Pair of Linear Equations in Two Variables

EXERCISE-3.1

P1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Sol. Let the present age of Aftab be x.

And, present age of his daughter = y

Seven years ago,

Age of Aftab = x − 7

Age of his daughter = y − 7

According to the question, (x – 7) = 7(y – 7)

x – 7  = 7y – 49

x – 7y = – 42 ———– (1)

After 3 years,

Age of Aftab = x + 3

Age of his daughter = y + 3

According to the question,

(x + 3) = 3(y + 3)

x + 3   = 3y + 9

x – 3y = 6 ————– (2)

Therefore, the algebraic representation is x – 7y = – 42

x – 3y = 6

x – 7y = – 42,

x = – 42 + 7y

The solution table is

x	− 7	0	7
y	5	6	7

For x – 3y = 6

x = 6 + 3y

The solution table is

x	6	3	0
y	0	− 1	− 2

P2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Sol. Let the cost of a bat be Rs x.

And, cost of a ball = Rs y

According to the question, the algebraic representation is 3x + 6y = 3900

x + 2y = 1300

For 3x + 6y = 3900,

$x=\frac{3900–6y}{3}$

The solution table is

x	300	100	− 100
y	500	600	700

For x + 2y = 1300

x = 1300 − 2y

The solution table is

x	300	100	− 100
y	500	600	700

The graphical representation is as follows.

P3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Sol. Let the cost of 1 kg of apples be Rs x.

And, cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation is

2x + y = 160

4x + 2y = 300

For 2x + y = 160,

The solution table is     

x	50	60	70
y	60	40	20

For 4x + 2y = 300,

$y=\frac{300–4x}{2}$

The solution table is

x	70	80	75
y	10	−10	0

EXERCISE-3.2

P1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Sol. (i) Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

x − y = 4

For x + y = 10,

x = 10 − y

x	5	4	6
y	5	6	4

For x − y = 4,

x = 4 + y

x	5	4	3
y	1	0	−1

(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.

According to the question, the algebraic representation is 5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,

$x=\frac{50–7y}{5}$

x	3	10	− 4
y	5	0	10

7x + 5y = 46

$x=\frac{46–5y}{7}$

x	8	3	− 2
y	− 2	5	12

From the figure, it can be observed that these lines intersect each other at point (3, 5).

Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (3, 5).

Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

P2. On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0   

(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0

Sol. (i) 5x − 4y + 8 = 0

7x + 6y − 9 = 0

Comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we obtain

a₁ = 5,     b₁ = – 4,    c₁ = 8

a₂ = 7,     b₂ = 6,   c₂ = – 9

$\frac{a_1}{a_2}=\frac{5}{7}$

$\frac{b_1}{b_2}=\frac{–4}{6}=\frac{–2}{3}$

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$,

Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

Comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we obtain

a₁ = 9,     b₁ = 3,   c₁ = 12

a₂ = 18,   b₂ = 6,   c₂ = 24

$\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},$

Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.

(iii) 6x − 3y + 10 = 0

2x − y + 9 = 0

Comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we obtain

a₁ = 6,     b₁ =–3,   c₁ = 10

a₂ = 2,     b₂ = –1,  c₂ = 9

$\frac{a_1}{a_2}=\frac{6}{2}=\frac{3}{1}$

$\frac{b_1}{b_2}=\frac{–3}{–1}=\frac{3}{1},\frac{c_1}{c_2}=\frac{10}{9}$

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$,

Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

P3. On comparing the ratios $\frac{{\mathbf{a}}_{\mathbf{1}}}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{,}\mathbf{ }\frac{{\mathbf{b}}_{\mathbf{1}}}{{\mathbf{b}}_{\mathbf{2}}}$ and $\frac{{\mathbf{c}}_{\mathbf{1}}}{{\mathbf{c}}_{\mathbf{2}}}$, find out whether the following pair of linear equations are consistent, or inconsistent.

i) 3x + 2y = 5; 2x – 3y = 7

ii) 2x – 3y = 8; 4x – 6y = 9

iii) $\frac{\mathbf{3}}{\mathbf{2}}\mathit{x}\mathbf{+}\frac{\mathbf{5}}{\mathbf{3}}\mathit{y}\mathbf{=}\mathbf{7}$; 9x – 10y = 14

iv) 5x – 3y = 11; –10x + 6y = – 22

v) $\frac{\mathbf{4}}{\mathbf{3}}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{8}\mathbf{;}$ 2x + 3y = 12

Sol. (i) 3x + 2y = 5;  2x − 3y = 7

$\frac{a_1}{a_2}=\frac{3}{2}, \frac{b_1}{b_2}=\frac{2}{–3}, \frac{c_1}{c_2}=\frac{5}{7}$

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

These linear equations are intersecting each other at one point and thus have only one possible solution

Hence, the pair of linear equations is consistent.

(ii) 2x − 3y = 8

4x − 6y = 9

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{–3}{–6}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{8}{9}$

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) $\frac{3}{2}x+\frac{5}{3}y=7$; 9x − 10y = 14

9x − 10y = 14

4x − 6y = 9

$\frac{a_1}{a_2}=\frac{\frac{3}{2}}{9}=\frac{1}{6}, \frac{b_1}{b_2}=\frac{\frac{5}{3}}{–10}=\frac{–1}{6}, \frac{c_1}{c_2}$ $=\frac{7}{14}=\frac{1}{2}$

Since $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv) 5x − 3 y = 11

− 10x + 6y = − 22

$\frac{a_1}{a_2}=\frac{5}{–10}=\frac{–1}{2},\frac{b_1}{b_2}=\frac{–3}{6}=\frac{–1}{2}$, $\frac{c_1}{c_2}=\frac{11}{–22}=\frac{–1}{2}$

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) $\frac{4}{3}x+2y=8$

2x + 3y = 12

$\frac{a_1}{a_2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{2}{3}$, $\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$,

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

P4. Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically:

(i) x + y = 5,     2x + 2y = 10

(ii) x – y = 8,     3x – 3y = 16

(iii) 2x + y – 6 = 0,     4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0,     4x – 4y – 5 = 0

Sol. (i) x + y = 5,    2x + 2y = 10

$\frac{a_1}{a_2}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}$

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

x + y = 5

x = 5 − y

x	4	3	2
y	1	2	3

and, 2x + 2y = 10

$x=\frac{10–2y}{2}$

x	4	3	2
y	1	2	3

From the figure, it can be observed that these lines are overlapping each other.

Therefore, infinite solutions are possible for the given pair of equations.

Hence, the graphic representation is as follows.

