Solve the given equation and verify the answer. n 4 - 5 = n 6 + 1 2

Given: n 4 - 5 = n 6 + 1 2 n 4 - n 6 = 1 2 + 5 [Transposing n 6 to the L.H.S. and 5 to the R.H.S.] ⇒ 3 n - 2 n 12 = 1 + 10 2 ⇒ n 12 = 11 2 ⇒ n 12 × 12 = 11 2 × 12 [Multiplying both the sides of the equation by 12 ] ⇒ n = 66 So, n = 66 is the solution of the given equation. Verification: Substituting n = 66 in the equation, we get, L.H.S. = 66 4 - 5 = 33 2 - 5 = 33 - 10 2 = 23 2 R.H.S. = 66 6 + 1 2 = 11 + 1 2 = 23 2 We can see that, L.H.S. = R.H.S. Hence, verified.

Solve the given equation and verify the answer: 2 m 3 + 8 = m 2 - 1

Given, 2 m 3 + 8 = m 2 - 1 or, 2 m 3 - m 2 = - 1 - 8 [Transposing m 2 to the L.H.S. and 8 to the R.H.S.] o r , 4 m - 3 m 6 = - 9 o r , m 6 = - 9 or, m 6 × 6 = - 9 × 6 [Multiplying both the sides by 6 ] or, m = - 54 So, m = - 54 is the solution of the given equation. Verification: Substituting x = − 54 in the equation, we get, L.H.S: 2 ( - 54 ) 3 + 8 = - 108 3 + 8 = - 36 + 8 = - 28 R.H.S: - 54 2 - 1 = - 27 - 1 = - 28 So, L.H.S. = R.H.S. Hence, verified.

Solve the given equation and verify the answer: 2 x 5 - 3 2 = x 2 + 1

Given, 2 x 5 - 3 2 = x 2 + 1 or, 2 x 5 - x 2 = 1 + 3 2 [Transposing x 2 to the L.H.S. and 3 2 to R.H.S.] o r , 4 x - 5 x 10 = 2 + 3 2 o r , - x 10 = 5 2 o r , - x 10 × ( - 10 ) = 5 2 × ( - 10 ) [Multiplying both the sides by − 10 ] or, x = - 25 So, x = - 25 is the solution of the given equation. Verification: Substituting x = − 25 in the equation, we get L.H.S. : 2 ( - 25 ) 5 - 3 2 = - 50 5 - 3 2 = - 10 - 3 2 = - 23 2 R.H.S. : - 25 2 + 1 = - 23 2 So, L.H.S. = R.H.S. Hence, verified.

Solve the given equation and verify the answer: x - 3 5 - 2 = 2 x 5

Given, x - 3 5 - 2 = 2 x 5 or, x 5 - 3 5 - 2 = 2 x 5 or, - 3 5 - 2 = 2 x 5 - x 5 [Transposing x 5 to the R.H.S.] or, - 3 - 10 5 = x 5 or, - 13 5 = x 5 or, - 13 5 × ( 5 ) = x 5 × ( 5 ) [Multiplying both the sides by 5 ] or, x = − 13 So, x = - 13 is the solution of the given equation. Verification: Substituting x = − 13 in the equation, we get, L.H.S. : - 13 - 3 5 - 2 = - 16 5 - 2 = - 16 - 10 5 = - 26 5 R.H.S. : 2 × ( - 13 ) 5 = - 26 5 So, L.H.S. = R.H.S. Hence, verified.

Solve the given equation and verify the answer: 3 x 10 - 4 = 14

Given, 3 x 10 - 4 = 14 or, 3 x 10 - 4 + 4 = 14 + 4 [Adding 4 on both the sides] or, 3 x 10 = 18 or, 3 x 10 × 10 = 18 × 10 [Multiplying both the sides by 10 ] or, 3 x = 180 or, 3 x 3 = 180 3 [Dividing both the sides by 3 ] or, x = 60 So, x = 60 is the solution of the given equation. Verification: Substituting x = 60 in the equation, we get, L.H.S. : 3 × 60 10 - 4 = 180 10 - 4 = 18 - 4 = 14 = R.H.S. So, L.H.S. = R.H.S. Hence, verified.

E M B I B E

Mathematics For Class 6>Chapter 9 - Linear Equations in One Variable>EXERCISE 9B

R S Aggarwal Solutions for Chapter: Linear Equations in One Variable, Exercise 2: EXERCISE 9B

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Chapter: Linear Equations in One Variable, Exercise 2: EXERCISE 9B

Attempt the practice questions on Chapter 9: Linear Equations in One Variable, Exercise 2: EXERCISE 9B with hints and solutions to strengthen your understanding. Mathematics For Class 6 solutions are prepared by Experienced Embibe Experts.