In this chapter, we provide RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable Exercise 9.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable Exercise 9.1 pdf, free RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable Exercise 9.1 book pdf download. Now you will get step by step solution to each question.

### 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^{2 }+ 5x + 6 + x^{2} – 5x +6 – 2x^{2} – 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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