(ii) x − y = 8

3x − 3y = 16

$\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{–1}{–3}=\frac{1}{3}$, $\frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}$

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii)  2x + y − 6 = 0

4x − 2y − 4 = 0

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\frac{b_1}{b_2}=\frac{1}{–2}=\frac{–1}{2}$, $\frac{c_1}{c_2}=\frac{–6}{–4}=\frac{3}{2}$

Since $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y − 6 = 0

y = 6 − 2x

x	0	1	2
y	6	4	2

And 4x − 2y − 4 = 0

$y=\frac{4x–4}{2}$

x	1	2	3
y	0	2	4

From the figure, it can be observed that these lines are intersecting each other at the only point i.e., (2, 2) and it is the solution for the given pair of equations.

Hence, the graphic representation is as follows.

(iv) 2x − 2y − 2 = 0

4x − 4y − 5 = 0

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{–2}{–4}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{2}{5}$

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$,

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

P5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Sol. Let the width of the garden be x and length be y.

According to the question,

y − x = 4 ———–  (1)

y + x = 36 ———–   (2)

y − x = 4

y = x + 4

x	0	8	12
y	4	12	16

y + x = 36

y = 36 – x           

x	0	36	16
y	36	0	20

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20).

Therefore, the length and width of the given garden is 20 m and 16 m respectively.

P6. Given the linear equation 2x + 3y − 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

Sol. (i) Intersecting lines:

For this condition,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

The second line such that it is intersecting the given line is

2x + 4y – 6  = 0

as $\frac{a_1}{a_2}=\frac{2}{2}=1, \frac{b_1}{b_2}=\frac{3}{4}$, and $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

(ii) Parallel lines:

For this condition,

$\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{ \mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$

Hence, the second line can be

4x + 6y − 8 = 0 as

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$, $\frac{c_1}{c_2}=\frac{–8}{–8}=1$

and clearly, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

(iii) Coincident lines:

For coincident lines,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence, the second line can be

6x + 9y − 24 = 0 as

$\frac{a_1}{a_2}=\frac{2}{6}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{3}{9}=\frac{1}{3}$, $\frac{c_1}{c_2}=\frac{–8}{–24}=\frac{1}{3}$

and clearly, $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{ \mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$

P7. Draw the graphs of the equations x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Sol. x − y + 1 = 0

x = y − 1

x	0	1	2
y	1	2	3

3x + 2y − 12 = 0

$x=\frac{12–2y}{3}$

x	4	2	0
y	0	3	6

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and  x-axis at (−1, 0) and (4, 0).

Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

EXERCISE-3.3

P1. Solve the following pair of linear equations by the substitution method.

(i) x + y = 14;  x – y = 4

(ii) s – t = 3; $\frac{\mathbf{s}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{t}}{\mathbf{2}}\mathbf{=}\mathbf{6}$

(iii) 3x – y = 13;  9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3

(v) $\sqrt{\mathbf{2}\mathbf{x}}\mathbf{+}\sqrt{\mathbf{3}\mathbf{y}}\mathbf{=}\mathbf{0}$; $\sqrt{\mathbf{3}\mathbf{x}}\mathbf{–}\sqrt{\mathbf{8}\mathbf{y}}\mathbf{=}\mathbf{0}$

(vi) $\frac{\mathbf{3}\mathbf{x}}{\mathbf{2}}\mathbf{–}\frac{\mathbf{5}\mathbf{y}}{\mathbf{3}}\mathbf{=}\mathbf{–}\mathbf{2}$; $\frac{\mathbf{x}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{13}}{\mathbf{6}}$

Sol. (i) x + y = 14 ………….. (1)

x − y = 4 ………….. (2)

From (1), we obtain

x = 14 − y ………….. (3)

Substituting this value in equation (2), we obtain

(14 – y) – y = 4

14 – 2y = 4

2y = 10

y = 5  ………….. (4)

Substituting this in equation (3), we obtain x = 9

$\therefore$ x = 9, y = 5

(ii) s − t = 3   ………….. (1)

$\frac{s}{3}+\frac{t}{2}=6$ ………….. (2)

From (1), we obtain

s = t + 3 ………….. (3)

Substituting this value in equation (2), we obtain

$\frac{t+3}{3}+\frac{t}{2}=6$

2t + 6 + 3t = 36 $\Rightarrow$ 5t = 30

t = 6 ^____________ (4)

Substituting this in equation (3), we obtain

s = 9

$\therefore$ s = 9, t = 6

(iii) 3x − y = 3   ………….. (1)

9x − 3y = 9   ………….. (2)

From (1), we obtain

y = 3x − 3 ………….. (3)

Substituting this value in equation (2), we obtain

9x – 3 (3x – 3) = 9

9x – 9x + 9 = 9 $\Rightarrow$ 9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by

y = 3x − 3

Therefore, one of its possible solutions is x = 1, y = 0.

(iv) 0.2x + 0.3y = 1.3       ……….. (1)

0.4x + 0.5y = 2.3       ……….. (2)

From equation (1), we obtain

$x=\frac{1.3–0.3y}{0.2}$………….. (3)

Substituting this value in equation (2), we obtain

$0.4\left(\frac{1.3–0.3\mathrm{y}}{0.2}\right)+0.5\mathrm{y}=2.3$

2.6 – 0.6y + 0.5y = 2.3

2.6 – 2.3 = 0.1y

0.3 = 0.1y

y = 3 ………….. (4)

Substituting this value in equation (3), we obtain

$x=\frac{1.3–0.3\times 3}{0.2}$ $=\frac{1.3–0.9}{0.2}=\frac{0.4}{0.2}=2$

$\therefore$ x = 2, y = 3

(v)  $\sqrt{2}x+\sqrt{3}y=0$………….. (1)

$\sqrt{3}x–\sqrt{8}y=0$………….. (2)

From equation (1), we obtain

$x=\frac{–\sqrt{3y}}{\sqrt{2}}$ ………….. (3)

Substituting this value in equation (2), we obtain

$\sqrt{3}\left(–\frac{\sqrt{3}\mathrm{y}}{\sqrt{2}}\right)–\sqrt{8}\mathrm{y}=0$

$\Rightarrow –\frac{3y}{\sqrt{2}}–2\sqrt{2}y=0$

$\mathrm{y}\left(–\frac{3}{\sqrt{2}}–2\sqrt{2}\right)=0$

y = 0 ………….. (4)

Substituting this value in equation (3), we obtain

x = 0

$\therefore$ x = 0, y = 0

(vi) $\frac{3x}{2}–\frac{5y}{3}=–2$………….. (1)

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$………….. (2)

From equation (1), we obtain

9x – 10y = –12

$x=\frac{–12+10y}{9}$………….. (3)

Substituting this value in equation (2), we obtain

$\frac{\frac{–12+10\mathrm{y}}{9}}{3}+\frac{\mathrm{y}}{2}=\frac{13}{6}$

$\frac{–12+10y}{27}+\frac{y}{2}=\frac{13}{6}$

$\frac{–24+20\mathrm{y}+27\mathrm{y}}{54}=\frac{13}{6}$

47y = 117 + 24

47y = 141

y = 3 ………….. (4)

Substituting this value in equation (3), we obtain

$x=\frac{–12+10\times 3}{9}=\frac{18}{9}=2$

Hence,    x = 2, y = 3

P2. Solve 2x + 3y = 11 and 2x − 4y = − 24 and hence find the value of ‘m’ for which y = mx + 3.

