# Class-8 Linear Equations In One Variable

Introduction to Linear Equations

Rules to Solve Linear Equations

Procedure to Solve Linear Equation

## Introduction to Linear Equations

An equation is a statement of equality of two algebraic expressions involving one more variable is known as Linear equation.

Some examples of linear equations are given below.

a) p − 7 = 10 − 2p

b) 4m − 15 = 5

c) ^{4a}⁄_{5} − 2 = 2a + 5

d) y − ^{2}⁄_{5} + 2y − ^{3}⁄_{5} = ^{2}⁄_{5}

All the above 4 linear equations are having only one variable and having power 1.

## Rules to Solve Linear Equations

We must follow certain rules to find out the value of a variable of a given linear equation and the rules are given below.

- We can add same number to both sides of an equation
- We can subtract same number from both sides of an equation
- We can multiply same nonzero number to both sides of an equation
- We can divide same nonzero to both sides of an equation

## Rules of Transposition

A term can be transposed from one side of the equation to the other side by changing its sign. Let's see some examples.

**Example 1.** 5b − 3 = 12

The above linear equation can be written as

=> 5b = 12 + 3

Transposing −3 from LHS to RHS by changing it's sign to +3.

**Example 2.** 5q + 5 = 19 − 2q

The above linear equation can be written as

=> 5q + 2q = 19 − 5

Here we have transpose 5 from LHS to RHS by changing it's sign to −5.

Similarly, we have transpose −2q from RHS to LHS by changing it's sign to +2q.

## Procedure to Solve Linear Equations

1. Simplify both sides by removing group symbols and collecting like terms

2. Remove fractions by multiplying both sides by appropriate factor(LCM of denominator for fraction)

3. Arrange all the variable terms on one side and all constant terms on the other side

4. Make the coefficient of the variable as 1

Let's see some examples.

**Example 1.** Solve 3m + 5 = 25 − 2m.

**Solution.** 3m + 5 = 25 − 2m

=> 3m + 2m = 25 − 5

=> 5m = 20

=> m = 20 ÷ 5

=> m = 4

**Example 2.** Solve 2(p − 1) = p + 12.

**Solution.** 2(p − 1) = p + 12

=> 2p − 2 = p + 12

=> 2p − p = 12 + 2

=> p = 14

**Example 3.** Solve 5n − ^{4}⁄_{5} = 20.

**Solution.** 5n − ^{4}⁄_{5} = 20

Multiply both sides by 5.

=> 25n − 4 = 100

=> 25n = 100 + 4

=> 25n = 104

=> n = 104 ÷ 25

=> n = ^{204}⁄_{5}

**Example 4.** Solve ^{(2x+3)}⁄_{5} + 2 = ^{x+2}⁄_{2}

**Solution.** ^{(2x+3)}⁄_{5} + 2 = ^{x+2}⁄_{2}

Multiplying both side by 10, (LCM of 5 and 2) we get

=> 10 × ^{(2x+3)}⁄_{5} + 10 × 2 = 10 × ^{x+2}⁄_{2}

=> 4x + 6 + 20 = 5x + 10

=> 4x − 5x = 10 − 26

=> −x = −16

=> x = 16

## Linear Equations Word Problem

Problem stated in words are known as word problems. Solving word problem consists two steps. First step is translating the words of the problem into an algebraic equation. Second step is solving the equation. Let's see some examples.

**Example 1.** If 7 is added to three times of a number, then the result is 28. Find the number.

**Solution.** Let's assume the number is p.

As per the word problem given, linear equation will be

3p + 7 = 28

=> 3p = 28 − 7

=> 3p = 21

=> p = 21 ÷ 3

=> p = 7

Hence, the number is 7.

**Example 2.** Find the three consecutive odd numbers whose sum is 105.

**Solution.** Let's consider the smallest, odd number is equal to m.

Next two odd numbers are (m+2) and (m+4).

According to the given word problem following linear equation can be formed.

m + m + 2 + m + 4 = 105

=> 3m + 4 = 105

=> 3m = 99

=> m = 99 ÷ 3

=> m = 33

Hence, the required consecutive odd numbers are 33, 35 and 37.

**Example 3.** The cost of 3 notebooks and 5 similar pens is Rs. 460. If the notebook cost is Rs. 20 more than the pen, then find the cost of each.

**Solution.** Let's consider cost of the pen = q

Then, cost of the notebook = q + 20

So, the linear equation will be

3(q + 20) + 5q = 460

=> 3q + 60 + 5q = 460

=> 8q + 60 = 460

=> 8q = 400

=> q = 400 ÷ 8

=> q = 50

Hence, cost of the pen and notebook is Rs. 50 and Rs. 70 respectively.

**Example 4.** Denominator of a number is 9 less than its numerator. If 4 is added to the numerator, it becomes twice the denominator. Find the fraction.

**Solution.** Let the numerator of the number be x.

Then, denominator = (x − 9)

Then, fraction = ^{x}⁄_{x-9}

According to question if 4 added to numerator , it becomes twice the denominator.

So, x + 4 = 2(x − 9)

=> x − 2x = -18 − 4

=> −x = −22

=> x = 22

Hence, fraction = ^{x}⁄_{x-9} = ^{22}⁄_{22-9} = ^{22}⁄_{13}

**Example 5.** After 15 years Prakash shall be 4 times as old as he was 3 years ago. Find his present age.

**Solution.**Let Prakash's present age be 'm' years.

After 15 yrs Prakash's age = (m + 15)years

And 3 yrs ago Prakash's age = (m − 3)

According to the question,

(m + 15) = 4(m − 3)

=> m − 4m = −12 − 15

=> −3m = −27

=> m = ^{-27}⁄_{-3} = 9 yrs

Hence, Prakash's present age is 9 years.

## Class-8 Linear Equations Test

## Class-8 Linear Equations Worksheets

Linear Equations Worksheet - 1

Linear Equations Worksheet - 2

Linear Equations Worksheet - 3

## Answer Sheet

**Linear-Equations-Answer**Download the pdf

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