# RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable Exercise 9.3

RD Sharma Solutions for Class 8 Maths Exercise 9.3 Chapter 9 is provided here. In Exercise 9.3, we shall discuss the concept of cross-multiplication method for solving equations. RD Sharma Solutions which is designed by our experts at BYJUâ€™S to help students understand the concepts clearly. On regular practice, students can top their exams. Students can download the pdf easily from the links provided below.

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1. (2x-3) / (3x+2) = -2/3

Solution:

We have,

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

Let us perform cross-multiplication we get,

3(2x – 3) = -2(3x + 2)

6x â€“ 9 = -6x â€“ 4

When rearranged,

6x + 6x = 9 â€“ 4

12x = 5

x = 5/12

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(2(5/12) – 3) / (3(5/12) + 2) = -2/3

((5/6)-3) / ((5/4) + 2) = -2/3

((5-18)/6) / ((5+8)/4) = -2/3

(-13/6) / (13/4) = -2/3

(-13/6) Ã— (4/13) = -2/3

-4/6 = -2/3

-2/3 = -2/3

Hence, the given equation is verified

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

Solution:

We have,

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

Let us perform cross-multiplication we get,

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

10 â€“ 5y = 3y + 21

When rearranged,

10 â€“ 21 = 3y + 5y

8y = – 11

y = -11/8

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(2 â€“ (-11/8)) / ((-11/8) + 7) = 3/5

((16+11)/8) / ((-11+56)/8) = 3/5

(27/8) / (45/8) = 3/5

(27/8) Ã— (8/45) = 3/5

27/45 = 3/5

3/5 = 3/5

Hence, the given equation is verified

3. (5x – 7) / (3x) = 2

Solution:

We have,

(5x – 7) / (3x) = 2

Let us perform cross-multiplication we get,

5x â€“ 7 = 2(3x)

5x â€“ 7 = 6x

5x â€“ 6x = 7

-x = 7

x = -7

Now let us verify the given equation,

(5x – 7) / (3x) = 2

By substituting the value of â€˜xâ€™ we get,

(5(-7) – 7) / (3(-7)) = 2

(-35 – 7) / -21 = 2

-42/-21 = 2

2 = 2

Hence, the given equation is verified

4. (3x+5) / (2x + 7) = 4

Solution:

We have,

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

Let us perform cross-multiplication we get,

3x + 5 = 4(2x+7)

3x + 5 = 8x + 28

3x â€“ 8x = 28 â€“ 5

-5x = 23

x = -23/5

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(3(-23/5) + 5) / (2(-23/5) + 7) = 4

(-69/5 + 5) / (-46/5 + 7) = 4

(-69+25)/5 / (-46+35)/5 = 4

-44/5 / -11/5 = 4

-44/5 Ã— 5/-11 = 4

44/11 = 4

4 = 4

Hence, the given equation is verified

5. (2y + 5) / (y + 4) = 1

Solution:

We have,

(2y + 5) / (y + 4) = 1

Let us perform cross-multiplication we get,

2y + 5 = y + 4

2y â€“ y = 4 â€“ 5

y = -1

Now let us verify the given equation,

(2y + 5) / (y + 4) = 1

By substituting the value of â€˜yâ€™ we get,

(2(-1) + 5) / (-1 + 4) = 1

(-2+5) / 3 = 1

3/3 = 1

1 = 1

Hence, the given equation is verified

6. (2x + 1) / (3x – 2) = 5/9

Solution:

We have,

(2x + 1) / (3x – 2) = 5/9

Let us perform cross-multiplication we get,

9(2x + 1) = 5(3x – 2)

18x + 9 = 15x â€“ 10

18x â€“ 15x = -10 â€“ 9

3x = -19

x = -19/3

Now let us verify the given equation,

(2x + 1) / (3x – 2) = 5/9

By substituting the value of â€˜xâ€™ we get,

(2(-19/3) + 1) / (3(-19/3) – 2) = 5/9

(-38/3 + 1) / (-57/3 – 2) = 5/9

(-38 + 3)/3 / (-57 – 6)/3 = 5/9

-35/3 / -63/3 = 5/9

-35/3 Ã— 3/-63 = 5/9

-35/-63 = 5/9

5/9 = 5/9

Hence, the given equation is verified

7. (1 â€“ 9y) / (19 â€“ 3y) = 5/8

Solution:

We have,

(1 â€“ 9y) / (19 â€“ 3y) = 5/8

Let us perform cross-multiplication we get,

8(1- 9y) = 5(19-3y)