Sol. 2x + 3y = 11         ………….. (1)

2x – 4y = –24       ………….. (2)

From equation (1), we obtain

$x=\frac{11–3y}{2}$          ………….. (3)

Substituting this value in equation (2), we obtain

$2\left(\frac{11–3y}{2}\right)–4y=–24$

11 – 3y – 4y = –24

–7y = –35

y = 5  ………….. (4)

Putting this value in equation (3), we obtain

$x=\frac{11–3\times 5}{2}=–\frac{4}{2}=–2$

Hence,  x = −2, y = 5

Also,

y = mx + 3

5 = –2m + 3

–2m = 2

m = –1

P3. Form the pair of linear equations for the following problems and find their solution by substitution method.

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

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

(v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Sol. (i) Let the first number be x and the other number be y such that y > x.

According to the given information,

y = 3x                 ………….. (1)

y – x = 26            ………….. (2)

On substituting the value of y from equation (1) into equation (2), we obtain

3x – x = 26

x = 13                 ………….. (3)

Substituting this in equation (1), we obtain

y = 39

Hence, the numbers are 13 and 39.

(ii) Let the larger angle be x and smaller angle be y.

We know that the sum of the measures of angles of a supplementary pair is always 180º.

According to the given information,

x + y = 180°  ………….. (1)

x – y = 18°    ………….. (2)

From (1), we obtain

x = 180º − y  ………….. (3)

Substituting this in equation (2), we obtain

180° – y – y = 18°

162° = 2y

81° = y    ………….. (4)

Putting this in equation (3), we obtain

x = 180º − 81º = 99º

Hence, the angles are 99º and 81º.

(iii) Let the cost of a bat and a ball be x and y respectively.

According to the given information,

7x + 6y = 3800  ………….. (1)

3x + 5y = 1750  ………….. (2)

From (1), we obtain

$y=\frac{3800–7x}{6}$ ………….. (3)

Substituting this value in equation (2), we obtain

$3x+5\left(\frac{3800–7x}{6}\right)=1750$

$3x+\frac{9500}{3}–\frac{35x}{6}=1750$

$3x–\frac{35x}{6}=1750–\frac{9500}{3}$

Let the cost of a bat and a ball be x and y respectively.

$\frac{18x–35x}{6}=\frac{5250–9500}{3}$

$–\frac{17x}{6}=\frac{–4250}{3}$

-17x = -8500

x = 500  ………….. (4)

Substituting this in equation (3), we obtain

$\mathrm{y}=\frac{3800–7\times 500}{6}$ $=\frac{300}{6}=50$

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

(iv) Let the fixed charge be Rs x and per km charge be Rs y.

According to the given information,

x + 10y = 105  ………….. (1)

x + 15y = 155  ………….. (2)

From (3), we obtain

x = 105 – 10y ………….. (3)

Substituting this in equation (2), we obtain

105 – 10y + 15y = 155

5y = 50

y = 10 ………….. (4)

Putting this in equation (3), we obtain

x = 105 – 10 × 10

x = 5

Hence, fixed charge = Rs 5

And per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

(v) Let the fraction be $\frac{x}{y}$

According to the given information,

$\frac{x+2}{y+2}=\frac{9}{11}$

11x + 22 = 9y + 18

11x – 9y = – 4 ………….. (1)

$\frac{x+3}{y+3}=\frac{5}{6}$

6x + 18 = 5y + 15

6x – 5y = – 3 ………….. (2)

From equation (1), we obtain

$x=\frac{–4+9y}{11}$ ………….. (3)

Substituting this in equation (2), we obtain

$6\left(\frac{–4+9\mathrm{y}}{11}\right)–5\mathrm{y}=–3$

– 24 + 54y – 55y = – 33

– y = – 9

y = 9 ………….. (4)

Substituting this in equation (3), we obtain

$x=\frac{–4+81}{11}=7$

Hence, the fraction is $\frac{7}{9}$

(vi) Let the age of Jacob be x and the age of his son be y.

According to the given information,

(x + 5) = 3 (y + 5)

x – 3y = 10  ^{_______________} (1)

(x – 5) = 7(y – 5)

x – 7y = – 30 ^{_______________} (2)

From (1), we obtain

x = 3y + 10 ^____________ (3)

Substituting this value in equation (2), we obtain

3y + 10 – 7y = – 30

– 4y = – 40

y = 10 ^___________ (4)

Substituting this value in equation (3), we obtain

x = 3 × 10 + 10 = 40

Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.

EXERCISE-3.1

P1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x – 2y + 7

(iv) $\frac{\mathbf{x}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{2}\mathbf{y}}{\mathbf{3}}\mathbf{=}\mathbf{–}\mathbf{1}$ and $\mathit{x}\mathbf{–}\frac{\mathbf{y}}{\mathbf{3}}\mathbf{=}\mathbf{3}$

Sol. (i) By elimination method

x + y = 5 ————— (1)

2x − 3y = 4 ————— (2)

Multiplying equation (1) by 2, we obtain

2x + 2y =10 ————— (3)

Subtracting equation (2) from equation (3), we obtain

5y = 6

$y=\frac{6}{5}$————— (4)

Substituting the value in equation (1), we obtain

$x=5–\frac{6}{5}=\frac{19}{5}$

$\therefore x=\frac{19}{5}, y=\frac{6}{5}$

By substitution method

From equation (1), we obtain

x = 5 − y ————— (5)

Putting this value in equation (2), we obtain

2(5 − y) – 3y = 4

−5y = −6

$y=\frac{6}{5}$

Substituting the value in equation (5), we obtain

$x=5–\frac{6}{5}=\frac{19}{5}$

$\therefore x=\frac{19}{5}, y=\frac{6}{5}$

(ii) By elimination method

3x + 4y = 10  ———- (1)

2x − 2y = 2 ———- (2)

Multiplying equation (2) by 2, we obtain

4x − 4y = 4 ———- (3)

Adding equation (1) and (3), we obtain

7x = 14

x = 2       —————- (4)

Substituting in equation (1), we obtain

6 + 4y = 10

4y = 4 $\Rightarrow$y = 1

Hence, x = 2, y = 1

By substitution method

From equation (2), we obtain

From equation (2), we obtain

Putting this value in equation (1), we obtain

3(1 + y) + 4y = 10

7y = 7  $\Rightarrow$   y = 1

Substituting the value in equation (5), we obtain

x = 1 + 1 = 2

$\therefore$  x = 2, y = 1

(iii) By elimination method

3x − 5y − 4 = 0 —————- (1)

