0                                      ...(i)
And, (2k − 1)x + (k − 1)y = (2k + 1)
⇒ (2k − 1)x + (k − 1)y − (2k + 1) = 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 = (2k − 1), b2 = (k − 1), c2 = −(2k + 1)
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.  $\frac{3}{\left(2k-1\right)}=\frac{1}{\left(k-1\right)}\ne \frac{-1}{-\left(2k+1\right)}$
$\frac{3}{\left(2k-1\right)}=\frac{1}{\left(k-1\right)}$ and $\frac{1}{\left(k-1\right)}\ne \frac{-1}{-\left(2k+1\right)}$
⇒ 3(k − 1) = 2k − 1 and  −(2k + 1) ≠ −(k − 1)
⇒ 3k − 3 = 2k − 1 and −2k − 1 ≠ −k + 1
k = 2 and k ≠ −2
Thus, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ holds when k is equal to 2.
Hence, the given system of equations has no solution when the value of k is 2.

#### Question 11:

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

The given system of equations:
(3k + 1)x + 3y − 2 = 0                            ...(i)
And, (k2 + 1)x + (k − 2)y − 5 = 0             ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = (3k + 1), b1= 3, c1 = −2 and a2 = (k2 + 1), b2 = (k − 2), c2 = −5
In order that the given system has no solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
i.e.  $\frac{3k+1}{{k}^{2}+1}=\frac{3}{\left(k-2\right)}\ne \frac{-2}{-5}$
$\frac{3k+1}{{k}^{2}+1}=\frac{3}{\left(k-2\right)}$ and $\frac{3}{\left(k-2\right)}\ne \frac{2}{5}$
⇒ (k − 2) (3k + 1) = 3(k2 + 1) and 15 ≠ 2(k − 2)
⇒ 3k2 + k − 6k − 2 = 3k2 + 3 and 15 ≠ 2k − 4
⇒ −5k = 5 and 2k ≠ 19
k = −1 and $k\ne \frac{19}{2}$
Thus, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ holds when k is equal to −1.
Hence, the given system of equations has no solution when k is equal to −1.

#### Question 12:

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

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

#### Question 13:

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

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

#### Question 14:

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

The given system of equations:
x − 2y = 3
⇒ x − 2y − 3 = 0           ....(i)
And, 3x + ky = 1
⇒ 3x + ky − 1 = 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 = 3, b2 = k, c2 = −1
For a unique solution, we must have:
$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$ i.e.  $\frac{1}{3}\ne \frac{-2}{k}$
This happens when k ≠ −6.
Thus, for all real values of k other than −6, the given system of equations will have a unique solution.

#### Question 15:

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

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

#### Question 16:

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

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

#### Question 17:

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

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

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

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

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

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

#### Question 18:

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

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

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

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

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

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

#### Question 19:

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

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

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

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

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

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

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

#### Question 20:

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

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

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

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

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

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

#### Question 21:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$x+\left(k+1\right)y=5,\phantom{\rule{0ex}{0ex}}\left(k+1\right)x+9y=\left(8k-1\right).$

The given system of equations:
x + (k + 1)y = 5
⇒ x + (k + 1)y − 5 = 0                    ...(i)
And, (k + 1)x + 9y = (8k − 1)
⇒ (k + 1)x + 9y − (8k − 1) = 0        ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 1, b1= (k + 1), c1 = −5 and a2 = (k + 1), b2 = 9, c2 = −(8k − 1)
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{1}{\left(k+1\right)}=\frac{\left(k+1\right)}{9}=\frac{-5}{-\left(8k-1\right)}$
$⇒\frac{1}{\left(k+1\right)}=\frac{\left(k+1\right)}{9}=\frac{5}{\left(8k-1\right)}$

Now, we have the following three cases:
Case I:
$\frac{1}{\left(k+1\right)}=\frac{\left(k+1\right)}{9}\phantom{\rule{0ex}{0ex}}$
$⇒{\left(k+1\right)}^{2}=9\phantom{\rule{0ex}{0ex}}$
$⇒\left(k+1\right)=±3$

Case II:
$\frac{k+1}{9}=\frac{5}{8k-1}\phantom{\rule{0ex}{0ex}}$
$⇒\left(k+1\right)\left(8k-1\right)=45\phantom{\rule{0ex}{0ex}}$
$⇒8{k}^{2}-k+8k-1=45\phantom{\rule{0ex}{0ex}}$
$⇒8{k}^{2}+7k-46=0\phantom{\rule{0ex}{0ex}}$
$⇒8{k}^{2}+23k-16k-46=0\phantom{\rule{0ex}{0ex}}$
$⇒k\left(8k+23\right)-2\left(8k+23\right)=0\phantom{\rule{0ex}{0ex}}$
$⇒\left(k-2\right)\left(8k+23\right)=0\phantom{\rule{0ex}{0ex}}$

Case III:
$\frac{1}{\left(k+1\right)}=\frac{5}{\left(8k-1\right)}\phantom{\rule{0ex}{0ex}}$
$⇒\left(8k-1\right)=5\left(k+1\right)\phantom{\rule{0ex}{0ex}}$
$⇒\left(8k-1\right)=5k+5\phantom{\rule{0ex}{0ex}}$
$⇒3k=6⇒k=2$

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

#### Question 22:

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

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

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

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

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

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

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

#### Question 23:

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

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

#### Question 24:

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

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

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

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

#### Question 25:

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

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

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

#### Question 26:

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

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

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

#### Question 27:

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

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0                ...(i)
And, a(x + y) − b(xy) = 3a + b − 2
⇒ ax + aybx + by = 3a + b − 2
⇒ (ab)x + (a + b)y − (3a + b − 2) = 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 = (ab), b2 = (a + b), c2 = −(3a + b − 2)
For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(a-b\right)}=\frac{3}{\left(a+b\right)}=\frac{7}{\left(3a+b-2\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

⇒ 2(a + b) = 3(ab) and 3(3a + b − 2) =7(a + b)
⇒ 2a + 2b = 3a − 3b and 9a + 3b − 6 = 7a + 7b
a = 5b        ...(iii)
And, 2a − 4b = 6
or a − 2b = 3         ....(iv)
On substituting a = 5b in (iv), we get:
5b − 2b = 3
⇒ 3b = 3
b = 1
On substituting b = 1 in (iii), we get:
a = 5
a = 5 and b = 1

#### Question 28:

Find the value of k for which the system of equations

has a nonzero solution.

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

#### Question 1:

5 chairs and 4 tables together cost Rs 2800 while 4 chairs and 3 tables together cost Rs 2170. Find the cost of a chair and that of a table.

Let the cost of each chair be Rs. x and that of each table be Rs. y .
Then, we have:
5x + 4y = 2800                      ...(i)
4x + 3y = 2170                      ...(ii)
On multiplying (i) by 3 and (ii) by 4, we get:
15x + 12y = 8400                   ...(iii)
16x + 12y = 8680                   ...(iv)
Subtracting (iii) from (iv), we get:
(16x − 15x) = (8680 − 8400)
x = 280
On substituting x = 280 in (i), we get:
(5 × 280) + 4y = 2800
⇒ 4y = 2800 − 1400
⇒ 4y = 1400
y = 350
Hence, the cost of each chair is Rs. 280 and the cost of each table is Rs. 350.

#### Question 2:

37 pens and 53 pencils together cost Rs 820 while 53 pens and 37 pencils together cost Rs 980. Find the cost of a pen and that of a pencil.

Let the cost of each pen be Rs. x and that of each pencil be Rs. y.
Then, we have:
37x + 53y = 820                     ...(i)
53x + 37y = 980                     ...(ii)
Adding (i) and (ii), we get:
90x + 90y = 1800
⇒ 90(x + y) = 1800
x + y = 20                          ...(iii)
Subtracting (i) from (ii), we get:
16x − 16y = 160
⇒ 16(xy) = 160
xy = 10                            ...(iv)
On adding (iii) and (iv), we get:
2x = 30
x = 15
On substituting x = 15 in (iii), we get:
15 + y = 20
y = (20 − 15) = 5
Hence, the cost of each pen is Rs. 15 and the cost of each pencil is Rs. 5.

#### Question 3:

A lady has only 20-paisa coins and 25-paisa coins in her purse. If she has 50 coins in all totalling Rs 11.50, how many coins of each kind does she have?

Let the number of 20-paisa coins be x and that of 25-paisa coins be y.
Then, we have:
x + y = 50                          ....(i)
20x + 25y = 1150               ....(ii)         [Since Re. 1 = 100 paisa]
On multiplying (i) by 20, we get:
20x + 20y = 1000              ....(iii)
Subtracting (iii) from (ii), we get:
(25y − 20y) = (1150 − 1000)
⇒ 5y = 150
y = 30
On substituting y = 30 in (i), we get:
x + 30 = 50
⇒ x = (50 − 30) = 20
Hence, the number of 20-paisa coins is 20 and the number of 25-paisa coins is 30.

#### Question 4:

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

Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137                  ...(i)
xy = 43                    ...(ii)
On adding (i) and (ii), we get:
2x = 180 ⇒ x = 90
On substituting x = 90 in (i), we get:
90 + y = 137
y = (137 − 90) = 47
Hence, the required numbers are 90 and 47.

#### Question 5:

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

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

#### Question 6:

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

Let the first number be x and the second number be y.
Then, we have:
3x + y = 142                              ....(i)
4xy = 138                              ....(ii)
On adding (i) and (ii), we get:
7x = 280
x = 40
On substituting x = 40 in (i), we get:
3 × 40 + y = 142
y = (142 − 120) = 22
y = 22
Hence, the first number is 40 and the second number is 22.

#### Question 7:

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

Let the greater number be x and the smaller be y.
Then, we have:
2x − 45 = y  or  2xy = 45             .... (i)
2y − 21 = x  or  −x + 2y = 21           ....(ii)
On multiplying (i) by 2, we get:
4x − 2y = 90                                      ....(iii)
On adding (ii) and (iii), we get:
3x = (90 + 21) = 111
x = 37
On substituting x = 37 in (i), we get:
2 × 37 − y = 45
⇒ 74 − y = 45
y = (74 − 45) = 29
Hence, the greater number is 37 and the smaller number is 29.

#### Question 8:

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

We know:
Dividend = Divisor × Quotient + Remainder

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

#### Question 9:

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

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

#### Question 10:

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

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

#### Question 11:

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

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
x + y = 12                 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 18
⇒ 10y + x − 10xy = 18
⇒ 9y − 9x = 18
yx = 2                ....(ii)
On adding (i) and (ii), we get:
2y = 14
y = 7
On substituting y = 7 in (i), we get:
x + 7 = 12
x = (12 − 7) = 5
Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.

#### Question 12:

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

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
10x + y = 7(x + y)
10x + y = 7x + 7y  or  3x − 6y = 0       ....(i)
Number obtained on reversing its digits = (10y + x)
(10x + y) − 27 = 10y + x
⇒ 10xx + y − 10y = 27
⇒ 9x − 9y = 27
⇒ 9(xy) = 27
xy = 3                        ....(ii)
On multiplying (ii) by 6, we get:
6x − 6y = 18                      ....(iii)
On subtracting (i) from (iii), we get:
3x = 18
x = 6
On substituting x = 6 in (i), we get:
3 × 6 − 6y = 0
⇒ 18 − 6y = 0
⇒ 6y = 18
y = 3
Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63
Hence, the required number is 63.

#### Question 13:

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

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
x + y = 15                  ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 9
⇒ 10y + x − 10xy = 9
⇒ 9y − 9x = 9
yx = 1                ....(ii)
On adding (i) and (ii), we get:
2y = 16
y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15
x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.

#### Question 14:

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

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

Again, we have:
10x + y + 18 = 10y + x
⇒ 9x − 9y = −18
xy = −2                 ....(ii)
On subtracting (ii) from (i), we get:
x = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = 1
y = 6 − 1 = 5
Required number = (10x + y) = 10 × 3 + 5 = 30 + 5 = 35
Hence, the required number is 35.

#### Question 15:

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

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

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y
∴ 10x + y = (x + y) × 6 + 0
⇒ 10x − 6x + y − 6y = 0
⇒ 4x − 5y = 0                     ....(i)
Number obtained on reversing its digits = (10y + x)
∴ 10x + y − 9 = 10y + x
⇒ 9x − 9y = 9
⇒ 9(xy) = 9
x y = 1                        ....(ii)
On multiplying (ii) by 5, we get:
5x − 5y = 5                        ....(iii)
On subtracting (i) from (iii), we get:
x = 5
On substituting x = 5 in (i), we get:
4 × 5 − 5y = 0
⇒ 20 − 5y = 0
y = 4
∴ Number = (10x + y) = 10 × 5 + 4 = 50 + 4 = 54
Hence, the required number is 54.

#### Question 16:

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

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

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

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

#### Question 17:

A two-digit number is such that the product of its digit is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Let the tens and the units digits of the required number be x and y​, respectively.
Then, we have:
xy = 14                            ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) + 45 = 10y + x
⇒ 9x − 9y = −45
⇒ 9(yx) = 45
yx = 5                      ....(ii)

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

y + x = 9                     ....(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2y = 9 + 5 = 14
y = 7
On substituting y = 7 in (ii), we get:
7 − x = 5
x = (7 − 5) = 2
∴ Number = (10x + y) = 10 × 2 + 7 = 20 + 7 = 27
Hence, the required number is 27.

#### Question 18:

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

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

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

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

#### Question 19:

A two-digit number is four times the sum of its digits and twice the product of its digits. Find the number.

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
∴ 10x + y = 4(x + y)
⇒ 10x + y = 4x + 4y
⇒ 6x − 3y = 0
⇒ 2xy = 0  or  y = 2x           ....(i)
Again, we have:
10x + y = 2xy                          ....(ii)
On substituting y = 2x in (ii), we get:
10x + 2x = 2x(2x)
12x = 4x2
4x = 12
x = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = 0
⇒  y = 6
∴ Number = (10x + y) = 10 × 3 + 6 = 30 + 6 = 36
Hence, the required number is 36.

#### Question 20:

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

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

#### Question 21:

The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes $\frac{1}{2}$. Find the fraction.

Let the required fraction be $\frac{x}{y}$.
Then, we have:
x + y = 2y − 3
xy = −3                       ...(i)
And, $\frac{x-1}{y-1}=\frac{1}{2}$
⇒ 2(x − 1) = 1(y − 1)
⇒ 2x − 2 = y − 1
⇒ 2xy = 1                       ...(ii)
On subtracting (i) from (ii), we get:
x = (1 + 3) = 4
On substituting x = 4 in (i), we get:
4 − y = −3
y = (4 + 3) = 7
∴​ ​x = 4 and y = 7
Hence, the fraction is $\frac{4}{7}$.

#### Question 22:

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

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

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

#### Question 23:

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

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

#### Question 24:

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

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

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

#### Question 25:

A fraction becomes $\frac{1}{3}$, if 2 is added to both of its numerator and denominator. If 3 is added to both of its numerator and denominator, then it becomes $\frac{2}{5}$. Find the fraction.

Let the required fraction be $\frac{x}{y}$.
Then, we have:
$\frac{x+2}{y+2}=\frac{1}{3}$
⇒ 3(x + 2) = 1(y + 2)
⇒ 3x + 6 = y + 2
⇒ 3xy = −4                   ....(i)

Again, $\frac{x+3}{y+3}=\frac{2}{5}$
⇒ 5(x + 3) = 2(y + 3)
⇒ 5x + 15 = 2y  + 6
⇒ 5x − 2y = −9                 ....(ii)
On multiplying (i) by 2, we get:
6x − 2y = −8                     ....(iii)
On subtracting (ii) from (iii), we get:
x = (−8 + 9) = 1
On substituting x = 1 in (i), we get:
3 × 1 − y = −4
⇒ 3 − y = −4
y = (3 + 4) = 7
x = 1 and y = 7
Hence, the required fraction is $\frac{1}{7}$.

#### Question 26:

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

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

#### Question 27:

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

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

#### Question 28:

Five years ago, A was thrice as old as B and ten years later A shall be twice as old as B. What are the present ages of A and B?

Let A's present age be x years.
Let B's present age be y years.
A's age 5 years ago = (x − 5) years
B's age 5 years ago = (y − 5) years
Then, we have:
(x − 5) = 3(y − 5)
x − 5 = 3y − 15
x − 3y = −10                ....(i)
A's age 10 years later = (x + 10) years
B's age 10 years later = (y + 10) years
Then, we have:
(x + 10) = 2(y + 10)
x + 10 = 2y + 20
x − 2y = 10                  ....(ii)
On subtracting (ii) from (i), we get:
y = −20
y = 20
On substituting y = 20 in (i), we get:
x − 3 × 20 = −10
x − 60 = −10
x = (−10 + 60) = 50
x = 50
Hence, A's present age is 50 years and B's present age is 20 years.

#### Question 29:

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

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

#### Question 30:

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

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

#### Question 31:

A man's age is three times the sum of the ages of his two sons. After 5 years, his age will be twice the sum of the ages of his two sons. Find the age of the man.

Let the man's present age be x years.
Let the sum of the ages of his two sons be y years.
Then, we have:
x = 3y                     ....(i)
5 years later:
Man's age = (x + 5)
Note that the age of each son is increased by 5 years.
∴ Sum of the ages of the two sons after 5 years = (y + 5 + 5) = (y + 10)
∴ x + 5 = 2(y + 10)
x + 5 = 2y + 20
x − 2y = 15         ....(ii)
On substituting x = 3y in (ii), we get:
3y − 2y = 15
y = 15
On substituting y = 15 in (i), we get:
x = (3 × 15) = 45
Hence, the age of the man is 45 years.

#### Question 32:

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

Let the man's present age be x years.
Let his son's present age be y years.
Ten years later:
(x + 10) = 2(y + 10)
x + 10 = 2y + 20
x − 2y = 10              .....(i)
Ten years ago:
(x − 10) = 4(y − 10)
x − 10 = 4y − 40
x − 4y = −30           .....(ii)
On subtracting (i) from (ii), we get:
−2y = −40
y = 20
On substituting y = 20 in (i), we get:
x − 2 × 20 = 10
x − 40 = 10
x = (10 + 40) = 50 years
Hence, the man's present age is 50 years and his son's present age is 20 years.

#### Question 33:

The monthly incomes of A and B are in the ratio of 5 : 4 and their monthly expenditures are in the ratio of 7 : 5. If each saves Rs 3000 per month, find the monthly income of each.

Let the monthly income and expenditure of A be Rs. 5x and Rs. 7y, respectively.
Let the monthly income and expenditure of B be Rs. 4x and Rs. 5y,​ respectively.
Then, we have:
5x − 7y = 3000                   ...(i)
4x − 5y = 3000                   ...(ii)
On multiplying (i) by 5 and (ii) by 7, we get:
25x − 35y = 15000              ...(iii)
28x − 35y = 21000               ...(iv)
On subtracting (iii) from (iv), we get:
3x = 6000
x = 2000
On substituting x = 2000 in (i), we get:
5 × 2000 − 7y = 3000
⇒ −7y = 3000 − 10000 = −7000
y = 1000
∴ Income of A = 5x = 5 × 2000 = Rs. 10000
∴ Income of B = 4x = 4 × 2000 = Rs. 8000

#### Question 34:

A man sold a chair and a table together for Rs 760, thereby marking a profit of 25% on chair and 10% on table. By selling them together for Rs 767.50, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.

Let the cost price of the chair be Rs. x and the cost price of the table be Rs. y.
Then, we have:
If the profit is 25%, then the selling price of the chair = Rs. = Rs. $\left(x+\frac{25x}{100}\right)$ = Rs. $\left(\frac{125}{100}x\right)$
If the profit is 10%, then the selling price of the table = Rs. = Rs. $\left(y+\frac{10y}{100}\right)$ = Rs. $\left(\frac{110}{100}y\right)$
Given:
Selling price of the chair and the table = Rs. 760
⇒  Rs. $\left(\frac{125}{100}x\right)$ + Rs. $\left(\frac{110}{100}y\right)$ = Rs. 760
$\frac{25}{20}x+\frac{22}{20}y=760$
⇒ 25x + 22y = 15200            .....(i)

Again, we have:
If the profit is 10%, then the selling price of the chair = Rs. = Rs. $\left(x+\frac{10x}{100}\right)$ = Rs. $\left(\frac{110}{100}x\right)$
If the profit is 25%, then the selling price of the table = Rs. = Rs. $\left(y+\frac{25y}{100}\right)$ = Rs. $\left(\frac{125}{100}y\right)$
Selling price of the chair and the table = Rs. 767.50
⇒  Rs. $\left(\frac{110}{100}x\right)$ + Rs. $\left(\frac{125}{100}y\right)$ = Rs. 767.50
$\frac{22}{20}x+\frac{25}{20}y=767.50$
⇒ 22x + 25y = 15350            .....(ii)

On adding (i) and (ii), we get:
47x + 47y = 30550
⇒ 47(x + y) = 30550
⇒ (x + y) = 650                     ....(iii)
On subtracting (ii) from (i), we get:
3x − 3y = −150
⇒ 3(xy) = −150
⇒ (x y) = −50                    ....(iv)
On adding (iii) and (iv), we get:
2x = (650 − 50) = 600
x = 300
On substituting x = 300 in (iii), we get:
300 + y = 650
y = (650 − 300) = 350
Hence, the cost price of the chair is Rs. 300 and the cost price of the table is Rs. 350.

