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Class 8 RD Sharma Solutions – Chapter 9 Linear Equation In One Variable – Exercise 9.3 | Set 2

  • Last Updated : 07 Apr, 2021

Chapter 9 Linear Equation In One Variable – Exercise 9.3 | Set 1

Question 13. (7x – 2)/(5x – 1) = (7x + 3)/(5x + 4)

Solution:

Given:

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(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)



(7x – 2) / (5x – 1) – (7x +3)/(5x + 4) = 0

By taking LCM as (5x – 1) (5x + 4)

((7x-2) (5x+4) – (7x+3)(5x-1)) / (5x – 1) (5x + 4) = 0

After cross-multiplying we will get,

(7x-2) (5x+4) – (7x+3)(5x-1) = 0

Now after simplification,

35x2 + 28x – 10x – 8 – 35x2 + 7x – 15x + 3 = 0

10x – 5 = 0

10x = 5

x = 5/10

x = 1/2

Now verify the given equation by substituting x = 1/2,

(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)

x = 1/2

(7(1/2) – 2) / (5(1/2) – 1) = (7(1/2) + 3) /(5(1/2) + 4)

(7/2 – 2) / (5/2 – 1) = (7/2 + 3) / (5/2 + 4)

((7-4)/2) / ((5-2)/2) = ((7+6)/2) / ((5+8)/2)

(3/2) / (3/2) = (13/2) / (13/2)



1 = 1

Here, L.H.S. = R.H.S., 

Thus the given equation is verified.

Question 14. ((x+1)/(x+2))2 = (x+2)/(x + 4)

Solution:

Given:

((x+1)/(x+2))2 = (x+2) / (x + 4)

(x+1)2 / (x+2)2 – (x+2) / (x + 4) = 0

By taking LCM as (x+2)2 (x+4)

((x+1)2 (x+4) – (x+2) (x+2)2) / (x+2)2 (x+4) = 0

After cross-multiplying we will get,

(x+1)2 (x+4) – (x+2) (x+2)2 = 0

Now expand the equation as follows,

(x2 + 2x + 1) (x + 4) – (x + 2) (x2 + 4x + 4) = 0

x3 + 2x2 + x + 4x2 + 8x + 4 – (x3 + 4x2 + 4x + 2x2 + 8x + 8) = 0

x3 + 2x2 + x + 4x2 + 8x + 4 – x3 – 4x2 – 4x – 2x2 – 8x – 8 = 0

-3x – 4 = 0

x = -4/3

Now verify the given equation by substituting x = -4/3,

((x+1)/(x+2))2 = (x+2) / (x + 4)

x = -4/3



(x+1)2 / (x+2)2 = (x+2) / (x + 4)

(-4/3 + 1)2 / (-4/3 + 2)2 = (-4/3 + 2) / (-4/3 + 4)

((-4+3)/3)2 / ((-4+6)/3)2 = ((-4+6)/3) / ((-4+12)/3)

(-1/3)2 / (2/3)2 = (2/3) / (8/3)

1/9 / 4/9 = 2/3 / 8/3

1/4 = 2/8

1/4 = 1/4

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 15. ((x+1)/(x-4))2 = (x+8)/(x-2)

Solution:

Given:

((x+1)/(x-4))2 = (x+8)/(x-2)

(x+1)2 / (x-4)2 – (x+8) / (x-2) = 0

By taking LCM as (x-4)2 (x-2)

((x+1)2 (x-2) – (x+8) (x-4)2) / (x-4)2 (x-2) = 0

After cross-multiplying we will get,

(x+1)2 (x-2) – (x+8) (x-4)2 = 0

After expansion we get,

(x2 + 2x + 1) (x-2) – ((x+8) (x2 – 8x + 16)) = 0

x3 + 2x2 + x – 2x2 – 4x – 2 – (x3 – 8x2 + 16x + 8x2 – 64x + 128) = 0

x3 + 2x2 + x – 2x2 – 4x – 2 – x3 + 8x2 – 16x – 8x2 + 64x – 128 = 0

45x – 130 = 0

x = 130/45

x = 26/9

Now verify the given equation by substituting x = 26/9

((x+1)/(x-4))2 = (x+8)/(x-2)

(x+1)2 / (x-4)2 = (x+8) / (x-2)

x = 26/9

(26/9 + 1)2 / (26/9 – 4)2 = (26/9 + 8) / (26/9 – 2)

((26+9)/9)2 / ((26-36)/9)2 = ((26+72)/9) / ((26-18)/9)