8 â€“ 72y = 95 â€“ 15y

8 â€“ 95 = 72y â€“ 15y

57y = -87

y = -87/57

= -29/19

Now let us verify the given equation,

(1 â€“ 9y) / (19 â€“ 3y) = 5/8

By substituting the value of â€˜yâ€™ we get,

(1 â€“ 9(-29/19)) / (19 â€“ 3(-29/19)) = 5/8

(19+261)/19 / (361+87)/19 = 5/8

280/19 Ã— 19/448 = 5/8

280/ 448 = 5/8

5/8 = 5/8

Hence, the given equation is verified

8. 2x / (3x + 1) = 1

Solution:

We have,

2x / (3x + 1) = 1

Let us perform cross-multiplication we get,

2x = 1(3x + 1)

2x = 3x + 1

2x â€“ 3x = 1

-x = 1

x = -1

Now let us verify the given equation,

2x / (3x + 1) = 1

By substituting the value of â€˜xâ€™ we get,

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

-2 /(-3+1) = 1

-2/-2 = 1

1 = 1

Hence, the given equation is verified

9. y â€“ (7 â€“ 8y)/9y – (3 + 4y) = 2/3

Solution:

We have,

y â€“ (7 â€“ 8y)/9y – (3 + 4y) = 2/3

(y â€“ 7 + 8y) / (9y â€“ 3 – 4y) = 2/3

(-7 + 9y) / (5y – 3) = 2/3

Let us perform cross-multiplication we get,

3(-7 + 9y) = 2(5y – 3)

-21 + 27y = 10y â€“ 6

27y â€“ 10y = 21 â€“ 6

17y = 15

y = 15/17

Now let us verify the given equation,

y â€“ (7 â€“ 8y)/9y – (3 + 4y) = 2/3

By substituting the value of â€˜yâ€™ we get,

15/17 â€“ (7-8(15/17))/ 9(15/17) â€“ (3 + 4(15/17)) = 2/3

15/17 â€“ (7 â€“ 120/17) / 135/17 â€“ (3 + 60/17) = 2/3

15/17 â€“ ((119-120)/17) / 135/17 â€“ ((51+60)/17) = 2/3

15/17 â€“ (-1/17) / 135/17 – (111/17) = 2/23

((15 + 1)/17) / ((135-111)/17) = 2/3

16/17 / 24/17 = 2/3

16/24 = 2/3

2/3 = 2/3

Hence, the given equation is verified

10. 6/ 2x â€“ (3 â€“ 4x) = 2/3

Solution:

We have,

6/ 2x â€“ (3 â€“ 4x) = 2/3

6/(2x â€“ 3 + 4x) = 2/3

6/(6x – 3) = 2/3

Let us perform cross-multiplication we get,

3(6) = 2(6x – 3)

18 = 12x â€“ 6

12x = 18 + 6

12x = 24

x = 24/12

= 2

Now let us verify the given equation,

6/ 2x â€“ (3 â€“ 4x) = 2/3

6/(6x – 3) = 2/3

By substituting the value of â€˜xâ€™ we get,

6/ (6(2) – 3) = 2/3

6/(12-3) = 2/3

6/9 = 2/3

2/3 = 2/3

Hence, the given equation is verified

11. 2/3x â€“ 3/2x = 1/12

Solution:

We have,

2/3x â€“ 3/2x = 1/12

By taking LCM for 2 and 3 which is 6

4-9/6x = 1/12

-5/6x = 1/12

By cross-multiplying we get,

12(-5) = 1 (6x)

-60 = 6x

x = -60/6

= -10

Now let us verify the given equation,

2/3x â€“ 3/2x = 1/12

By substituting the value of â€˜xâ€™ we get,

2/3(-10) â€“ 3/2(-10) = 1/12

2/-30 â€“ 3/-20 = 1/12

-4+6/60 = 1/12

5/60 = 1/12

1/12 = 1/12

Hence, the given equation is verified

12. (3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)

Solution:

We have,

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

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

By taking LCM as (4x + 2) (4x + 7)

((3x + 5) (4x + 7) â€“ (3x + 4) (4x + 2)) / (4x + 2) (4x + 7) = 0

By cross-multiplying we get,

(3x + 5) (4x + 7) â€“ (3x + 4) (4x + 2) = 0

(3x + 5) (4x + 7) â€“ (3x + 4) (4x + 2) = 0

12x2 + 21x + 20x + 35 â€“ 12x2 â€“ 6x â€“ 16x â€“ 8 = 0

19x + 35 â€“ 8 = 0

19x = -27

x = -27/19

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(3(-27/19) +5) / (4(-27/19) + 2) = (3(-27/19) + 4) / (4(-27/19) + 7)

(-81/19 + 5) / (-108/19 + 2) = (-81/19 + 4) / (-108/19 + 7)