#### Question 35:

On selling a TV at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs 3250. But,  if he sells TV at 10% gain and the fridge at 5% loss, he gains Rs 1500. Find the actual cost price of TV and that of the fridge.

Let the cost price of the TV be Rs. x and the cost price of the fridge be Rs. y.
Then, we have:
When the gain on the TV is 5%:
Rs.  = Rs. $\left(\frac{5x}{100}\right)$ = Rs. $\left(\frac{x}{20}\right)$
When the gain of the fridge is 10%:
Rs. = Rs. $\left(\frac{10y}{100}\right)$ = Rs. $\left(\frac{y}{10}\right)$
Given:
Gain on the TV and the fridge = Rs. 3250
Rs. $\left(\frac{x}{20}\right)$ + Rs. $\left(\frac{y}{10}\right)$ = Rs. 3250
$\left(\frac{x+2y}{20}\right)=3250$
⇒ (x + 2y) = 65000               ....(i)

Again, we have:
When the gain on the TV is 10%:
Rs. = Rs. $\left(\frac{10x}{100}\right)$ = Rs. $\left(\frac{x}{10}\right)$
When the loss on the fridge is 5%:
Rs. = Rs. $\left(\frac{5y}{100}\right)$ = Rs. $\left(\frac{y}{20}\right)$
Gain on the TV and the fridge = Rs. 1500
Rs. $\left(\frac{x}{10}\right)$ − Rs. $\left(\frac{y}{20}\right)$ = Rs.1500
$\left(\frac{2x-y}{20}\right)=1500$
⇒ (2xy) = 30000               ....(ii)
On multiplying (ii) by 2, we get:
4x − 2y = 60000                   ....(iii)
On adding (i) and (iii), we get:
5x = 125000
x = 25000
On substituting x = 25000 in (i), we get:
25000 + 2y = 65000
⇒ 2y = 40000
y = 20000
Hence, the cost price of the TV is Rs. 25000 and the cost price of the fridge is Rs. 20000.

#### Question 36:

A man invested an amount at 12% per annum simple interest and another amount at 10% per annum simple interest. He received an annual interest of Rs 1145. But, if he had interchanged the amounts invested, he would have received Rs 90 less. What amounts did he invest at the different rates?

Let the amounts invested at 12% per annum and 10% per annum be Rs. x and Rs. y​, respectively.
Then, we have:
Simple interest on Rs. x at 12% p.a. for 1 year = Rs. $\left(\frac{x×12×1}{100}\right)$ = Rs. $\left(\frac{3x}{25}\right)$
Simple interest on Rs. y at 10% p.a. for 1 year = Rs. $\left(\frac{y×10×1}{100}\right)$ = Rs. $\left(\frac{y}{10}\right)$
Given:
Total simple interest = Rs. 1145
Rs. $\left(\frac{3x}{25}\right)$ + Rs. $\left(\frac{y}{10}\right)$ = Rs. 1145
$\left(\frac{6x+5y}{50}\right)=1145$
⇒ (6x + 5y) = 57250              ....(i)

Again, we have:
​Simple interest on Rs. x at 10% p.a. for 1 year = Rs. $\left(\frac{x×10×1}{100}\right)$ = Rs. $\left(\frac{x}{10}\right)$
Simple interest on Rs. y at 12% p.a. for 1 year = Rs. $\left(\frac{y×12×1}{100}\right)$ = Rs. $\left(\frac{3y}{25}\right)$
Given:
Total simple interest = Rs. (1145 − 90) = Rs. 1055
Rs. $\left(\frac{x}{10}\right)$ + Rs. $\left(\frac{3y}{25}\right)$ = Rs. 1055
$\left(\frac{5x+6y}{50}\right)=1055$
⇒ (5x + 6y) = 52750              ....(ii)
On multiplying (i) by 6 and (ii) by 5, we get:
36x + 30y = 343500              ....(iii)
25x + 30y = 263750              ....(iv)
On subtracting (iv) from (iii), we get:
11x = (343500 − 263750) = 79750
x = 7250
On substituting x = 7250 in (i), we get:
6 × 7250 + 5y = 57250
⇒ 43500 + 5y = 57250
⇒ 5y = (57250 − 43500) = 13750
y = 2750
∴ Amount invested at 12% = Rs. 7250
And, amount invested at 10% = Rs. 2750

#### Question 37:

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

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

#### Question 38:

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

Let X and Y be the cars starting from points A and B, respectively and let their speeds be x km/h and y km/h, respectively.
Then, we have the following cases:
Case I: When the two cars move in the same direction
In this case, let the two cars meet at point M. Distance covered by car X in 7 hours = 7x km
Distance covered by car Y in 7 hours = 7y km
∴ AM = (7x) km and BM = (7y) km
⇒ (AM − BM) = AB
⇒ (7x − 7y) = 70
⇒ 7(xy) = 70
⇒ (xy) = 10                     ....(i)

Case II: When the two cars move in opposite directions
In this case, let the two cars meet at point N. Distance covered by car X in 1 hour = x km
Distance covered by car Y in 1 hour = y km
∴ AN = x km and BN = y km
⇒ AN + BN = AB
x + y = 70                        ....(ii)
On adding (i) and (ii), we get:
2x = 80
x = 40
On substituting x = 40 in (i), we get:
40 − y = 10
y = (40 − 10) = 30
Hence, the speed of car X is 40 km/h and the speed of car Y is 30 km/h.

#### Question 39:

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

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

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

#### Question 40:

A man travels 370 km, partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.

