## RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2

**Other Exercises**

- RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.1
- RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2
- RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.3
- RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.4

**Solve each of the following equations and also check your result in each case :**

**Question 1.**

**\(\frac { 2x + 5 }{ 3 }\) = 3x – 10**

**Solution:**

\(\frac { 2x + 5 }{ 3 }\) = \(\frac { 3x – 10 }{ 1 }\)

By cross multiplication

⇒ 2x + 5 = 3 (3x – 10)

⇒ 2x + 5 = 9x – 30

⇒ 5 + 30 = 9x – 2x (By transposition)

⇒ 35 = 7x

⇒ x = 5

**Question 2.**

**\(\frac { a – 8 }{ 3 }\) = \(\frac { a – 3 }{ 2 }\)**

**Solution:**

**Question 3.**

**\(\frac { 7y + 2 }{ 5 }\) = \(\frac { 6y – 5 }{ 11 }\)**

**Solution:**

**Question 4.**

**x – 2x + 2 – \(\frac { 16 }{ 3 }\) x + 5 = 3 – \(\frac { 7 }{ 2 }\) x.**

**Solution:**

**Question 5.**

**\(\frac { 1 }{ 2 }\) x + 7x – 6 = 7x + \(\frac { 1 }{ 4 }\)**

**Solution:**

**Question 6.**

**\(\frac { 3 }{ 4 }\) x + 4x = \(\frac { 7 }{ 8 }\) + 6x – 6**

**Solution:**

**Question 7.**

**\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10**

**Solution:**

\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10

**Question 8.**

**\(\frac { 6x + 1 }{ 2 }\) + 1 = \(\frac { 7x – 3 }{ 3 }\)**

**Solution:**

**Question 9.**

**\(\frac { 3a – 2 }{ 3 }\) + \(\frac { 2a + 3 }{ 2 }\) = a + \(\frac { 7 }{ 6 }\)**

**Solution:**

**Question 10.**

**x – \(\frac { x – 1 }{ 2 }\) = 1 – \(\frac { x – 2 }{ 3 }\)**

**Solution:**

**Question 11.**

**\(\frac { 3x }{ 4 }\) – \(\frac { x – 1 }{ 2 }\) = \(\frac { x – 2 }{ 3 }\)**

**Solution:**

**Question 12.**

**\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)**

**Solution:**

\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)

**Question 13.**

**Solution:**

**Question 14.**

**Solution:**

**Question 15.**

**Solution:**

**Question 16.**

**0.18 (5x – 4) = 0.5x + 0.8**

**Solution:**

**Question 17.**

**Solution:**

**Question 18.**

**Solution:**

**Question 19.**

**Solution:**

**Question 20.**

**Solution:**

**Question 21.**

**Solution:**

**Question 22.**

**Solution:**

**Question 23.**

**Solution:**

**Question 24.**

**(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)**

**Solution:**

(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)

⇒ (9x² + 6x – 24x – 16) – (8x² + 4x – 22x – 11) = x² + 7x – 3x – 21

⇒ 9x² + 6x – 24x – 16 – 8x² – 4x + 22x + 11 = x² + 4x – 21

⇒ 9x² – 8x² – x² + 6x – 24x + 22x – 4x – 4x = -21 + 16 – 11

⇒ 28x – 32x = -32 + 16

⇒ -4x = -16

⇒ x = 4

Verification:

L.H.S. = (3x – 8) (3x + 2) – (4x – 11) (2x + 1)

= (3 x 4 – 8) (3 x 4 + 2) – (4 x 4 – 11) (2 x 4 + 1)

= (12 – 8) (12 + 2) – (16 – 11) (8 + 1)

= 4 x 14 – 5 x 9 = 56 – 45 = 11

R.H.S. = (x – 3) (x + 7) = (4 – 3) (4 + 7) = 1 x 11 = 11

L.H.S. = R.H.S.

**Question 25.**

**[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92**

**Solution:**

[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92

⇒ (2x + 3 + x + 5)² + (2x + 3 – x – 5)² = 10x² + 92

⇒ (3x + 8)² + (x – 2)² = 10x² + 92

⇒ 9x² + 2 x 3x x 8 + 64 + x² – 2 x x x 2 + 4 = 10x² + 92

⇒ 9x² + 48x + 64 + x² – 4x + 4 = 10x² + 92

⇒ 9x² + x² – 10x² + 48x – 4x = 92 – 64 – 4

⇒ 44x = 24

Hope given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2 are helpful to complete your math homework.

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