 # RS Aggarwal Class 10 Solutions Chapter 3 - Linear Equations In Two Variables - Ex 3D(3.4)

## RS Aggarwal Class 10 Chapter 3 - Linear Equations In Two Variables - Ex 3D(3.4) Solutions Free PDF

The RS Aggarwal Solutions Class 10 Chapter 3 – Linear Equations In Two Variables is the best solutions manual available on our BYJU’S website from the exam point of view. The more you practice and solve different types of questions, the faster you can solve difficult questions asked in board and other competitive exams. It helps you to learn various mathematics tricks and shortcuts for quick and easy calculations. The solutions will end all your worries and doubts and help you to fetch good marks.

You are required to go through RS Aggarwal Class 10 Solutions for Chapter 3 thoroughly before the final exams to score well and enhance your problem-solving abilities. The solutions are primarily designed for CBSE students and are based on the latest syllabus prescribed as per the CCE guidelines by CBSE Board.

## Download PDF of RS Aggarwal Class 10 Chapter 3 – Linear Equations In Two Variables – Ex 3D (3.4)

Question 1: 3x – 5y – 19 = 0, -7x + 3y + 1= 0

Solution:

The given equations are.

3x – 5y – 19 = 0 …..(1)

-7x + 3y + 1= 0 …..(2)

Multiplying (1) by 3 and (2) by 5, we get,

9x – 15y = 57 …..(3)

-35x + 15y = -5 …..(4)

Adding (3) and (4), we get

-26x = 52 $\Rightarrow$ x = -2

Substituting x = -2 in (1), we get

(3)(-2) – 5y = 19 $\Rightarrow$ -6 – 5y = 19

-5y = 19 + 6 $\Rightarrow$ -5y = 25

Y = -5

∴ x = -2 and y = -5

Question 2: 4x – 3y = 8, 6x – y = (29/3)

Solution:

The given equations are.

4x – 3y = 8 …..(1)

6x – y = (29/3) …..(2)

Multiplying (1) by 1 and (2) by 3, we get,

4x – 3y = 8 …..(3)

18x – 3y = 29 …..(4)

Subtracting (3) from (4), we get

14x = 21 $\Rightarrow x = \frac{21}{14} = \frac{3}{2}$

Substituting $x = \frac{3}{2}$ in (1), we get

$4 \times \frac{3}{2} – 3y = 8 \Rightarrow 6 – 3y = 8 \Rightarrow y = \frac{-2}{3}$

$x = \frac{3}{2} \, and \, y = \frac{-2}{3}$

Question 3: $2x – \frac{3y}{4} = 3$, $5x = 2y + 7$

Solution:

The given equations are.

$2x – \frac{3y}{4} = 3$ …..(1)

$5x = 2y + 7$ …..(2)

Multiplying (1) by 2 and (2) by 3/4, we get,

$4x – \frac{3y}{2} = 6$ …..(3)

$\frac{15x}{4} – \frac{3y}{2} = \frac{21}{4}$ …..(4)

Subtracting (3) from (4), we get

$\frac{-1}{4} x = \frac{-3}{4}$

-x = -3 $\rightarrow$ x = 3

Substituting x = 3 in (1), we get

$2 \times 3 – \frac{3y}{4} = 3 \Rightarrow \frac{3y}{4} = 3 – 6 \Rightarrow y = \frac{-3 \times 4}{-3} \, = \, 4$

∴ Solution is x = 3 and y =4

Question 4: 11x – 5y + 23 = 0, 7x – 2y – 20 = 0

Solution:

The given equations are.

11x – 5y + 23 = 0 …..(1)

7x – 2y – 20 = 0 …..(2)

Multiplying (1) by 2 and (2) by 15, we get,

22x + 30y = -46 …..(3)

105x – 30y = 300 …..(4)

Adding (3) and (4), we get

127x = 254 $\Rightarrow x = \frac{254}{127} = 2$

Substituting x = 2 in (1), we get

(11)(2) + 15y = -23

15y = -23 – 22 $\Rightarrow$ 15y = -45

Y = -3

∴ Solution is x = 2 and y = -3

Question 5: 2x – 5y + 8 = 0, x – 4y + 7 = 0

Solution:

The given equations are.