(35/9)2 / (-10/9)2 = (98/9) / (8/9)

(35/-10)2 = (98/8)

(7/2)2 = 49/4

49/4 = 49/4

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 16. (9x-7)/(3x+5) = (3x-4)/(x+6)

Solution:

Given:

(9x-7)/(3x+5) = (3x-4)/(x+6)

(9x-7)/(3x+5) – (3x-4)/(x+6) = 0

By taking LCM as (3x+5) (x+6)

((9x-7) (x+6) – (3x-4) (3x+5)) / (3x+5) (x+6) = 0

After cross-multiplying we will get,

(9x-7) (x+6) – (3x-4) (3x+5) = 0

Upon expansion we will get,

9x2 + 54x – 7x – 42 – (9x2 + 15x – 12x – 20) = 0

44x – 22 = 0

44x = 22

x = 22/44

= 2/4

x = 1/2

Now verify the given equation by substituting x =1/2,

(9x-7)/(3x+5) = (3x-4)/(x+6)

x = 1/2

(9(1/2) – 7) / (3(1/2) + 5) = (3(1/2) – 4) / ((1/2) + 6)

(9/2 – 7) / (3/2 + 5) = (3/2 – 4) / (1/2 + 6)

((9-14)/2) / ((3+10)/2) = ((3-8)/2) / ((1+12)/2)

-5/2 / 13/2 = -5/2 / 13/2

-5/13 = -5/13

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 17. (x+2)/(x+5) = x/(x+6)

Solution:

Given:

(x+2)/(x+5) = x/(x+6)

(x+2)/(x+5) – x/(x+6) = 0

By taking LCM as (x+5) (x+6)

((x+2) (x+6) – x(x+5)) / (x+5) (x+6) = 0

After cross-multiplying we will get,

(x+2) (x+6) – x(x+5) = 0

Upon expansion we will get

x2 + 8x + 12 – x2 – 5x = 0

3x + 12 = 0

3x = -12

x = -12/3

x = -4

Now verify the given equation by substituting x = -4,

(x+2)/(x+5) = x/(x+6)

x = -4

(-4 + 2) / (-4 + 5) = -4 / (-4 + 6)

-2/1 = -4 / (2)



-2 = -2

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 18. 2x – (7-5x)/9x – (3+4x) = 7/6

Solution:

Given:

2x – (7-5x) / 9x – (3+4x) = 7/6

(2x – 7 + 5x) / (9x – 3 – 4x) = 7/6

(7x – 7) / (5x – 3) = 7/6

After cross-multiplying we will get,

6(7x – 7) = 7(5x – 3)

42x – 42 = 35x – 21

42x – 35x = -21 + 42

7x = 21

x = 21/7

x = 3

Now verify the given equation by substituting

2x – (7-5x) / 9x – (3+4x) = 7/6

(7x – 7) / (5x – 3) = 7/6

x = 3

(7(3) -7) / (5(3) – 3) = 7/6

(21-7) / (15-3) = 7/6

14/12 = 7/6

7/6 = 7/6

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 19. (15(2-x) – 5(x+6))/(1-3x) = 10

Solution:

Given:

15(2-x) – 5(x+6) / (1-3x) = 10

(30-15x) – (5x + 30) / (1-3x) = 10

After cross-multiplying we will get,

(30-15x) – (5x + 30) = 10(1- 3x)

30- 15x – 5x – 30 = 10 – 30x

30- 15x – 5x – 30 + 30x = 10

10x = 10

x = 10/10

x = 1

Now verify the given equation by substituting x =1,

(15(2-x) – 5(x+6)) / (1-3x) = 10

x = 1

(15(2-1) – 5(1+6)) / (1- 3) = 10

(15 – 5(7))/-2 = 10

(15-35)/-2 = 10

-20/-2 = 10

10 = 10

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 20. (x+3)/(x-3) + (x+2)/(x-2) = 2

Solution:

Given:

(x+3)/(x-3) + (x+2)/(x-2) = 2

By taking LCM as (x-3) (x-2)

((x+3)(x-2) + (x+2) (x-3)) / (x-3) (x-2) = 2

After cross-multiplying we will get,

(x+3)(x-2) + (x+2) (x-3) = 2 ((x-3) (x-2))

Upon expansion we will get,

x2 + 3x – 2x – 6 + x2 – 3x + 2x – 6 = 2(x2 – 3x – 2x + 6)

