Solve the following equation and check your result. 3 4 ( 7 x - 1 ) - 2 x - 1 - x 2 = x + 3 2

We have: 3 4 ( 7 x − 1 ) − 2 x − 1 − x 2 = x + 3 2 ⇒ 3 4 ( 7 x − 1 ) − 2 x + 1 − x 2 − x = 3 2 ⇒ 3 × 7 4 x − 3 4 − 2 x + 1 2 − x 2 − x = 3 2 ⇒ 21 4 x − 2 x − x 2 − x = 3 2 + 3 4 − 1 2 ( By transposition ) ⇒ 21 x − 8 x − 2 × x − 4 x 4 = 1 + 3 4 ⇒ 7 x 4 = 7 4 ⇒ x = 1 ∴ x = 1 is a solution of the given equation. To check: Substituting x = 1 in the given equation, we get LHS = 3 4 ( 7 x − 1 ) − 2 x − 1 − x 2 = 3 4 × 6 − 2 = 9 2 − 2 = 5 2 RHS = x + 3 2 = 1 + 3 2 = 5 2 ∴ LHS = RHS Hence, x = 1 is a solution of the given equation.

Solve the following equation and check your result. x + 2 6 - 11 - x 3 - 1 4 = 3 x - 4 12

We have: x + 2 6 - 11 - x 3 - 1 4 = 3 x - 4 12 ⇒ x + 2 6 − 11 - x 3 + 1 4 = 3 x − 4 12 ⇒ x + 2 6 − 11 - x 3 − 3 x − 4 12 = − 1 4 ⇒ 2 ( x + 2 ) − 4 ( 11 − x ) − 1 ( 3 x − 4 ) 12 = − 1 4 ⇒ 2 x + 4 − 44 + 4 x − 3 x + 4 12 = − 1 4 ⇒ 3 x − 36 = − 1 4 × 12 ⇒ 3 x = − 3 + 36 ⇒ x = 33 3 = 11 ∴ x = 11 is a solution of the given equation. To check: Substituting x = 11 in the given equation, we get LHS = x + 2 6 − 11 − x 3 − 1 4 = 13 6 - 0 - 1 4 = 13 6 + 1 4 = 13 × 2 + 3 12 = 29 12 RHS = 3 x − 4 12 = 33 − 4 12 = 29 12 ∴ LHS = RHS Hence, x = 11 is a solution of the given equation.

Solve the following equation and check your result. 9 x + 7 2 - x - x - 2 7 = 36

We have: 9 x + 7 2 - x - x - 2 7 = 36 ⇒ 9 x + 7 2 − x + x − 2 7 = 36 ⇒ 7 ( 9 x + 7 ) − 14 × x + 2 × ( x − 2 ) 14 = 36 ⇒ 63 x + 49 − 14 x + 2 x − 4 14 = 36 ⇒ 51 x + 45 = 36 × 14 ⇒ 51 x = 504 − 45 ⇒ x = 459 51 = 9 Therefore, x = 9 is a solution of the given equation. To check: Substituting x = 9 in the given equation, we get LHS = 9 × 9 + 7 2 − 9 − 9 − 2 7 = 88 2 − 9 + 7 7 = = 44 - 9 + 1 = 36 RHS = 36 ∴ LHS = RHS Hence, x = 9 is a solution of the given equation.

Solve the following equation and check your result. x + 2 x - 2 = 7 3

We have: x + 2 x − 2 = 7 3 ⇒ 3 x + 6 = 7 x − 14 ⇒ 3 x - 7 x = - 14 - 6 ⇒ 4 x = 20 ⇒ x = 20 4 = 5 To check: Substituting x = 5 in the given equation, we get LHS = 5 + 2 5 - 2 = 7 3 RHS = 7 3 ∴ LHS = RHS Hence, x = 5 is a solution of the given equation.

Solve the following equation and check your result. 2 x + 5 3 x + 4 = 3

We have: 2 x + 5 3 x + 4 = 3 ⇒ ( 2 x + 5 ) = 3 × ( 3 x + 4 ) ⇒ 2 x + 5 = 9 x + 12 ⇒ 2 x - 9 x = 12 - 5 ⇒ - 7 x = 7 ⇒ x = - 1 To check: Substituting x = - 1 in the given equation, we get LHS = 2 × ( − 1 ) + 5 3 × ( − 1 ) + 4 = - 2 + 5 - 3 + 4 = 3 RHS = 3 ∴ LHS = RHS Hence, x = - 1 is a solution of the given equation.

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MATHEMATICS FOR CLASS 7>Chapter 7 - Linear Equations in One Variable>EXERCISE 7A

R S Aggarwal Solutions for Chapter: Linear Equations in One Variable, Exercise 1: EXERCISE 7A

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Chapter: Linear Equations in One Variable, Exercise 1: EXERCISE 7A

Attempt the practice questions on Chapter 7: Linear Equations in One Variable, Exercise 1: EXERCISE 7A with hints and solutions to strengthen your understanding. MATHEMATICS FOR CLASS 7 solutions are prepared by Experienced Embibe Experts.