((-81+95)/19) / ((-108+38)/19) = ((-81+76)/19) / ((-108+133)/19)

14/19 / -70/19 = -5/19 / 25/19

-14/70 = -5/25

-1/5 = -1/5

Hence, the given equation is verified

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

Solution:

We have,

(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

By cross-multiplying we get,

(7x-2) (5x+4) – (7x+3)(5x-1) = 0

Upon simplification,

35x2 + 28x â€“ 10x â€“ 8 â€“ 35x2 + 7x â€“ 15x + 3 = 0

10x â€“ 5 = 0

10x = 5

x = 5/10

= 1/2

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(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

Hence, the given equation is verified

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

Solution:

We have,

((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

By cross-multiplying we get,

(x+1)2 (x+4) – (x+2) (x+2)2 = 0

Let us expand the equation

(x2 + 2x + 1) (x + 4) â€“ (x + 2) (x2 + 4x + 4) = 0

x3 + 2x2 + x + 4x2 + 8x + 4 â€“ (x3 + 4x2 + 4x + 2x2 + 8x + 8) = 0

x3 + 2x2 + x + 4x2 + 8x + 4 â€“ x3 â€“ 4x2 â€“ 4x â€“ 2x2 â€“ 8x â€“ 8 = 0

-3x â€“ 4 = 0

x = -4/3

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(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

Hence, the given equation is verified

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

Solution:

We have,

((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) â€“ (x+8) (x-4)2) / (x-4)2 (x-2) = 0

By cross-multiplying we get,

(x+1)2 (x-2) â€“ (x+8) (x-4)2 = 0

Upon expansion we get,

(x2 + 2x + 1) (x-2) â€“ ((x+8) (x2 â€“ 8x + 16)) = 0

x3 + 2x2 + x â€“ 2x2 â€“ 4x â€“ 2 â€“ (x3 â€“ 8x2 + 16x + 8x2 â€“ 64x + 128) = 0

x3 + 2x2 + x â€“ 2x2 â€“ 4x â€“ 2 â€“ x3 + 8x2 â€“ 16x â€“ 8x2 + 64x â€“ 128 = 0

45x â€“ 130 = 0

x = 130/45

= 26/9

Now let us verify the given equation,

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

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

By substituting the value of â€˜xâ€™ we get,

(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

Hence, the given equation is verified

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

Solution:

We have,

(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

By cross-multiplying we get,

(9x-7) (x+6) – (3x-4) (3x+5) = 0

Upon expansion we get,

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

44x â€“ 22 = 0

44x = 22

x = 22/44

= 2/4

= 1/2

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(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

Hence, the given equation is verified

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

Solution:

We have,

(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) â€“ x(x+5)) / (x+5) (x+6) = 0

By cross-multiplying we get,

(x+2) (x+6) â€“ x(x+5) = 0

Upon expansion,

x2 + 8x + 12 â€“ x2 â€“ 5x = 0

3x + 12 = 0

3x = -12

x = -12/3

= -4

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

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

-2/1 = -4 / (2)

-2 = -2

Hence, the given equation is verified

18. 2x â€“ (7-5x) / 9x â€“ (3+4x) = 7/6

Solution:

We have,

2x â€“ (7-5x) / 9x â€“ (3+4x) = 7/6

(2x â€“ 7 + 5x) / (9x â€“ 3 â€“ 4x) = 7/6

(7x â€“ 7) / (5x â€“ 3) = 7/6

By cross-multiplying we get,

6(7x – 7) = 7(5x – 3)

42x â€“ 42 = 35x â€“ 21

42x â€“ 35x = -21 + 42

7x = 21

x = 21/7

= 3

Now let us verify the given equation,

2x â€“ (7-5x) / 9x â€“ (3+4x) = 7/6

(7x â€“ 7) / (5x â€“ 3) = 7/6

By substituting the value of â€˜xâ€™ we get,

(7(3) -7) / (5(3) – 3) = 7/6

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

14/12 = 7/6

7/6 = 7/6

Hence, the given equation is verified

19. (15(2-x) â€“ 5(x+6)) / (1-3x) = 10

Solution:

We have,

15(2-x) â€“ 5(x+6) / (1-3x) = 10

(30-15x) â€“ (5x + 30) / (1-3x) = 10

By cross-multiplying we get,

(30-15x) â€“ (5x + 30) = 10(1- 3x)