9x = 2y + 7 = 0

9x − 2y − 7 = 0 —————- (2)

Multiplying equation (1) by 3, we obtain

9x − 15y − 12 = 0 ————- (3)

Subtracting equation (3) from equation (2), we obtain

13y = − 5

$y=\frac{–5}{13}$—————- (4)

Substituting in equation (1), we obtain

$3x+\frac{25}{13}–4=0$

$3x=\frac{27}{13}$

$x=\frac{9}{13}$

$\therefore \mathrm{x}=\frac{9}{13}, \mathrm{y}=\frac{–5}{13}$

(iii) By substitution method

From equation (1), we obtain

$x=\frac{5y+4}{3}$ ————–  (5)

Putting this value in equation (2), we obtain

$9\left(\frac{5y+4}{3}\right)–2y–7=0$

13y = – 5

$y=–\frac{5}{13}$

Substituting the value in equation (5), we obtain

$x=\frac{5\left(\frac{–5}{13}\right)+4}{3}$

$x=\frac{9}{13}$

$\therefore \mathrm{x}=\frac{9}{13}, \mathrm{y}=\frac{–5}{13}$

(iv) By elimination method

$\frac{x}{2}+\frac{2y}{3}=–1$

3x + 4y = – 6  ——————– (1)

$x–\frac{y}{3}=3$

3x – y = 9     ——————– (2)

Subtracting equation (2) from equation (1), we obtain

5y = – 15

y = – 3 ——————– (3)

Substituting this value in equation (1), we obtain

3x – 12 = – 6

3x = 6

x = 2

Hence, x = 2, y = −3

(iv) By substitution method

From equation (2), we obtain

$x=\frac{y+9}{3}$ ——————– (5)

Substituting the value in equation (1)

$3\left(\frac{\mathrm{y}+9}{3}\right)+4\mathrm{y}=–6$

5y = −15

y = −3

Substituting the value in equation (5), we obtain

$x=\frac{–3+9}{3}=2$

∴ x = 2, y = −3

P2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Sol. Let the fraction be $\frac{x}{y}$.

According to the given information,

$\frac{x+1}{y–1}=1$

$\Rightarrow$ x – y = –2  ——————– (1)

$\frac{x}{y+1}=\frac{1}{2}$

$\Rightarrow$ 2x – y = 1 ——————– (2)

Subtracting equation (1) from equation (2), we obtain

x = 3 ——————– (3)

Substituting this value in equation (1), we obtain

3 – y = – 2

– y = – 5

y =  5

Hence, the fraction is $\frac{3}{5}$.

(ii) Let present age of Nuri = x

and present age of Sonu = y

According to the given information,

(x – 5) = 3(y – 5)

x – 3y = – 10 ——————– (1)

(x + 10) = 2(y + 10)

x – 2y = 10 ——————– (2)

Subtracting equation (1) from equation (2), we obtain

y = 20 ——————– (3)

Substituting it in equation (1), we obtain

x – 60 = – 10

x = 50

Hence, age of Nuri = 50 years

And, age of Sonu = 20 years

(iii) Let the unit digit and tens digits of the number be x and y respectively. Then, number = 10y + x

Number after reversing the digits = 10x + y

According to the given information,

x + y = 9 ——————– (1)

9(10y + x) = 2(10x + y)

88y − 11x = 0

− x + 8y =0 ——————–(2)

Adding equation (1) and (2), we obtain

9y = 9

y = 1 ——————– (3)

Substituting the value in equation (1), we obtain

x = 8

Hence, the number is 10y + x = 10 × 1 + 8 = 18

(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.

According to the given information,

x + y = 25            ——————–(1)

50x + 100y = 2000 ——————– (2)

Multiplying equation (1) by 50, we obtain

50x + 50y = 1250 ——————– (3)

Subtracting equation (3) from equation (2), we obtain

50y = 750

y = 15

Substituting in equation (1), we have x = 10

Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.

According to the given information,

x + 4y = 27 ——————– (1)

x + 2y = 21 ——————– (2)

Subtracting equation (2) from equation (1), we obtain

2y = 6

y = 3       ——————– (3)

Substituting in equation (1), we obtain

x + 12 = 27

x = 15

Hence, fixed charge = Rs 15

and Charge per day = Rs 3

EXERCISE-3.5

P1. Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0; 3x – 9y – 2 = 0

(ii) 2x + y = 5; 3x + 2y = 8

(iii) 3x – 5y = 20; 6x – 10y = 40

(iv) x – 3y – 7 = 0; 3x – 3y – 15 = 0

Sol.   

(i) x – 3y – 3 = 0

3x – 9y – 2 = 0

$\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{–3}{–9}=\frac{1}{3}$, $\frac{c_1}{c_2}=\frac{–3}{–2}=\frac{3}{2}$

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, the given sets of lines are parallel to each other.

Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x +y = 5

3x + 2y = 8

$\frac{a_1}{a_2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{1}{2}$, $\frac{c_1}{c_2}=\frac{–5}{–8}$

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

$\frac{\mathrm{x}}{{\mathrm{b}}_1{\mathrm{c}}_2–{\mathrm{b}}_2{\mathrm{c}}_1}=\frac{\mathrm{y}}{{\mathrm{c}}_1{\mathrm{a}}_2–{\mathrm{c}}_2{\mathrm{a}}_1}=\frac{1}{{\mathrm{a}}_1{ \mathrm{b}}_2–{\mathrm{a}}_2{ \mathrm{b}}_1}$

$\frac{\mathrm{x}}{–8–(–10)}=\frac{\mathrm{y}}{–15+16}=\frac{1}{4–3}$

$\frac{x}{2}=\frac{y}{1}=1$

$\frac{x}{2}=1, \frac{y}{1}=1c$

$\therefore$ x = 2, y = 1

(iii) 3x – 5y = 20

6x – 10y = 40

$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$,

$\frac{b_1}{b_2}=\frac{–5}{10}=\frac{1}{2}$,

$\frac{c_1}{c_2}=\frac{–20}{–40}=\frac{1}{2}$

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

$\frac{a_1}{a_2}=\frac{1}{3}$,

$\frac{b_1}{b_2}=\frac{–3}{–3}=1$,

$\frac{c_1}{c_2}=\frac{–7}{–15}=\frac{7}{15}, \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

$\frac{\mathrm{x}}{45–\left(21\right)}=\frac{\mathrm{y}}{–21–(–15)}=\frac{1}{–3–\left(9\right)}$

$\frac{x}{24}=\frac{y}{–6}=\frac{1}{6}$

$\frac{x}{24}=\frac{1}{6}$ and $\frac{y}{–6}=\frac{1}{6}$

$\therefore$ x = 4 and y = –1

P2. (i) For which values of a and b will the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Sol. 2x + 3y – 7 = 0