2x2 – 12 = 2x2 – 10x + 12

2x2 – 2x2 + 10x = 12 + 12

10x = 24

x = 24/10

x = 12/5



Now verify the given equation by substituting x = 12/5,

(x+3)/(x-3) + (x+2)/(x-2) = 2

x = 12/5

(12/5 + 3)/(12/5 – 3) + (12/5 + 2)/(12/5 – 2) = 2

((12+15)/5)/((12-15)/5) + ((12+10)/5)/((12-10)/5) = 2

(27/5)/(-3/5) + (22/5)/(2/5) = 2

-27/3 + 22/2 = 2

((-27×2) + (22×3))/6 = 2

(-54 + 66)/6 = 2

12/6 = 2

2 = 2

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 21. ((x+2)(2x-3) – 2x2 + 6)/(x-5) = 2

Solution:

We have,

((x+2) (2x-3) – 2x2 + 6)/(x-5) = 2

After cross-multiplying we will get,

(x+2) (2x-3) – 2x2 + 6) = 2(x-5)

2x2 – 3x + 4x – 6 – 2x2 + 6 = 2x – 10

x = 2x – 10

x – 2x = -10

-x = -10

x = 10

Now verify the given equation by substituting x = 10

((x+2) (2x-3) – 2x2 + 6)/(x-5) = 2

x = 10

((10+2) (2(10) – 3) – 2(10)2 + 6)/ (10-5) = 2

(12(17) – 200 + 6)/5 = 2

(204 – 194)/5 = 2

10/5 = 2

2 = 2

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 22. (x2 – (x+1)(x+2))/(5x+1) = 6

Solution:

Given:

(x2 – (x+1) (x+2))/(5x+1) = 6

After cross-multiplying we will get,

(x2 – (x+1) (x+2)) = 6(5x+1)

x2 – x2 – 2x – x – 2 = 30x + 6

-3x – 2 = 30x + 6

30x + 3x = -2 – 6

33x = -8

x = -8/33

Now verify the given equation by substituting x = -8/33

(x2 – (x+1) (x+2))/(5x+1) = 6

x = -8/33

((-8/33)2 – ((-8/33)+1) (-8/33 + 2))/(5(-8/33)+1) = 6

(64/1089 – ((-8+33)/33) ((-8+66)/33)) / (-40+33)/33) = 6

(64/1089 – (25/33) (58/33)) / (-7/33) = 6

(64/1089 – 1450/1089) / (-7/33) = 6

((64-1450)/1089 / (-7/33)) = 6

-1386/1089 × 33/-7 = 6

1386 × 33 / 1089 × -7 = 6

6 = 6

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 23. ((2x+3) – (5x-7))/(6x+11) = -8/3

Solution:

Given:

((2x+3) – (5x-7))/(6x+11) = -8/3

After cross-multiplying we will get,

3((2x+3) – (5x-7)) = -8(6x+11)

3(2x + 3 – 5x + 7) = -48x – 88

3(-3x + 10) = -48x – 88

-9x + 30 = -48x – 88

-9x + 48x = -88 – 30

39x = -118

x = -118/39

Now verify the given equation by substituting x = -118/39

((2x+3) – (5x-7))/(6x+11) = -8/3

x = -118/39



((2(-118/39) + 3) – (5(-118/39) – 7)) / (6(-118/39) + 11) = -8/3

((-336/39 + 3) – (-590/39 – 7)) / (-708/39 + 11) = -8/3

(((-336+117)/39) – ((-590-273)/39)) / ((-708+429)/39) = -8/3

(-219+863)/39 / (-279)/39 = -8/3

644/-279 = -8/3

-8/3 = -8/3

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 24. Find the positive value of x for which the given equation is satisfied:

(i) (x2 – 9)/(5+x2) = -5/9

Solution:

Given:

(x2 – 9)/(5+x2) = -5/9

After cross-multiplying we will get,

9(x2 – 9) = -5(5+x2)

9x2 – 81 = -25 – 5x2

9x2 + 5x2 = -25 + 81

14x2 = 56

x2 = 56/14

x2 = 4

x = √4

x = 2

(ii) (y2 + 4)/(3y2 + 7) = 1/2

Solution:

Given:

(y2 + 4)/(3y2 + 7) = 1/2

After cross-multiplying we will get,

2(y2 + 4) = 1(3y2 + 7)

2y2 + 8 = 3y2 + 7

3y2 – 2y2 = 7 – 8

y2 = -1

y = √-1

= 1




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