2x – 5y + 8 = 0 …..(1)

x – 4y + 7 = 0 …..(2)

Multiplying (1) by 4 and (2) by 5, we get,

8x – 20y = -32 …..(3)

5x – 20y = -35 …..(4)

Subtracting (3) from (4), we get

-3x = -3 $\Rightarrow$ x = 1

Substituting x = 1 in (1), we get

(2)(1) – 5y = -8

-5y = -8 – 2 $\Rightarrow$ -5y = -10

y = 2

∴ Solution is x = 1 and y = 2

Question 6: 7 – 4x = 3y $\Rightarrow$ -4x – 3y = -7, 2x + 3y = -1

Solution:

The given equations are.

7 – 4x = 3y  $\Rightarrow$ -4x – 3y = -7 …..(1)

2x + 3y = -1 …..(2)

Subtracting (2) from (1), we get

2x = 8 $\Rightarrow$ x = 4

Substituting x = 4 in (1), we get

(4)(4) + 3y = 7

3y =7 – 16 $\Rightarrow$ 3y = -9

y = -3

∴ Solution is x = 4 and y = -3

Question 7: 2x + 5y = (8/3), 3x – 2y = (5/6)

Solution:

The given equations are.

2x + 5y = (8/3) …..(1)

3x – 2y = (5/6) …..(2)

Multiplying (1) by 2 and (2) by 5, we get,

4x + 10y = (16/3)  …..(3)

15x – 10y = (25/6) …..(4)

Adding (3) and (4), we get

19x = (57/6)  $\Rightarrow x = \frac{57}{6 \times 19} = \frac{1}{2}$

Substituting x = ½ in (1), we get

(4)(1/2)+ 10y = (16/3)

10y = (16/3) – 2 $\Rightarrow 10y = (10/3)$

$y = \frac{10}{3 \times 10} = \frac{1}{3}$

∴ Solution is x = (½) and y = (⅓)

Question 8: $\frac{x}{3} + \frac{y}{4} = 11$, $\frac{5x}{6} – \frac{y}{3} = -7$

Solution:

The given equations are:

$\frac{x}{3} + \frac{y}{4}$ = 11

$\frac{5x}{6} – \frac{y}{3}$ = -7

$\frac{x}{3} + \frac{y}{4}$ = 11 (by taking LCM)

$\frac{4x + 3y}{12}$ = 11

4x + 3y = 132 ……..(1)

$\frac{5x}{6} – \frac{y}{3}$ = -7 (by taking LCM)

$\frac{5x – 2y}{6} = -7$

5x – 2y = -42 ……..(2)

Multiplying (1) by 2 and (2) by 3, we get,

8x + 6y = 264 …..(3)

15x – 6y = -126 …..(4)

Adding (3) and (4), we get

23x = 138 $\Rightarrow$ x = 6

Substituting x = 6 in (1), we get

(4)(6) + 3y = 132

3y = 132 – 24 $\Rightarrow$ 3y = 36

y = 36

∴ Solution is x = 6 and y = 36

Question 9: 7(y + 3) – 2(x + 2) = 14, 4(y – 2) + 3(x – 3) = 2

Solution:

The given equations are.

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2

7(y + 3) – 2( x + 2) = 14

$\Rightarrow$ 7y + 21 – 2x – 4 = 14

$\Rightarrow$ 7y – 2x = 14 + 4 -21

$\Rightarrow$ – 2x + 7y = -3 ……..(1)

4(y – 2) + 3(x – 3) = 2

$\Rightarrow$ 4y – 8 + 3x – 9 = 2

$\Rightarrow$ 4y+ 3x = 2 + 8 + 9

$\Rightarrow$ 3x + 4y = 19 ……..(2)

Multiplying (1) by 4 and (2) by 7, we get,

-8x + 28y = -12 …..(3)

21x + 28y = 133 …..(4)

Subtracting (3) from (4), we get

29x = 145 $\Rightarrow$ x = 5

Substituting x = 5 in (1), we get

(-2)(5) + 7y = -3

7y = -3 + 10 $\Rightarrow$ 7y = 7

y = 1

∴ Solution is x = 5 and y = 1

Question 10: 6x + 5y = 7x + 2y + 1 = 2(x + 6y – 1)

Solution:

The given equations are.