(a – b) x + (a + b) y – (3a + b – 2) = 0

$\frac{a_1}{a_2}=\frac{2}{a–b}, \frac{b_1}{b_2}=\frac{3}{a+b},$ $\frac{c_1}{c_2}=\frac{–7}{–(3a+b–2)}=\frac{7}{(3a+b–2)}$

For infinitely many solutions,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{2}{a–b}=\frac{7}{3a+b–2}$

6a + 2b – 4 = 7a – 7b

a – 9b = –4 ^_________ (1)

$\frac{2}{a–b}=\frac{3}{a+b}$

2a + 2b = 3a – 3b

a – 5b = 0 ^_________ (2)

Subtracting (1) from (2), we obtain

4b = 4

b = 1

Substituting this in equation (2), we obtain

a – 5 × 1 = 0

a = 5

Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

(ii) 3x + y – 1 = 0

(2k – 1) x + (k – 1)y – 2k – 1 = 0

$\frac{a_1}{a_2}=\frac{3}{2k–1}$, $\frac{b_1}{b_2}=\frac{1}{k–1}$, $\frac{c_1}{c_2}=\frac{–1}{–2k–1}=\frac{1}{2k+1}$

For no solution,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

$\frac{3}{2k–1}=\frac{1}{k–1}\ne \frac{1}{2k+1}$

$\Rightarrow \frac{3}{2k–1}=\frac{1}{k–1}$

$\Rightarrow$ 3k – 3 = 2k – 1

k = 2

Hence, for k = 2, the given equation has no solution.

P3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Sol. 8x + 5y = 9  ^{______________} (i)

3x + 2y = 4  ^{______________} (ii)

From equation (ii), we obtain

$x=\frac{4–2y}{3}$^{______________} (iii)

Substituting this value in equation (i), we obtain

$8\left(\frac{4–2y}{3}\right)+5y=9$

32 – 16y + 15y = 27

– y  = – 5

y = 5 ^__________ (iv)

Substituting this value in equation (ii), we obtain

3x + 10 = 4

x = – 2

Hence,    x = – 2, y = 5

Again, by cross-multiplication method, we obtain

8x + 5y – 9 = 0

3x + 2y – 4 = 0

$\frac{x}{–20–(–18)}=\frac{y}{–27–(–32)}=\frac{1}{16–15}$

$\frac{x}{–2}=\frac{y}{5}=\frac{1}{1}$

$\frac{x}{–2}=1$ and $\frac{y}{5}=1$

x = –2  and  y = 5

P4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many Ps were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Sol. (i) Let x be the fixed charge of the food and y be the charge for food per day.

According to the given information,

x + 20y = 1000   ……… (1)

x + 26y = 1180   ……… (2)

Subtracting equation (1) from equation (2), we obtain

6y = 180

y = 30

Substituting this value in equation (1), we obtain

x + 20 × 30 = 1000

x = 1000 – 600

x = 400

Hence, fixed charge = Rs 400

And charge per day = Rs 30

(ii) Let the fraction be $\frac{x}{y}$

According to the given information,

$\frac{x–1}{y}=\frac{1}{3}$

$\Rightarrow$ 3x – y = 3  ……… (1)

$\frac{x}{y+8}=\frac{1}{4}$

$\Rightarrow$ 4x – y = 8  ……… (2)

Subtracting equation (1) from equation (2), we obtain

x = 5  ……… (3)

Putting this value in equation (1), we obtain

15 – y = 3 $\Rightarrow$ y = 12

Hence, the fraction is $\frac{5}{12}$

(iii) Let the number of right answers and wrong answers be x and y respectively.

According to the given information,

3x – y = 40  ……… (1)

4x – 2y = 50

$\Rightarrow$ 2x – y = 25  ……… (2)

Subtracting equation (2) from equation (1), we obtain

x = 15 ……… (3)

Substituting this in equation (2), we obtain

30 – y = 25

y = 5

Therefore, number of right answers = 15

And number of wrong answers = 5

Total number of questions =20

(iv) Let the speed of 1^st car and 2^nd car be u km/h and v km/h.

Respective speed of both cars while they are travelling in same direction = (u – v)km/h

Respective speed of both cars while they are travelling in opposite directions

i.e., travelling towards each other = (u +v) km/h

According to the given information,

5(u – v) = 100

$\Rightarrow$ u – v = 20  ……… (1)

1(u + v) = 100 ……… (2)

Adding both the equations, we obtain

2u = 120

u = 60 km/h ……… (3)

Substituting this value in equation (2), we obtain v = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) Let length and breadth of rectangle be x unit and y unit respectively.

Area = xy

According to the question,

(x – 5) (y + 3) = xy – 9

$\Rightarrow$ 3x – 5y – 6 = 0  ……… (1)

(x + 3) (y + 2) = xy + 67

$\Rightarrow$ 2x + 3y – 61 = 0 ……… (2)

By cross-multiplication method, we obtain

$\frac{x}{305–(–18)}=\frac{y}{–12–(–183)}=\frac{1}{9–(–10)}$

x = 17, y = 19

Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.

EXERCISE-3.6

P1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) $\frac{1}{2x}+\frac{1}{3y}=2; \frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

(ii) $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2; \frac{4}{\sqrt{x}}–\frac{9}{\sqrt{y}}=–1$

(iii) $\frac{4}{x}+3y=14; \frac{3}{x}–4y=23$

(iv) $\frac{5}{x–1}+\frac{1}{y–2}=2; \frac{6}{x–1}–\frac{1}{y–2}=1$

(v) $\frac{7x–2y}{xy}=5; \frac{8x+7y}{xy}=15$

(vi) $6x+3y=6xy; 2x+4y=5xy$

(vii) $\frac{10}{x+y}+\frac{2}{x–y}=4;$ $\frac{15}{x+y}=\frac{5}{x–y}=–2$

(viii) $\frac{1}{3x+y}+\frac{1}{3x–y}=\frac{3}{4}$; $\frac{1}{2(3x+y)}–\frac{1}{2(3x–y)}=\frac{–1}{8}$

Sol. (i) $\frac{1}{2x}+\frac{1}{3y}=2$

$\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

Let $\frac{1}{x}=p$ and $\frac{1}{y}=q$, then the equations change as follows.