Let the speed of the train be x km/h and the speed of the car be y km/h.
Then, we have:
Time taken to cover 250 km by train = $\left(\frac{250}{x}\right)$ hrs
Time taken to cover 120 km by car = $\left(\frac{120}{y}\right)$ hrs  (∵ Total distance = 370 km)
Total time taken = 4 hrs
$\frac{250}{x}+\frac{120}{y}=4$
$\frac{125}{x}+\frac{60}{y}=2$
⇒ 125u + 60v = 2             ...(i)           where
Again, we have:
Time taken to cover 130 km by train = $\left(\frac{130}{x}\right)$ hrs
Time taken to cover 240 km by car = $\left(\frac{240}{y}\right)$ hrs  (∵ Total distance = 370 km)
Total time taken = 4 hours 18 minutes = $\left(4+\frac{18}{60}\right)$ hrs = $\left(4+\frac{3}{10}\right)$ hrs = $\left(\frac{43}{10}\right)$ hrs
$\frac{130}{x}+\frac{240}{y}=\frac{43}{10}$
$\left(\frac{1300}{x}+\frac{2400}{y}\right)=43$
⇒ 1300u + 2400v = 43      ...(ii)          Here,
On multiplying (i) by 40, we get:
5000u + 2400v = 80         ...(iii)
On subtracting (ii) from (iii), we get:
3700u = 37
u = $\frac{37}{3700}=\frac{1}{100}$
On substituting $u=\frac{1}{100}$ in (i), we get:
$125×\frac{1}{100}+60v=2$
$\frac{5}{4}+60v=2$
$60v=\left(2-\frac{5}{4}\right)=\frac{3}{4}$
v = $\left(\frac{3}{4×60}\right)=\frac{1}{80}$
$x=\frac{1}{u}=\frac{1}{1}{100}}=100$
$y=\frac{1}{v}=\frac{1}{1}{80}}=80$

∴ Speed of the train = 100 km/h
And, speed of the car = 80 km/h

#### Question 41:

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

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

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

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

#### Question 42:

Taxi charges in a city consists of fixed charges per day and the remaining depending upon the distance travelled in kilometres. If a person travels 110 km, he pays Rs 1130, and for travelling 200 km, he pays Rs 1850. Find the fixed charges per day and the rate per km.

Let the fixed taxi charges per day be Rs. x and the charges for travelling one km be Rs. y.
It is given that for travelling 110 km, a person pays Rs. 1130.
∴ Rs. x + Rs. 110y = Rs. 1130
⇒ (x + 110y) = 1130            ....(i)
Again, it is given that for travelling 200 km, a person pays Rs. 1850.
∴ Rs. x + Rs. 200y = Rs. 1850
⇒ (x + 200y) = 1850            ....(ii)
On subtracting (i) from (ii), we get:
90y = 720
y = 8
On substituting y = 8 in (i), we get:
x + 110 × 8 = 1130
x = (1130 − 880) = 250
Hence, the fixed charges per day is Rs. 250 and the rate per km is Rs. 8.

#### Question 43:

A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs 3500 as hostel charges whereas a student B, who takes food for 28 days, pays Rs 3800 as hostel charges. Find the fixed charges and the cost of the food per day.

Let Rs. x be the part of monthly hostel charges that is fixed.
Let Rs. be the food charge for one day.
Then, we have:
For student A:
He takes food for 25 days and pays Rs. 3500.
∴ Rs. x + Rs. 25y = Rs. 3500
x + 25y = 3500            .....(i)
For student B:
He takes food for 28 days and pays Rs. 3800.
∴ Rs. x + Rs. 28y = Rs. 3800
⇒ x + 28y = 3800            .....(ii)
On subtracting (i) from (ii), we get:
3y = (3800 − 3500)
⇒ 3y = 300
y = 100
On substituting y = 100 in (i), we get:
x + 25 × 100 = 3500
x = 3500 − 2500
x = 1000
∴ Part of the monthly hostel charges that is fixed = Rs. 1000
And, food charge per day = Rs. 100

#### Question 44:

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

Let the length of the room be x metres and the breadth of the room be y metres.
Then, we have:
Area of the room = xy
According to the question, we have:
x = y + 3
xy = 3               ....(i)
And, (x + 3) (y − 2) = xy
xy − 2x + 3y − 6 = xy
⇒ 3y − 2x = 6           ....(ii)
On multiplying (i) by 2, we get:
2x − 2y = 6               ....(iii)
On adding (ii) and (iii), we get:
y = (6 + 6) = 12
On substituting y = 12 in (i), we get:
x − 12 = 3
x = (3 + 12) =15
Hence, the length of the room is 15 metres and its breadth is 12 metres.

#### Question 45:

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

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

Case 1: When the length is reduced by 5 m and the breadth is increased by 3 m:
New length = (x − 5) m
New breadth = (y + 3) m
∴ New area = (x − 5) (y + 3) sq. m
xy − (x − 5) (y + 3) = 8
xy − [xy − 5y + 3x − 15] = 8
xyxy + 5y − 3x + 15 = 8
⇒ 3x − 5y = 7                .....(i)

Case 2: When the length is increased by 3 m and the breadth is increased by 2 m:
New length = (x + 3) m
New breadth = (y + 2) m
∴ New area = (x + 3) (y + 2) sq. m
∴ (x + 3) (y + 2) − xy = 74
⇒ [xy + 3y + 2x + 6] − xy = 74
⇒ 2x + 3y = 68             .....(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 21              .....(iii)
10x + 15y = 340          .....(iv)
On adding (iii) and (iv), we get:
19x = 361
x = 19
On substituting x = 19 in (iii), we get:
9 × 19 − 15y = 21
⇒ 171 −15y = 21
⇒ 15y = (171 − 21) = 150
y = 10
Hence, the length is 19 m and the breadth is 10 m.

#### Question 46:

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

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

#### Question 47:

In a
If $3y-5x=30,$ show that the triangle is right-angled.

In ∆ ABC, we have:
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$     (Angle sum property of a triangle)
∴ ​x + 3x + y = 180
4x + y = 180               ....(i)
Again, 3y − 5x = 30 (Given)
⇒ −5x + 3y = 30         ....(ii)
On multiplying (i) by 3, we get:
12x + 3y = 540           ....(iii)
On subtracting (ii) from (iii), we get:
17x = (540 − 30) = 510
x = 30
On substituting x = 30 in (i), we get:
4 × 30 + y = 180
⇒ 120 + y = 180
y = (180 − 120) = 60
Thus, we have:

Since in the given triangle, one angle is 90°, it is a right-angled triangle.

#### Question 48:

Find the four angles of a cyclic quadrilateral ABCD in which

Given:
In a cyclic quadrilateral ABCD​, we have:
$\angle A=\left(x+y+10\right)°$
$\angle B=\left(y+20\right)°$
$\angle C=\left(x+y-30\right)°$
$\angle D=\left(x+y\right)°$
$\angle A+\angle C=180°$ 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
Therefore, we have:
$\angle A=\left(x+y+10\right)°=\left(40+60+10\right)°=110°$
$\angle B=\left(y+20°\right)=\left(60+20\right)°=80°$
$\angle C=\left(x+y-30\right)°=\left(40+60-30\right)°=70°$
$\angle D=\left(x+y\right)°=\left(40+60\right)°=100°$

#### Question 1:

The graphs of the equations  are two lines which are

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

The correct option is (b).

The given equations are as follows:

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

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

#### Question 2:

The graphs of the equations  are two which are

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

The correct option is (c).

The given equations are as follows:

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

#### Question 3:

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

The correct option is (a).