30- 15x â€“ 5x â€“ 30 = 10 â€“ 30x

30- 15x â€“ 5x â€“ 30 + 30x = 10

10x = 10

x = 10/10

= 1

Now let us verify the given equation,

(15(2-x) â€“ 5(x+6)) / (1-3x) = 10

By substituting the value of â€˜xâ€™ we get,

(15(2-1) â€“ 5(1+6)) / (1- 3) = 10

(15 â€“ 5(7))/-2 = 10

(15-35)/-2 = 10

-20/-2 = 10

10 = 10

Hence, the given equation is verified

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

Solution:

We have,

(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

By cross-multiplying we get,

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

Upon expansion,

x2 + 3x â€“ 2x â€“ 6 + x2 â€“ 3x + 2x â€“ 6 = 2(x2 â€“ 3x â€“ 2x + 6)

2x2 â€“ 12 = 2x2 â€“ 10x + 12

2x2 â€“ 2x2 + 10x = 12 + 12

10x = 24

x = 24/10

= 12/5

Now let us verify the given equation,

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

By substituting the value of â€˜xâ€™ we get,

(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

((-27Ã—2) + (22Ã—3))/6 = 2

(-54 + 66)/6 = 2

12/6 = 2

2 = 2

Hence, the given equation is verified

21. ((x+2) (2x-3) â€“ 2x2 + 6)/(x-5) = 2

Solution:

We have,

((x+2) (2x-3) â€“ 2x2 + 6)/(x-5) = 2

By cross-multiplying we get,

(x+2) (2x-3) â€“ 2x2 + 6) = 2(x-5)

2x2 â€“ 3x + 4x â€“ 6 â€“ 2x2 + 6 = 2x â€“ 10

x = 2x â€“ 10

x â€“ 2x = -10

-x = -10

x = 10

Now let us verify the given equation,

((x+2) (2x-3) â€“ 2x2 + 6)/(x-5) = 2

By substituting the value of â€˜xâ€™ we get,

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

(12(17) â€“ 200 + 6)/5 = 2

(204 â€“ 194)/5 = 2

10/5 = 2

2 = 2

Hence, the given equation is verified

22. (x2 â€“ (x+1) (x+2))/(5x+1) = 6

Solution:

We have,

(x2 â€“ (x+1) (x+2))/(5x+1) = 6

By cross-multiplying we get,

(x2 â€“ (x+1) (x+2)) = 6(5x+1)

x2 â€“ x2 â€“ 2x â€“ x â€“ 2 = 30x + 6

-3x â€“ 2 = 30x + 6

30x + 3x = -2 â€“ 6

33x = -8

x = -8/33

Now let us verify the given equation,

(x2 â€“ (x+1) (x+2))/(5x+1) = 6

By substituting the value of â€˜xâ€™ we get,

((-8/33)2 â€“ ((-8/33)+1) (-8/33 + 2))/(5(-8/33)+1) = 6

(64/1089 â€“ ((-8+33)/33) ((-8+66)/33)) / (-40+33)/33) = 6

(64/1089 â€“ (25/33) (58/33)) / (-7/33) = 6

(64/1089 â€“ 1450/1089) / (-7/33) = 6

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

-1386/1089 Ã— 33/-7 = 6

1386 Ã— 33 / 1089 Ã— -7 = 6

6 = 6

Hence, the given equation is verified

23. ((2x+3) â€“ (5x-7))/(6x+11) = -8/3

Solution:

We have,

((2x+3) â€“ (5x-7))/(6x+11) = -8/3

By cross-multiplying we get,

3((2x+3) â€“ (5x-7)) = -8(6x+11)

3(2x + 3 â€“ 5x + 7) = -48x â€“ 88

3(-3x + 10) = -48x â€“ 88

-9x + 30 = -48x â€“ 88

-9x + 48x = -88 â€“ 30

39x = -118

x = -118/39

Now let us verify the given equation,

((2x+3) â€“ (5x-7))/(6x+11) = -8/3

By substituting the value of â€˜xâ€™ we get,

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

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

(((-336+117)/39) â€“ ((-590-273)/39)) / ((-708+429)/39) = -8/3

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

644/-279 = -8/3

-8/3 = -8/3

Hence, the given equation is verified

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

(i) (x2 – 9)/(5+x2) = -5/9

Solution:

We have,

(x2 – 9)/(5+x2) = -5/9

By cross-multiplying we get,

9(x2 – 9) = -5(5+x2)

9x2 â€“ 81 = -25 â€“ 5x2

9x2 + 5x2 = -25 + 81

14x2 = 56

x2 = 56/14

x2 = 4

x = âˆš4

= 2

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

Solution:

We have,

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

By cross-multiplying we get,

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

2y2 + 8 = 3y2 + 7

3y2 â€“ 2y2 = 7 â€“ 8

y2 = -1

y = âˆš-1

= 1

#### 1 Comment

1. Jagmal Dahiya

that is a good maths for students