$\frac{p}{2}+\frac{q}{3}=2$

$\Rightarrow$ 3p + 2q – 12 = 0  ………. (1)

$\frac{p}{3}+\frac{q}{2}=\frac{13}{6}$

$\Rightarrow$ 2p + 3q – 13 = 0  ………. (2)

Using cross-multiplication method, we obtain

$\frac{\mathrm{p}}{–26–(–36)}=\frac{\mathrm{q}}{–24–(–39)}=\frac{1}{9–4}$

$\frac{\mathrm{p}}{10}=\frac{\mathrm{q}}{15}=\frac{1}{5}$

$\frac{p}{10}=\frac{1}{5}$ and $\frac{\mathrm{q}}{15}=\frac{1}{5}$

p = 2 and q = 3

$\frac{1}{x}=2$ and $y=\frac{1}{3}$

$x=\frac{1}{2}$ and $y=\frac{1}{3}$

(ii) $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$

$\frac{4}{\sqrt{x}}–\frac{9}{\sqrt{y}}=–1$

Putting  $\frac{1}{\sqrt{x}}=p$ and $\frac{1}{\sqrt{y}}=q$ in the given equations, we obtain

2p + 3q = 2 ^{_____________} (1)

4p – 9q = –1^{_____________} (2)

Multiplying equation (1) by 3, we obtain

6p + 9q = 6 ^{_____________} (3)

Adding equation (2) and (3), we obtain

10p = 5

$p=\frac{1}{2}$^{_______________} (4)

Putting in equation (1), we obtain

$2\times \frac{1}{2}+3\mathrm{q}=2$

3q = 1

$\mathrm{q}=\frac{1}{3}$

$p=\frac{1}{\sqrt{x}}=\frac{1}{2}$

$\sqrt{x}=2$

x = 4

and  $\mathrm{q}=\frac{1}{\sqrt{\mathrm{y}}}=\frac{1}{3}$

$\sqrt{\mathrm{y}}=3$

y = 9

Hence, x = 4, y = 9

(iii) $\frac{4}{x}+3y=14$,   $\frac{3}{x}–4y=23$

Substituting  $\frac{1}{x}=p$ in the given equations, we obtain

4p + 3y = 14

$\Rightarrow$ 4p + 3y – 14 = 0  ^____________ (1)

3p – 4y = 23

$\Rightarrow$ 3p – 4y – 23 = 0  ^____________ (2)

By cross-multiplication, we obtain

$\frac{\mathrm{p}}{–69–56}–\frac{\mathrm{y}}{–42–(–92)}=\frac{1}{–16–9}$

$\frac{p}{–125}=\frac{y}{50}=\frac{–1}{25}$

$\frac{p}{–125}=\frac{–1}{25}$ and $\frac{y}{50}=\frac{–1}{25}$

p = 5 and y = –2

$p=\frac{1}{x}=5$

$\mathrm{x}=\frac{1}{5}$     y = –2

(iv) $\frac{5}{x–1}+\frac{1}{y–2}=2$

$\frac{6}{x–1}–\frac{3}{y–2}=1$

Putting $\frac{1}{x–1}=p$ and $\frac{1}{y–2}=q$ in the given equation, we obtain

5p + q = 2    ………. (1)

6p – 3q = 1  ………. (2)

Multiplying equation (1) by 3, we obtain

15p + 3q = 6  ………. (3)

Adding (2) an (3), we obtain

21p = 7

$p=\frac{1}{3}$

Putting this value in equation (1), we obtain

$5\times \frac{1}{3}+\mathrm{q}=2$

$q=2–\frac{5}{3}=\frac{1}{3}$

$\mathrm{p}=\frac{1}{\mathrm{x}–1}=\frac{1}{3}$

$\Rightarrow$ x – 1 = 3

$\Rightarrow$ x = 4

$\mathrm{q}=\frac{1}{\mathrm{y}–2}=\frac{1}{3}$

y – 2 = 3

y = 5

$\therefore$  x = 4, y = 5

(v) $\frac{7x–2y}{xy}=5$………. (1)

$\frac{8x+7y}{xy}=15$

$\frac{8}{y}+\frac{7}{x}=15$………. (2)

Putting $\frac{1}{x}=p$ and $\frac{1}{y}=q$ in the given equation, we obtain

-2p + 7q = 5

$\Rightarrow$ -2p + 7q – 5 = 0………. (3)

7p + 8q = 15

$\Rightarrow$ 7p + 8q – 15 = 0 ………. (4)

By cross-multiplication method, we obtain

$\frac{\mathrm{p}}{–105–(–40)}=\frac{\mathrm{q}}{–35–30}=\frac{1}{–16–49}$

$\frac{p}{–65}=\frac{q}{–65}=\frac{1}{–65}$

$\frac{\mathrm{p}}{–65}=\frac{1}{–65} \mathrm{and} \frac{\mathrm{q}}{–65}=\frac{1}{–65}$

p = 1 and q = 1

$\mathrm{p}=\frac{1}{\mathrm{x}}=1 \mathrm{and} \mathrm{q}=\frac{1}{\mathrm{y}}=1$

x = 1  and  y = 1

(vi) 6x + 3y = 6xy

$\Rightarrow \frac{6}{y}+\frac{3}{x}=6$ ………. (1)

2x + 4y = 5xy

$\frac{2}{y}+\frac{4}{x}=5$ ………. (2)

Putting $\frac{1}{\mathrm{x}}=\mathrm{p} \mathrm{and} \frac{1}{\mathrm{y}}=\mathrm{q}$ in these equations, we obtain

3p + 6q – 6 = 0

4p + 2q – 5 = 0

By cross-multiplication method, we obtain

$\frac{\mathrm{p}}{–30–(–12)}=\frac{\mathrm{q}}{–24–(–15)}=\frac{1}{6–24}$

$\frac{p}{–18}=\frac{q}{–9}=\frac{1}{–18}$

$\frac{p}{–18}=\frac{1}{–18}$ and $\frac{\mathrm{q}}{–9}=\frac{1}{–18}$

p = 1 and $q=\frac{1}{2}$

$p=\frac{1}{x}=1, q=\frac{1}{y}=\frac{1}{2}$

x = 1

y = 2

(vii) $\frac{10}{x+y}+\frac{2}{x–y}=4$

$\frac{15}{x+y}–\frac{5}{x–y}=–2$

Putting $\frac{1}{x+y}=p$ and $\frac{1}{x–y}=q$ in the given equations, we obtain

10p + 2q = 4

$\Rightarrow$ 10p + 2q – 4 = 0  ………. (1)

15p – 5q =  -2

$\Rightarrow$ 15p – 5q + 2 = 0  ………. (2)

Using cross-multiplication method, we obtain

$\frac{\mathrm{p}}{4–20}=\frac{\mathrm{q}}{–60–\left(20\right)}=\frac{1}{–50–30}$

$\frac{p}{–16}=\frac{q}{–80}=\frac{1}{–80}$

$\frac{p}{–16}=\frac{1}{–80}\text{ and }\frac{q}{–80}=\frac{1}{–80}$

$p=\frac{1}{5}$ and q = 1

and  x – y = 1   ………. (4)

Adding equation (3) and (4), we obtain

2x = 6

x = 3  ………. (5)

Substituting in equation (3), we obtain

y = 2

Hence, x = 3, y = 2

(viii) $\frac{1}{3x+y}+\frac{1}{3x–y}=\frac{3}{4}$

$\frac{1}{2(3x+y)}–\frac{1}{2(3x–y)}=\frac{–1}{8}$

Putting  $\frac{1}{3x+y}=p\text{ and }\frac{1}{3x–y}=q$ in these equations, we obtain