The given system of equations can be written as follows:

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

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

#### Question 4:

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

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

The correct option is (d).

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

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

#### Question 5:

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

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

The correct option is (d).

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

#### Question 6:

The pair of equations  has

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

The correct option is (d).

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

#### Question 7:

The pair of equations  has

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

The correct option is (d).

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

#### Question 8:

The system has no solution, when

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

The correct option is (a).

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

#### Question 9:

The pair of equations  has

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

The correct option is (a).

The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = −5, c1 = 4 and a2 = 2, b2 = 1 and c2 = −8

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

#### Question 10:

The system has a unique solution only when

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

The correct option is (b).

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

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

#### Question 11:

If the pair of equations  has infinitely many solutions, then

(a) m = −1, n = 5
(b) m = 1, n = 5
(c) m = 5, n = 1
(d) m = 1, n = −1

The correct option is (c).

The given system of equations can be written as follows:
2x + 3y − 11 = 0 and (m + n)x + (2mn)y − 33 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 =−11 and a2 = (m + n), b2 = (2mn) and c2 = −33

For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(m+n\right)}=\frac{3}{\left(2m-n\right)}=\frac{1}{3}$

⇒ (m + n) = 6                  ....(i)
And, (2mn) = 9              ...(ii)
On adding (i) and (ii) we get:
3m = 15 ⇒ m = 5
On substituting m = 5 in (i) we get:
5 + n = 6 ⇒ n = (6 − 5) = 1
∴​ m = 5 and n = 1

#### Question 12:

For what value of k does the pair of equations  have an infinite number of solutions?

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

The correct option is (b).

The given system of equations can be written as follows:
5x + 2y − 2k = 0 and 2(k + 1)x + ky − (3k + 4) = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 5, b1 = 2, c1 = −2k and a2 = 2(k + 1), b2 = k and c2 = −(3k + 4)

For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$

⇒ 5k = 4(k + 1)                   ...(i)
And, 2(3k + 4) = 2k2             ...(ii)
From (i) we have:
5k = 4k + 4 ⇒ 5k − 4k = 4 ⇒ k = 4
From (ii) we have:
6k + 8 = 2k2k2 − 3k − 4 = 0 ⇒ (k − 4)( k + 1) = 0 ⇒ k = 4, −1
Hence, from (i) and (ii), we get:
k = 4

#### Question 13:

For what value of c does the pair of equations  have infinitely many solutions?

(a) c = 3
(b) c = −3
(c) c = −12
(d) not possible for any value of c

The correct option is (d).

The given system of equations can be written as follows:
cx − y − 2 = 0 and 6x − 2y − 3 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = c, b1 = −1, c1 = −2 and a2 = 6, b2 = −2 and c2 = −3

For an infinite number of solutions, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{c}{6}=\frac{1}{2}=\frac{2}{3}$
The above condition is not possible for any value of c.

#### Question 14:

For what value of k is the pair of equations  inconsistent?

(a) $k=\frac{3}{5}$
(b) $k=\frac{5}{3}$
(c) k = 16
(d) k = −16

The correct option is (c).

The given system of equations can be written as follows:
8x + 5y − 9 = 0 and kx + 10y − 15 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 8, b1 = 5, c1 = −9 and a2 = k, b2 = 10 and c2 = −15
$\frac{{a}_{1}}{{a}_{2}}=\frac{8}{k},\frac{{b}_{1}}{{b}_{2}}=\frac{5}{10}=\frac{1}{2}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-9}{-15}=\frac{3}{5}$
For inconsistency, we must have:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$
$\frac{8}{k}=\frac{1}{2}\ne \frac{3}{5}⇒k=\left(2×8\right)=16$
Hence, the pair of equations is inconsistent if the value of k is 16.

#### Question 15:

One equation of a pair of dependent equations is −5x + 2y = 4. The second equation can be

(a) $10x+4y+8=0$
(b) $-10x-4y+8=0$
(c) $-10x+4y=8$
(d) $10x-4y=8$

The correct option is (c).

One of the two dependent equations:
−5x + 2y = 4
Thus, by observation, the right option is as follows:
−10x + 4y = 8
⇒ 2(−5x + 2y = 4)

#### Question 16:

The pair of equations x = a and y = b graphically represents lines which are

(a) parallel
(b) coincident
(c) intersecting at (a, b)
(d) intersecting at (b, a)

The correct option is (c).

We have:
x = a is the line parallel to the y-axis at a distance a from it.
y = b is the line parallel to the x-axis at a distance b from it.
Thus, the two lines intersect at the point (a, b). Thus, x = a and y = b represent two lines intersecting at (a, b).

#### Question 17:

The pair of equations y = 0 and y = −5 has

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

The correct option is (d).

Clearly, y = 0 is the x-axis.
And, y = −5 is the line parallel to the x-axis, which is below the x-axis at a distance of 5 units.
The two lines do not meet anywhere.
Hence, no solution exists.

#### Question 18:

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

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

The correct option is (d).

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

#### Question 19:

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

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

The correct option is (a).

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

#### Question 20:

On solving we get:
(a) x = 3, y = 2
(b) x = 3, y = −2
(c) x = 2, y = 1
(d) x = 5, y = −3

The correct option is (c).
19x + 17y = 55                     ....(i)
17x + 19y = 53                     ....(ii)
On adding (i) and (ii), we get:
36x + 36y = 108
36(x + y) = 108
(x + y) = 3                            ....(iii)
On subtracting (ii) from (i), we get:
2x − 2y = 2
2(xy) = 2
⇒ (x y) = 1                         ....(iv)
Again, on adding (iii) and (iv), we get:
2x = 4 ⇒ x = 2
On substituting x = 2 in (iii), we get:
y = 1
∴​ x = 2 and  y = 1

#### Question 21:

On solving we get:
(a) x = 3, y = 2
(b) x = 6, y = 1
(c) x = −3, y = −2
(d) x = 6, y = 2

The correct option is (a).

$\frac{2}{3}x+5y=12$
2x + 15y = 36         ....(i)
$x+\frac{3}{2}y=6$
2x + 3y = 12           ....(ii)
On subtracting (ii) from (i), we get:
12y = (36 − 12) = 24 ⇒ y = 2
On substituting y = 2 in (i), we get:
2x + 30 = 36 ⇒ 2x = (36 − 30) = 6 ⇒ x = 3
∴​ x = 3 and y = 2

#### Question 22:

On solving we get:
(a) x = 3, y = 4
(b) x = 6, y = 2
(c) x = 0, y = 6
(d) x = 3, y = 4

The correct option is (b).

2x + 3y = 18          ....(i)
x − 2y = 2              ....(ii)
On multiplying (ii) by 2, we get:
2x − 4y = 4            ....(iii)
On subtracting (iii) from (i), we get:
7y = 14 ⇒ y = 2
On substituting y = 2 in (i), we get:
2x + 6 = 18 ⇒ 2x = 12 ⇒ x = 6
∴​ x = 6 and  y = 2

#### Question 23:

On solving we get:
(a) x = 5, y = 1
(b) x = 8, y = 2
(c) x = 4, y = $\frac{1}{2}$
(d) x = 9, y = 3

The correct option is (b).