6x + 5y = 7x + 2y + 1 = 2(x + 6y – 1)

Therefore, we have,

6x + 5y = 2(x + 6y -1)

$\Rightarrow$ 6x + 5y = 2x + 12y -2

$\Rightarrow$ 6x – 2x + 5y – 12y = -2

$\Rightarrow$ 4x – 7y = -2  ……..(1)

7x + 3y + 1= 2(x + 6y -1)

$\Rightarrow$ 7x + 3y + 1 = 2x + 12y -2

$\Rightarrow$ 7x – 2x + 3y – 12y = -2 -1

$\Rightarrow$ 5x -9y = -3 ……..(2)

Multiplying (1) by 9 and (2) by 7, we get,

36x – 63y = -18 …..(3)

35x – 63y = -21 …..(4)

Subtracting (4) from (3), we get

x = 3

Substituting x = 3 in (1), we get

(4)(3) – 7y = -2

-7y = -2 – 12 $\Rightarrow$ -7y = -14

y = 2

∴ Solution is x = 3 and y = 2

Question 11: $\frac{x + y – 8}{2} = \frac{x + 2y – 14}{3} = \frac{3x + y – 12}{11}$

Solution:

The given equations are.

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

Therefore, we have,

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

By cross multiplication, we get

11x + 11y – 88 = 6x + 2y – 24

11x – 6x + 11y – 2y = -24 + 88

5x + 9y = 64 ……..(1)

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

11x + 22y – 154 = 9x + 3y – 36

11x – 9x + 22y – 3y = -36 + 154

2x + 19y = 118 ……..(2)

Multiplying (1) by 19 and (2) by 9, we get,

95x + 171y = 1216 …..(3)

18x + 171y = 1062 …..(4)

Subtracting (4) from (3), we get

77x = 154 $\Rightarrow$ x = 2

Substituting x = 2 in (1), we get

(5)(2) + 9y = 64

$\Rightarrow$ 9y = 54

y = 6

∴ Solution is x = 2 and y = 6

Question 12: 0.8x + 0.3y = 3.8, 0.4x – 0.5y = 0.6

Solution:

The given equations are.

0.8x + 0.3y = 3.8 ……..(1)

0.4x – 0.5y = 0.6 ……..(2)

Multiplying each one of the equation by 10, we get

8x + 3y = 38 ……..(3)

4x – 5y = 6 ……..(4)

Multiplying (3) by 5 and (4) by 3, we get,

40x + 15y = 190 …..(5)

12x – 15y = 18 …..(6)

Adding (5) and (6), we get

52x = 208 $\Rightarrow x = \frac{208}{52} = 4$

Substituting x = 4 in (3), we get

(8)(4) + 3y = 38

3y = 38 – 32 $\Rightarrow$ 3y = 6

y = 2

∴ Solution is x = 4 and y = 2

Question 13: 0.05x + 0.2y = 0.07, 0.3x – 0.1y = 0.03

Solution:

The given equations are.

0.05x + 0.2y = 0.07 ……..(1)

0.3x – 0.1y = 0.03 ……..(2)

Multiplying each one of the equation by 100, we get

5x + 20y = 7 ……..(3)

30x – 10y = 3 ……..(4)

Multiplying (3) by 10 and (4) by 20, we get,

50x + 200y = 70 …..(5)

600x – 200y = 60 …..(6)

Adding (5) and (6), we get

650x = 130 $\Rightarrow x = \frac{130}{650} = \frac{1}{5} = 0.2$

Substituting x = 0.2 in (3), we get

(5)(0.2) + 20y = 7

1 + 20y = 7 $\Rightarrow$ 20y = 6

y = 0.3

∴ Solution is x = 0.2 and y = 0.3

Question 14: mx – ny = m2 + n2, x + y = 2m

Solution:

mx – ny = m2 + n2 ……..(1)

x + y = 2m ……..(2)

Multiplying (1) by 1 and (2) ny n

mx – ny = m2 + n2 ……..(3)

nx + ny = 2mn ……..(4)