$p+q=\frac{3}{4}$………. (1)

$\frac{p}{2}–\frac{q}{2}=\frac{–1}{8}$

$p–q=\frac{–1}{4}$………. (2)

Adding (1) and (2), we obtain

$2p=\frac{3}{4}–\frac{1}{4}$

$2p=\frac{1}{2}$

$p=\frac{1}{4}$

Substituting in (2), we obtain

$\frac{1}{4}–\mathrm{q}=\frac{–1}{4}$

$\mathrm{q}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

$p=\frac{1}{3x+y}=\frac{1}{4}$

3x + y = 4 ………. (3)

$q=\frac{1}{3x–y}=\frac{1}{2}$

3x – y = 2 ………. (4)

Adding equations (3) and (4), we obtain

6x = 6

x = 1 ………. (5)

Substituting in (3), we obtain

3(1) + y = 4

y = 1

Hence,    x = 1, y = 1

P2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Sol. (i)  Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.

Speed of Ritu while rowing

Upstream = (x – y) km/h

Downstream = (x + y) km/h

According to question,    2(x + y) = 20

$\Rightarrow$ x + y = 10 ………. (1)

2(x – y) = 4

$\Rightarrow$ x – y = 2 ………. (2)

Adding equation (1) and (2), we obtain

2x = 12 $\Rightarrow$ x = 6

Putting this in equation (1), we obtain

y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii) Let the number of days taken by a woman and a man be x and y respectively.

Therefore, work done by a woman in

1 day = $\frac{1}{x}$

Work done by a man in 1 day = $\frac{1}{y}$

According to the question,

$4\left(\frac{2}{x}+\frac{5}{y}\right)=1$

$\frac{2}{x}+\frac{5}{y}=\frac{1}{4}$

$3\left(\frac{3}{x}+\frac{6}{y}\right)=1$

$\frac{3}{x}+\frac{6}{y}=\frac{1}{3}$

Putting $\frac{1}{x}=p\text{ and }\frac{1}{y}=q$in these equations, we obtain $2p+5q=\frac{1}{4}$

$\Rightarrow$ 8p + 20q = 1

$3p+6q=\frac{1}{3}$

$\Rightarrow$ 9p + 18q = 1

By cross-multiplication, we obtain

$\frac{\mathrm{p}}{–20–(–18)}=\frac{\mathrm{q}}{–9(–8)}=\frac{1}{144–180}$

$\frac{p}{–2}=\frac{q}{–1}=\frac{1}{–36}$

$\frac{p}{–2}=\frac{–1}{36}\text{ and }\frac{q}{–1}=\frac{1}{–36}$

$\mathrm{p}=\frac{1}{18}\text{ and }\mathrm{q}=\frac{1}{36}$

$p=\frac{1}{x}=\frac{1}{18}\text{ and }q=\frac{1}{y}=\frac{1}{36}$

x = 18

y = 36

Hence, number of days taken by a woman= 18

Number of days taken by a man   = 36

(iii) Let the speed of train and bus be u km/h and v km/h respectively.

According to the given information,

$\frac{60}{\mathrm{u}}+\frac{240}{\mathrm{v}}=4$……….(1)

$\left(\because \mathrm{t}=\frac{\text{ distance }}{\text{ speed }}\right)$

$\frac{100}{\mathrm{u}}+\frac{200}{\mathrm{v}}=\frac{25}{6}$   $\left(\because \left(4+\frac{10}{60}\right)\mathrm{hrs}\right)$ ………. (2)

Putting $\frac{1}{u}=p$ and $\frac{1}{v}=q$ in these equations, we obtain

60p + 240q = 4 ………. (3)

100p + 200q

600p + 1200q = 25 ………. (4)

Multiplying equation (3) by 10, we obtain

600p + 2400q = 40 ………. (5)

Subtracting equation (4) from (5), we obtain

1200 q = 15

$\mathrm{q}=\frac{15}{1200}=\frac{1}{80}$………. (6)

Substituting in equation (3), we obtain

60 p + 3 = 4 $\Rightarrow$ 60p = 1

$\mathrm{p}=\frac{1}{60}$

$\mathrm{p}=\frac{1}{\mathrm{u}}=\frac{1}{60}$ and $\mathrm{q}=\frac{1}{\mathrm{v}}=\frac{1}{80}$

u = 60 km/h   and   v = 80 km/h

Hence, speed of train = 60 km/h

Speed of bus = 80 km/h

EXERCISE-3.7

P1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju.

Sol. The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.

Let the age of Ani and Biju be x and y years respectively.

Therefore, age of Ani’s father, Dharam = 2 × x = 2x years

And age of Biju’s sister Cathy  years = $\frac{y}{2}$

By using the information given in the P,

Case (I) When Ani is older than Biju by 3 years,

x − y = 3 (i)

$2x–\frac{y}{2}=30$

4x − y = 60 (ii)

Subtracting (i) from (ii), we obtain

3x = 60 − 3 = 57

$\mathrm{x}=\frac{57}{3}=19$

Therefore, age of Ani = 19 years

And age of Biju = 19 − 3 = 16 years

Case (II) When Biju is older than Ani,

y − x = 3 (i)

$2x–\frac{y}{2}=30$

4x − y = 60 (ii)

Adding (i) and (ii), we obtain

3x = 63

x = 21

Therefore, age of Ani = 21 years

And age of Biju = 21 + 3 = 24 years

P2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II)

[Hint: x + 100 = 2 (y − 100), y + 10 = 6(x − 10)]

Sol. Let those friends were having Rs x and y with them.

Using the information given in the P, we obtain

x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300 (i)

And, 6(x − 10) = (y + 10)

6x − 60 = y + 10

6x − y = 70 (ii)

Multiplying equation (ii) by 2, we obtain

12x − 2y = 140 (iii)

Subtracting equation (i) from equation (iii), we obtain

11x = 140 + 300

11x = 440

x = 40

Using this in equation (i), we obtain

40 − 2y = −300

40 + 300 = 2y

2y = 340

y = 170

Therefore, those friends had Rs 40 and Rs 170 with them respectively.

P3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Sol. Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,

$\text{ Speed }=\frac{\text{ Distance travelled }}{\text{ Time taken to travel that distnace }}$

$x=\frac{d}{t}$

Or, d = xt (i)

Using the information given in the P, we obtain

$(x+10)=\frac{d}{(t–2)}$

(x + 10) (t – 2) = d

xt + 10t – 2x – 20 = d

By using equation (i), we obtain

− 2x + 10t = 20 (ii)

$(x–10)=\frac{d}{(t+3)}$

(x – 10) (t + 2) = d

xt – 10t + 3x – 30 = d

By using equation (i), we obtain

3x − 10t = 30 (iii)

Adding equations (ii) and (iii), we obtain

x = 50

Using equation (ii), we obtain

(−2) × (50) + 10t = 20

−100 + 10t = 20

10t = 120

t = 12 hours

From equation (i), we obtain

Distance to travel = d = xt = 50 × 12= 600 km

Hence, the distance covered by the train is 600 km

P4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Sol. Let the number of rows be x and number of students in a row be y.