The given equations are as follows:
$x+\frac{6}{y}=11\phantom{\rule{0ex}{0ex}}$
$3x+\frac{8}{y}=28$
Putting $\frac{1}{y}=u$ in the given equations, we get:
x + 6u = 11             ....(i)
3x + 8u  = 28          ....(ii)
On multiplying (i) by 3, we get:
3x + 18u = 33         ....(iii)
On subtracting (ii) from (iii), we get:
10u = 5 ⇒ u = $\frac{1}{2}$
Putting u = $\frac{1}{2}$ in (ii), we get:
3x + 4 = 28 ⇒ 3x = 24 ⇒ x = 8
∴​ x = 8 and  y = 2

#### Question 24:

On solving $\frac{2}{x}+\frac{3}{y}=3,\frac{3}{x}+\frac{6}{y}=5,$ we get:
(a) x = 2, y = 3
(b) x = $\frac{1}{2}$, y = $\frac{1}{3}$
(c) x = 1, y = 3
(d) x = 2, y = 6

The correct option is (c).
The given equations are as follows:
$\frac{2}{x}+\frac{3}{y}=3\phantom{\rule{0ex}{0ex}}$
$\frac{3}{x}+\frac{6}{y}=5$
Putting in the given equations, we get:
2u + 3v = 3             ...(i)
3u + 6v = 5             ...(ii)
On multiplying (i) by 2, we get:
4u + 6v = 6             ...(iii)
On subtracting (ii) from (iii), we get:
u = 1 ⇒ $\frac{1}{x}=1⇒x=1$
Putting u = 1 in (i), we get:
2 + 3v = 3 ⇒ 3v = 1 ⇒ v = $\frac{1}{3}$$\frac{1}{y}=\frac{1}{3}⇒y=3$
∴​ x = 1 and y = 3

#### Question 25:

On solving we get:
(a) x = 1, y = −1
(b) x = −1, y = 1
(c) x = −1, y = −1
(d) x = 0, y = $\frac{-2}{7}$

The correct option is (c).

The given equations are as follows:
3x + 4y = −7                    ...(i)
5x − 7y = 2                       ...(ii)
On multiplying (i) by 5 and (ii) by 3, we get:
15x + 20y = −35              ...(iii)
15x − 21y = 6                   ...(iv)
On subtracting (iii) from (iv), we get:
−41y = 41 ⇒ y = −1
On substituting the value of in (i), we get:
3x − 4 = − 7 ⇒ 3x = (−7 + 4)= −3 ⇒ x = −1
∴​ x = −1 and  y = −1

#### Question 26:

On solving we get:
(a) x = 1, y = −2
(b) x = −1, y = 2
(c) x = 0, y = $\frac{15}{11}$
(d) x = $\frac{-11}{3}$, y = 0

The correct option is (b).

The given equations are as follows:
3x − 4y = −11                 ...(i)
7x + 11y = 15                  ...(ii)
On multiplying (i) by 7 and (ii) by 3, we get:
21x − 28y = −77               ...(iii)
21x + 33y = 45                ...(iv)
On subtracting (iii) from (iv), we get:
61y = 122 ⇒ y = 2
On substituting y = 2 in (i), we get:
3x − 8 = −11 ⇒ 3x = (−11 + 8)= −3 ⇒ x = −1
∴​ x = −1 and y = 2

#### Question 27:

On solving we get:
(a) x = 2, y = −1
(b) x = −2, y = 1
(c) x = 1, y = 2
(d) x = 3, y = −2

The correct option is (a).

The given equations are as follows:
31x + 47y = 15             ....(i)
47x + 31y = 63             ....(ii)
On adding (i) and (ii), we get:
78x + 78y = 78
78(x + y) = 78
⇒ (x + y) = 1                 ....(iii)
On subtracting (i) from (ii), we get:
16x − 16y = 48
16(xy) = 48
⇒(x y) = 3                    ....(iv)
On adding (iii) and (iv), we get:
2x = 4 ⇒ x = 2
On substituting x = 2 in (iii), we get:
2 + y = 1 ⇒ y = −1
x = 2 and y = −1

#### Question 28:

In a
(a) 20°
(b) 40°
(c) 60°
(d) 80°

The correct option is (b).

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

#### Question 29:

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

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
∴​ $\angle B=\left(y+20°\right)=\left(60+20\right)°=80°$

#### Question 30:

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

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

The correct option is (d).

Let the tens and the units digits of the required number be x and y​, respectively.
Required number = (10x + y)
According to the question, we have:
x + y = 15                   ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) = (10x + y) + 9
⇒ 10y + x − 10xy = 9
⇒ 9y − 9x = 9
yx = 1                 ....(ii)
On adding (i) and (ii), we get:
2y = 16 ⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15 ⇒ x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.

#### Question 31:

In a given fraction, if 1 is subtracted from the numerator and 2 is added to the denominator, it becomes $\frac{1}{2}$. If 7 is subtracted from the numerator and 2 is subtracted from the denominator, it becomes $\frac{1}{3}$. The fraction is
(a) $\frac{13}{24}$
(b) $\frac{15}{26}$
(c) $\frac{16}{27}$
(d) $\frac{16}{21}$

The correct option is (b).

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, we have:
$\frac{x-7}{y-2}=\frac{1}{3}$
⇒ 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 = (30 − 4) = 26
∴ ​x = 15 and  y = 26
Hence, the required fraction is $\frac{15}{26}$.

#### Question 32:

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

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

#### Question 33:

Look at the statements given below:

I.
II.
III.
Which of the above statements are true?
(a) I only
(b) II only
(c) I and II
(d) I and III

Option (c) is correct.

For I:
3x + 4y = 5            ...(i)
2x + 3y = 9            ...(ii)
On multiplying (i) by 2 and (ii) by 3 and subtracting, we get:
y  = 17
On substituting y = 17 in (i), we get:
3x + 68 = 5
⇒ 3x = (5 − 68) = −63
x = −21
Thus, x = −21 and  y = 17

For II:
xy = −1                  ...(i)
3x − 5y = −1              ...(ii)
On multiplying (i) by 3 and subtracting, we get:
y  = −1
On substituting y = −1 in (i), we get:
x + 1 = −1
x = −2
Thus, x = −2 and  y = −1

For III:
$\frac{x}{a}+\frac{y}{b}=2$                  ...(i)
$\frac{x}{a}-\frac{y}{b}=4$                  ...(ii)
On adding (i) and (ii), we get:
$\frac{2x}{a}=6⇒x=3a$
Putting x = 3a in (i), we get:
y = −b
Thus, x = 3a and  y = −b
So, x = 2a and y = 2b is not true.

