RS Aggarwal Solutions Class 10 Ex 3D

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.

 


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