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Maths Query > