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

Exercise 9.3 Set 2 builds upon the foundations laid in previous exercises, presenting students with more advanced problems involving linear equations in one variable. This set focuses on word problems and real-life applications of linear equations, challenging students to translate verbal descriptions into mathematical equations and then solve them. The problems cover a wide range of scenarios, helping students see the practical applications of algebra in everyday situations.

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

Solution:

Given:

(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

Summary

Exercise 9.3 Set 2 challenges students to apply their understanding of linear equations to solve real-world problems. It emphasizes the importance of translating word problems into mathematical equations, a crucial skill in algebra. The problems cover various topics such as age calculations, number relationships, geometry, and speed-distance-time calculations. This exercise set helps students develop their problem-solving skills, logical thinking, and ability to connect mathematical concepts to practical situations. By working through these problems, students not only reinforce their algebraic skills but also learn to approach complex scenarios methodically.