Solve: 5 x + 7 2 = 3 2 x - 14

Given expression is, 5 x + 7 2 = 3 2 x - 14 Multiply both sides of the equation by 2 . We get 2 × 5 x + 7 2 = 2 × 3 2 x - 14 2 × 5 x + 2 × 7 2 = 2 × 3 2 x - 2 × 14 10 x + 7 = 3 x - 28 Subtracting 7 on both sides, we get 10 x = 3 x - 28 - 7 10 x - 3 x = - 28 - 7 (transposing 3 x to L.H.S) 7 x = - 28 - 7 7 x = - 35 x = - 35 7 x = - 5 Hence, the value of x = - 5 .

Solve: 6 x + 1 3 + 1 = x - 3 6

Given, 6 x + 1 3 + 1 = x - 3 6 Multiplying both sides of the equation by 6 , 6 6 x + 1 3 + 6 × 1 = 6 x - 3 6 2 6 x + 1 + 6 = x - 3 12 x + 2 + 6 = x - 3 (opening the brackets) 12 x + 8 = x - 3 12 x - x = - 3 - 8 (Transposing x to L.H.S and 8 to the R.H.S) or 11 x = - 11 or x = - 1 (required solution) Check: L.H.S = 6 - 1 + 1 3 + 1 = - 6 + 1 3 + 1 = - 5 3 + 3 3 = - 5 + 3 3 = - 2 3 R.H.S = - 1 - 3 6 = - 4 6 = - 2 3 L.H.S = R.H.S. (as required)

Solve: 5 x - 2 2 x - 7 = 2 3 x - 1 + 7 2

Given, 5 x - 2 2 x - 7 = 2 3 x - 1 + 7 2 Let us open the brackets, L.H.S = 5 x - 4 x + 14 = x + 14 - - - - - - ( 1 ) R.H.S = 6 x - 2 + 7 2 = 6 x - 4 2 + 7 2 = 6 x + 3 2 - - - - - - - ( 2 ) From equation 1 and 2 , we get x + 14 = 6 x + 3 2 14 = 6 x - x + 3 2 14 = 5 x + 3 2 14 - 3 2 = 5 x (transposing 3 2 ). 28 - 3 2 = 5 x 25 2 = 5 x x = 25 2 × 1 5 = 5 2 Therefore, the required solution is x = 5 2 . Check: L.H.S = 5 × 5 2 - 2 5 2 × 2 - 7 = 25 2 - 2 5 - 7 = 25 2 - 2 - 2 = 25 2 + 4 = 25 + 8 2 = 33 2 R.H.S = 2 5 2 × 3 - 1 + 7 2 = 2 15 2 - 2 2 + 7 2 = 2 × 13 2 + 7 2 = 26 + 7 2 = 33 2 = L.H.S. (as required)

Solve: x + 1 2 x + 3 = 3 8

Observe that the equation is not a linear equation, since the expression on its LHS is not linear. But we can put it into the form of a linear equation. We multiply both sides of the equation by 2 x + 3 . x + 1 2 x + 3 × 2 x + 3 = 3 8 × 2 x + 3 Notice that 2 x + 3 gets cancelled on the L.H.S We have then, x + 1 = 3 2 x + 3 8 We have now a linear equation which we know how to solve. Multiplying both sides by 8 8 x + 1 = 3 2 x + 3 or 8 x + 8 = 6 x + 9 or 8 x = 6 x + 9 - 8 or 8 x = 6 x + 1 or 8 x - 6 x = 1 or 2 x = 1 or x = 1 2 Therefore, the solution is x = 1 2 . Check: Numerator of L.H.S = 1 2 + 1 = 1 + 2 2 = 3 2 Denominator of L.H.S = 2 x + 3 = 2 × 1 2 + 3 = 1 + 3 = 4 L.H.S = numerator ÷ denominator = 3 2 ÷ 4 = 3 2 × 1 4 = 3 8 L.H.S = R.H.S.

E M B I B E

Mathematics 8 Part-I>Chapter 3 - Linear Equations in One Variable>Solved Examples

Karnataka Board Solutions for Chapter: Linear Equations in One Variable, Exercise 1: Solved Examples

Author:Karnataka Board

Karnataka Board Mathematics Solutions for Exercise - Karnataka Board Solutions for Chapter: Linear Equations in One Variable, Exercise 1: Solved Examples

Attempt the free practice questions on Chapter 3: Linear Equations in One Variable, Exercise 1: Solved Examples with hints and solutions to strengthen your understanding. Mathematics 8 Part-I solutions are prepared by Experienced Embibe Experts.

Questions from Karnataka Board Solutions for Chapter: Linear Equations in One Variable, Exercise 1: Solved Examples with Hints & Solutions

EASY

8th Karnataka Board

IMPORTANT

Mathematics 8 Part-I>Chapter 3 - Linear Equations in One Variable>Solved Examples>Q 1

Find the solution of $2 x - 3 = 7$

MEDIUM

8th Karnataka Board

IMPORTANT

Mathematics 8 Part-I>Chapter 3 - Linear Equations in One Variable>Solved Examples>Q 13

Solve: $5 x + \frac{7}{2} = \frac{3}{2} x - 14$

MEDIUM

8th Karnataka Board

IMPORTANT

Mathematics 8 Part-I>Chapter 3 - Linear Equations in One Variable>Solved Examples>Q 14

The digits of a two-digit number differ by $3$ . If the digits are interchanged, and the resulting number is added to the original number, we get $143$ . What can be the original number?

MEDIUM

8th Karnataka Board

IMPORTANT

Mathematics 8 Part-I>Chapter 3 - Linear Equations in One Variable>Solved Examples>Q 15

Arjun is twice as old as Shriya. Five years ago his age was three times Shriys's age. Find their present ages.

MEDIUM

8th Karnataka Board

IMPORTANT

Mathematics 8 Part-I>Chapter 3 - Linear Equations in One Variable>Solved Examples>Q 16

Solve: $\frac{6 x + 1}{3} + 1 = \frac{x - 3}{6}$