#### Question 34:

Assertion (A)
The system of equations is inconsistent when k = 6.
Reason (R)
The system of equations  is inconsistent when $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Option (a) is correct.
Reason (R) is clearly true.
For the given system to be inconsistent, we must have,

Also, $\frac{2}{4}=\frac{3}{k}\ne \frac{5}{7}⇒k=6$
So, Assertion (A) is true and, clearly, Reason (R) is a correct explanation of Assertion (A).

#### Question 35:

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

Option (b) is correct.
Reason (R) is clearly true.
For the given system to be inconsistent, we must have,

Also, $\frac{3}{6}=\frac{-2}{-4}\ne \frac{1}{3}⇒k=6$
So, Assertion (A) is true, but Reason (R) is not a correct explanation of Assertion (A).

#### Question 36:

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

Option (d) is correct.
Here, Reason (R) is clearly true.
We have a1 = 1, b1 = 3, c1 = −5 and a2 = 2, b2 = −6, c2 = 8
Thus, we have:

$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ is not true and so, Assertion (A) is false.
Thus, Assertion (A) is false and Reason (R) is true.

#### Question 37:

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

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

#### Question 38:

Match the following columns:

 Column I Column II (a) Solution of (p) (b) Solution of (q) (c) Solution of (r) (d) Solution of (s)

(a) = (r)
(b) = (p)
(c) = (s)
(d) = (q)

(a)
ax + by = ab                   ...(i)
bxay = a + b                   ...(ii)
On multiplying (i) by a and (ii) by b, we get:
a2x + aby = a2ab            ...(iii)
b2xaby = ab + b2            ...(iv)
On adding (iii) and (iv), we get:
(a2 + b2)x = a2 + b2
x = 1
On substituting x = 1 in (i), we get:
a + by = ab
by =b
y = −1
Hence, x = 1 and y = −1

(b)
3x − 2y = 10                      ...(i)
5x + 3y = 4                        ...(ii)
On multiplying (i) by 3 and (ii) by 2, we get:
9x − 6y = 30                      ...(iii)
10x + 6y = 8                      ...(iv)
On adding (iii) and (iv), we get:
19x = 38
x = 2
On substituting x = 2 in (i), we get:
6 − 2y = 10
⇒ −2y = 4
y = −2
Hence, x = 2 and y = −2

(c)
2xy = 5                        ...(i)
3x + 2y = 11                     ...(ii)
On multiplying (i) by 2, we get:
4x − 2y = 10                    ...(iii)
On adding (ii) and (iii), we get:
7x = 21
x = 3
On substituting x = 3 in (i), we get:
6 − y = 5
y = (6 − 5) = 1
Hence, x = 3 and y = 1

(d)
x + y = a + b                   ...(i)
ax by = a2b2            ...(ii)
On multiplying (i) by b, we get:
bx + by = ab + b2           ...(iii)
On adding (ii) and (iii), we get:
ax + bx = a2 + ab
x(a + b) = a(a + b)
x = a
On substituting x = a in (i), we get:
a + y = a + b
y = b

#### Question 1:

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

The correct option is (a).

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

#### Question 2:

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

The correct option is (d).

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

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

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

#### Question 3:

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

The correct option is (a).

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

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

#### Question 4:

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

The correct option is (d).

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

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

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

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

#### Question 5:

Show that the system of equations has a unique solution.

The given system of equations:

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

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

#### Question 6:

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

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

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

#### Question 7:

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

The given system of equations can be written as follows:

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

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

#### Question 8:

Solve the system of equations:

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

#### Question 9:

Show that the paths represented by the equations  are parallel.

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

#### Question 10:

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

Let the larger number be x and the smaller number be y.
Then, we have:
xy = 26                           ...(i)
x = 3y                                 ...(ii)
On substituting x = 3y in (i), we get:
3yy = 26 ⇒ 2y = 26 ⇒ y = 13
On substituting y = 13 in (i), we get:
x − 13 = 26 ⇒ x = 26 + 13 = 39
Hence, the required numbers are 39 and 13.

#### Question 11:

Solve:

The given equations are as follows:
23x + 29y = 98                              ....(i)
29x + 23y = 110                            ....(ii)
On adding (i) and (ii), we get:
52x + 52y = 208
x + y = 4                                   ....(iii)
On subtracting (i) from (ii), we get:
6x − 6y = 12
xy = 2                                   ....(iv)
On adding (iii) and (iv), we get:
2x = 6 ⇒ x = 3
On substituting x = 3 in (iii), we get:
3 + y = 4
y = 4 − 3 = 1
Hence, the required solution is x = 3 and y = 1.

#### Question 12:

Solve:

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

For equation (i), we have:

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

For equation (ii), we have:

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

#### Question 13:

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

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

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

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

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

#### Question 14:

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

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

#### Question 15:

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

Let the cost of each pencil be Rs. x and that of each pen be Rs. y.
Then, we have:
5x + 7y = 195                           ....(i)
7x + 5y = 153                           ....(ii)
Adding (i) and (ii), we get:
12x + 12y = 348
⇒ 12(x + y) = 348
x + y = 29                             ....(iii)
Subtracting (i) from (ii), we get:
2x − 2y = −42
⇒ 2(xy) = −42
xy = −21                          ....(iv)
On adding (iii) and (iv), we get:
2x = 8 ⇒ x = 4
On substituting x = 4 in (iii), we get:
4 + y = 29 ⇒ y = (29 − 4) = 25
Hence, the cost of each pencil is Rs. 4 and the cost of each pen is Rs. 25.

#### Question 16:

Solve the following system of equations graphically:

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

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

 x −1 2 5 y −1 1 3

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

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

#### Question 17:

Find the angles of a cyclic quadrilateral ABCD in which

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

#### Question 18:

Solve for x and y:

We have:

Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$:
35u + 14v 19 = 0                    ....(i)
14u + 35v 37 = 0                    ....(ii)
Here, a1 = 35, b1 = 14, c1 = 19, a2 = 14, b2 = 35, c2 = 37
By cross multiplication, we have: $\frac{u}{-518+665}=\frac{v}{-266+1295}=\frac{1}{1225-196}$
$\frac{u}{147}=\frac{v}{1029}=\frac{1}{1029}$
$u=\frac{147}{1029}=\frac{1}{7},v=\frac{1029}{1029}=1$
$\frac{1}{x+y}=\frac{1}{7},\frac{1}{x-y}=1$
∴ (x + y) = 7                          ....(iii)
And, (x − y) = 1                      ....(iv)

Again, the equations (iii) and (iv) can be written as follows:
x + y 7 = 0                               ....(v)
x y 1 = 0                               ....(vi)
Here, a1 =  1, b1 = 1, c1 = 7 , a2 = 1 , b2 = 1 , c2 = 1
By cross multiplication, we have: $\frac{x}{\left[1×\left(-1\right)-\left(-1\right)×\left(-7\right)\right]}=\frac{y}{\left[\left(-7\right)×1-\left(-1\right)×1\right]}=\frac{1}{\left[1×\left(-1\right)-1×1\right]}$

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

#### Question 19:

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

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

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

#### Question 20:

Solve:

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

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

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