Total students of the class = Number of rows × Number of students in a row = xy

Using the information given in the P,

Condition 1

Total number of students = (x − 1) (y + 3)

xy = (x − 1) (y + 3) = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3 (i)

Condition 2

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6 (ii)

Subtracting equation (ii) from (i),

(3x − y) − (3x − 2y) = 3 − (−6) − y + 2y = 3 + 6

y = 9

By using equation (i), we obtain

3x − 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Number of total students in a class = xy = 4 × 9 = 36

P5. In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.

Sol. Given that,

∠C = 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B)

3∠B = 2∠A + 2∠B

∠B = 2∠A

2 ∠A − ∠B = 0 … (i)

We know that the sum of the measures of all angles of a triangle is 180°. Therefore,

∠A + ∠B + ∠C = 180°

∠A + ∠B + 3 ∠B = 180°

∠A + 4 ∠B = 180° … (ii)

Multiplying equation (i) by 4, we obtain

8 ∠A − 4 ∠B = 0 … (iii)

Adding equations (ii) and (iii), we obtain

9 ∠A = 180°

∠A = 20°

From equation (ii), we obtain

20° + 4 ∠B = 180°

4 ∠B = 160°

∠B = 40°

∠C = 3 ∠B = 3 × 40° = 120°

Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

P6. Draw the graphs of the equations 5x − y = 5 and 3x − y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Sol. 5x − y = 5

Or, y = 5x − 5

The solution table will be as follows.

x	0	1	2
y	−5	0	5

3x − y = 3

Or, y = 3x − 3

The solution table will be as follows.

x	0	1	2
y	− 3	0	3

The graphical representation of these lines will be as follows.

It can be observed that the required triangle is ΔABC formed by these lines and y-axis.

The coordinates of vertices are A (1, 0), B (0, − 3), C (0, − 5).

P7. Solve the following pair of linear equations.

(i) px + qy = p − q

qx − py = p + q

(ii) ax + by = c

bx + ay = 1 + c

(iii) $\frac{x}{a}–\frac{y}{b}=0$

ax + by = a² + b²

(iv) (a − b) x + (a + b) y = a²− 2ab − b²

(a + b) (x + y) = a² + b²

(v) 152x − 378y = − 74

− 378x + 152y = − 604

Sol. (i) px + qy = p − q … (1)

qx − py = p + q … (2)

Multiplying equation (1) by p and equation (2) by q, we obtain

p²x + pqy = p² − pq … (3)

q²x − pqy = pq + q² … (4)

Adding equations (3) and (4), we obtain

p²x + q² x = p² + q²

(p² + q²) x = p² + q²

$\mathrm{x}=\frac{{\mathrm{p}}^2+{\mathrm{q}}^2}{{\mathrm{p}}^2+{\mathrm{q}}^2}=1$

From equation (1), we obtain

p (1) + qy = p − q

qy = − q

y = − 1

(ii) ax + by = c … (1)

bx + ay = 1 + c … (2)

Multiplying equation (1) by a and equation (2) by b, we obtain

a²x + aby = ac … (3)

b²x + aby = b + bc … (4)

Subtracting equation (4) from equation (3),

(a² − b²) x = ac − bc − b

$x=\frac{c(a–b)–b}{a^2–b^2}$

From equation (1), we obtain

ax + by = c

$a\left\{\frac{c(a–b)–b}{a^2–b^2}\right\}+by=c$

$\frac{ac(a–b)–ab}{a^2–b^2}+by=c$

$by=c–\frac{ac(a–b)–ab}{a^2–b^2}$

$by=\frac{a^{2}c–b^{2}c–a^{2}c+abc+ab}{a^2–b^2}$

$\text{ by }=\frac{abc–b^{2}c+ab}{a^2–b^2}$

$by=\frac{bc(a–b)+ab}{a^2–b^2}$

$y=\frac{c(a–b)+a}{a^2–b^2}$

(iii) $\frac{x}{a}–\frac{y}{b}=0$

Or, bx − ay = 0 … (1)

ax + by = a² + b² … (2)

Multiplying equation (1) and (2) by b and a respectively, we obtain

b²x − aby = 0 … (3)

a²x + aby = a³ + ab² … (4)

Adding equations (3) and (4), we obtain

b²x + a²x = a³ + ab²

x (b² + a²) = a (a² + b²)

x = a

By using (1), we obtain

b (a) − ay = 0

ab − ay = 0

ay = ab $\Rightarrow$ y = b

(iv) (a − b) x + (a + b) y = a²− 2ab − b² … (1)

(a + b) (x + y) = a² + b²

(a + b) x + (a + b) y = a² + b² … (2)

Subtracting equation (2) from (1), we obtain

(a − b) x − (a + b) x = (a² − 2ab − b²) − (a² + b²)

(a − b − a − b) x = − 2ab − 2b²

− 2bx = − 2b (a + b)

x = a + b

Using equation (1), we obtain

(a − b) (a + b) + (a + b) y = a² − 2ab − b²

a² − b² + (a + b) y = a²− 2ab − b²

(a + b) y = − 2ab

$y=\frac{–2ab}{a+b}$

(v) 152x − 378y = − 74

76x − 189y = − 37

$x=\frac{189y–37}{76}$… (1)

− 378x + 152y = − 604

− 189x + 76y = − 302 … (2)

Substituting the value of x in equation (2), we obtain

$–189\left(\frac{189y–37}{76}\right)+76y=–302$

− (189)² y + 189 × 37 + (76)² y

= − 302 × 76

189 × 37 + 302 × 76

= (189)² y − (76)² y

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

From equation (1), we obtain

$x=\frac{189\left(1\right)–37}{76}$

$\mathrm{x}=\frac{189–37}{76}=\frac{152}{76}$

x = 2

P8. ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

Sol. We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

Therefore, ∠A + ∠C = 180

4y + 20 − 4x = 180

− 4x + 4y = 160

x − y = − 40 (i)

Also, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 (ii)

Multiplying equation (i) by 3, we obtain

3x − 3y = − 120 (iii)

Adding equations (ii) and (iii), we obtain

− 7x + 3x = 180 − 120

− 4x = 60

x = −15

By using equation (i), we obtain

x − y = − 40

−15 − y = − 40

y = −15 + 40 = 25

∠A = 4y + 20 = 4(25) + 20 = 120°

∠B = 3y − 5 = 3(25) − 5 = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = − 7x + 5 = − 7(−15) + 5 = 110°