Solve the following equation and check your answer: 1 3 - 2 x = 0

Given, 1 3 - 2 x = 0 Subtracting 1 3 from both sides, we get 1 3 - 2 × x - 1 3 = 0 - 1 3 - 2 × x = – 1 3 Multiplying both sides by - 1 , we get - 2 x × - 1 = – 1 3 × ( - 1 ) 2 x = 1 3 Dividing both sides by 2 , we get 2 x 2 = 1 3 2 x = 1 6 Verification: Substituting x = 1 6 in LHS , we get LHS = 1 3 – 2 × 1 6 = 1 3 - 1 3 = 0 , and RHS = 0 LHS = RHS Hence, verified.

Solve the following equation and check your answer: x - 3 5 – 2 = - 1

Given : x - 3 5 - 2 = - 1 Adding 2 to both sides, we get x – 3 5 – 2 + 2 = - 1 + 2 x – 3 5 = 1 Multiplying both sides by 5 , we get x – 3 5 × 5 = 1 × 5 x - 3 = 5 Adding 3 to both sides, we get x - 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS , we get = 8 – 3 5 - 2 = 5 5 - 2 = 1 - 2 = - 1 = RHS LHS = RHS Hence, verified

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Mathematics for Class 7>Chapter 8 - Linear Equations in One Variable>EXERCISE

R. D. Sharma Solutions for Chapter: Linear Equations in One Variable, Exercise 2: EXERCISE

Author:R. D. Sharma

R. D. Sharma Mathematics Solutions for Exercise - R. D. Sharma Solutions for Chapter: Linear Equations in One Variable, Exercise 2: EXERCISE

Attempt the free practice questions on Chapter 8: Linear Equations in One Variable, Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. Mathematics for Class 7 solutions are prepared by Experienced Embibe Experts.