Adding (3) and (4), we get

mx + nx = m2 + n2 + 2mn

x(m + n) (m + n)2

$x = \frac{(m + n)^{2}}{m + n} = m + n$

Putting x = m + n in (1), we get

m(m + n) – ny =  m2 + n2

m2 + mn – ny = m2 + n2

-ny = m2 + n2 – m2 – mn

-ny = n2 – nm

-y = $\frac{n(n – m)}{n}$

-y = (n – m)

y = (m – n)

∴ The solution is x = (m + n) and y = (m -n)

Question 15:  $\frac{bx}{a} – \frac{ay}{b} + a + b = 0$

Solution:

$\frac{bx}{a} – \frac{ay}{b} + a + b = 0$

By taking LCM, we get

$\frac{b^{2}x – a^{2}y + a^{2}b + b^{2}a}{ab} = 0$

b2x – a2y = -a2b – b2a ……..(1)

bx – ay = -2ab ……..(2)

Multiplying (1) by 1 and (2) by ‘a’,

b2x – a2y = -a2b – b2a ……..(3)

abx – a2y = -2a2b ……..(4)

Subtracting (3) from (4)

(ab – b2)x = -2a2b + a2b + ab2

b(a – b)x = -a2b + ab2 = -ab(a – b)

$x = \frac{-ab(a – b)}{b(a – b)}$

x = -a

Putting x = -a, in (1), we get

b2(-a) – a2y = -a2b – b2a

-ab2 – a2y = -a2b – b2a

-a2y = -a2b – b2a + ab2

-a2y = -a2b $\Rightarrow y = \frac{-a^{2}b}{-a^{2}} = b$

∴ Solution is x = -a, y = b.

Question 16: $\frac{x}{a} + \frac{y}{b} = 2$

Solution:

$\frac{x}{a} + \frac{y}{b} = 2$

By taking LCM, we get

$\frac{bx + ay}{ab} = 2$

bx – ay = 2ab ……..(1)

ax – by = (a2 – b2) ……..(2)

Multiplying (1) by ‘b’ and (2) by ‘a’,

b2x + bay = 2ab2 ……..(3)

a2x – bay = a(a2 – b2)  ……..(4)

Adding (3) and (4), we get

b2x + a2x = 2ab2 + a(a2 – b2)

x(b2 + a2) = 2ab2 + a3 – ab2

x(b2 + a2) = a(b2 + a2)

$x = \frac{a(b^{2} + a^{2})}{(b^{2} + a^{2})} = a$

x = a

Putting x = a, in (1), we get

(b)(a) + ay = 2ab

ay = 2ab – ab $\Rightarrow$ ay = ab or y = b

∴ Solution is x = a, y = b.

Question 17: $\frac{bx}{a} – \frac{ay}{b} = a^{2} + b^{2}$

Solution:

$\frac{bx}{a} – \frac{ay}{b} = a^{2} + b^{2}$

By taking LCM, we get

$\frac{b^{2}x – a^{2}y}{ab} = a^{2} + b^{2}$

b2x + a2y = ab(a2 + b2) ……..(1)

x + y = 2ab ……..(2)

Multiplying (1) by 1 and (2) by ‘a2’,

b2x + a2y = a3b + b3a ……..(3)

a2x + a2y = 2a3b ……..(4)

Subtracting (4) from (3)

b2x – a2x = a3b + ab3– 2a3b

x(b2 – a2) = ab3 – a3b

x(b2– a2) = ab(b2 – a2)

$x = \frac{ab(b^{2} – a^{2}}{b^{2} – a^{2}} = ab$

x = ab

Putting x = ab, in (3), we get

b2(ab) + a2y = a3b + b3a

ab3 + a2y = a3b + b3a

a2y = a3b + b3a – ab3

a2y = a3b $\Rightarrow y = \frac{-a^{3}b}{a^{2}} = ab$

∴ Solution is x = ab, y = ab.

### Key Features of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3D (3.4)

• It enables you to have a good grasp of basic mathematical concepts from the very beginning itself.
• It provides all the solutions to the exercise questions mentioned in the Class 1o RS Aggarwal textbook.
• All the solutions are explained in a proper step by step manner for easy understanding
• The RS Aggarwal Class 10 Maths solution is the perfect reference material for Class 10 CBSE students.

#### Practise This Question

When a current carrying coil has a continuously varying current, and is placed close to another loop of wire, current will be induced in the wire.