# RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable Exercise 9.1

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### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 9.1 Chapter 9 Linear Equation in One Variable

#### EXERCISE 9.1 PAGE NO: 9.5

Solve each of the following equations and also verify your solution:

1. 9 ¼ = y – 1 1/3

Solution:

We have,

9 ¼ = y – 1 1/3

37/4 = y – 4/3

Upon solving we get,

y = 37/4 + 4/3

By taking LCM for 4 and 3 is 12

y = (37×3)/12 + (4×4)/12

= 111/12 + 16/12

= (111 + 16)/12

= 127/12

∴ y = 127/12

Verification-

RHS = y – 1 1/3

= 127/12 – 4/3

= (127 – 16)/12

= 111/12

= 37/4

= 9 ¼

= LHS

2. 5x/3 + 2/5 = 1

Solution:

We have,

5x/3 + 2/5 = 1

5x/3 = 1 – 2/5 (by taking LCM)

= (5-2)/5

By using cross-multiplication we get,

5x/3 = 3/5

5x = (3×3)/5

x = 9/(5×5)

= 9/25

∴ x = 9/25

Verification-

LHS = 5x/3 + 2/5

= 5/3 × 9/25 + 2/5

= 3/5 + 2/5

= (3 + 2)/5

= 5/5

= 1

= RHS

3. x/2 + x/3 + x/4 = 13

Solution:

We have,

x/2 + x/3 + x/4 = 13

let us take LCM for 2, 3 and 4 which is 12

(x×6)/12 + (x×4)/12 + (x×3)/12 = 13

6x/12 + 4x/12 + 3x/12 = 13

(6x+4x+3x)/12 = 13

13x/12 = 13

By using cross-multiplication we get,

13x = 12×13

x = 156/13

= 12

∴ x = 12

Verification-

LHS = x/2 + x/3 + x/4

= 12/2 + 12/3 + 12/4

= 6 + 4 + 3

= 13

= RHS

4. x/2 + x/8 = 1/8

Solution:

We have,

x/2 + x/8 = 1/8

let us take LCM for 2 and 8 which is 8

(x×4)/8 + (x×1)/8 = 1/8

4x/8 + x/8 = 1/8

5x/8 = 1/8

By using cross-multiplication we get,

5x = 8/8

5x = 1

x = 1/5

∴ x = 1/5

Verification-

LHS = x/2 + x/8

= (1/5)/2 + (1/5)/8

= 1/10 + 1/40

= (4 + 1)/40

= 5/40

= 1/8

= RHS

5. 2x/3 – 3x/8 = 7/12

Solution:

We have,

2x/3 – 3x/8 = 7/12

By taking LCM for 3 and 8 is 24

(2x×8)/24 – (3x×3)/24 = 7/12

16x/24 – 9x/24 = 7/12

(16x-9x)/24 = 7/12

7x/24 = 7/12

By using cross-multiplication we get,

7x×12 = 7×24

x = (7×24)/(7×12)

= 24/12

= 2

∴ x = 2

Verification-

LHS = 2x/3 – 3x/8

= 2(2)/3 – 3(2)/8

= 4/3 – 6/8

= 4/3 – 3/4

= (16 – 9)/ 12

= 7/12

= RHS

6. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

Solution:

We have,

(x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

Upon expansion we get,

x+ 5x + 6 + x2 – 5x +6 – 2x2 – 2x =0

-2x + 12 = 0

By dividing the equation using -2 we get,

x – 6 = 0

x = 6

∴ x = 6

Verification-

LHS = (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1)

= (6 + 2) (6 + 3) + (6 – 3) (6 – 2) – 2(6) (6 + 1)

= (8) (9) + (3) (4) – 12(7)

= 72 + 12 – 84

= 84 – 84

= 0

= RHS

7. x/2 – 4/5 + x/5 + 3x/10 = 1/5

Solution:

We have,

x/2 – 4/5 + x/5 + 3x/10 = 1/5

upon solving we get,

x/2 + x/5 + 3x/10 = 1/5 + 4/5

by taking LCM for 2, 5 and 10 which is 10

(x×5)/10 + (x×2)/10 + (3x×1)/10 = 5/5

5x/10 + 2x/10 + 3x/10 = 1

(5x+2x+3x)/10 = 1

10x/10 = 1

x = 1

∴ x = 1

Verification-

LHS = x/2 – 4/5 + x/5 + 3x/10

= ½ – 4/5 + 1/5 + 3(1)/10

= (5 – 8 + 2 + 3)/10

= (10 – 8)1/0

= 2/10

= 1/5

= RHS

8. 7/x + 35 = 1/10

Solution:

We have,

7/x + 35 = 1/10

7/x = 1/10 – 35

= ((1×1) – (35×10))/10

= (1 – 350)/10

7/x = -349/10

By using cross-multiplication we get,

x = -70/349

∴ x = -70/349

Verification-

LHS = 7/x + 35

= 7/(-70/349) + 35

= (-7 × 349)/70 + 35

= -349/10 + 35

= (-349 + 350)/ 10

= 1/10

= RHS

9. (2x-1)/3 – (6x-2)/5 = 1/3

Solution:

We have,

(2x-1)/3 – (6x-2)/5 = 1/3

By taking LCM for 3 and 5 which is 15

((2x-1)×5)/15 – ((6x-2)×3)/15 = 1/3

(10x – 5)/15 – (18x – 6)/15 = 1/3

(10x – 5 – 18x + 6)/15 = 1/3

(-8x + 1)/15 = 1/3

By using cross-multiplication we get,

(-8x + 1)3 = 15

-24x + 3 = 15

-24x = 15 – 3

-24x = 12

x = -12/24

= -1/2

∴ x = -1/2

Verification-

LHS = (2x – 1)/3 – (6x – 2)/5

= [2(-1/2) – 1]/3 – [6(-1/2) – 2]/5

= (- 1 – 1)/3 – (-3 – 2)/5

= – 2/3 – (-5/5)

= -2/3 + 1

= (-2 + 3)/3

= 1/3

RHS

10. 13(y – 4) – 3(y – 9) – 5(y + 4) = 0

Solution:

We have,

13(y – 4) – 3(y – 9) – 5(y + 4) = 0

Upon expansion we get,

13y – 52 – 3y + 27 – 5y – 20 = 0

13y – 3y – 5y = 52 – 27 + 20

5y = 45

y = 45/5

= 9

∴ y = 9

Verification-

LHS = 13(y – 4) – 3 (y – 9) – 5 (y + 4)

= 13 (9 – 4) – 3 (9 – 9) – 5 (9 + 4)

= 13 (5) – 3 (0) – 5 (13)

= 65 – 0 – 65

= 0

= RHS

11. 2/3(x – 5) – 1/4(x – 2) = 9/2

Solution:

We have,

2/3(x – 5) – 1/4(x – 2) = 9/2

Upon expansion we get,

2x/3 – 10/3 – x/4 + 2/4 = 9/2

2x/3 – 10/3 – x/4 + 1/2 = 9/2

2x/3 – x/4 = 9/2 + 10/3 – 1/2

By taking LCM for (3 and 4 is 12) (2 and 3 is 6)

(2x×4)/12 – (x×3)/12 = (9×3)/6 + (10×2)/6 – (1×3)/6

8x/12 – 3x/12 = 27/6 + 20/6 – 3/6

(8x-3x)/12 = (27+20-3)6

5x/12 = 44/6

By using cross-multiplication we get,

5x×6 = 44×12

30x = 528

x = 528/30

= 264/15

= 88/5

Verification-

LHS = 2/3 (x – 5) – ¼ (x – 2)

= 2/3 [(88/5) – 5] – ¼ [(88/5) – 2]

= 2/3 [(88 – 25)/5] – ¼ [(88 – 10)/5]

= 2/3 × 63/5 – ¼ × 78/5

= 42/5 – 39/10

= (84 – 39)/10

= 45/10

= 9/2

= RHS

All Chapter RD Sharma Solutions For Class 8 Maths

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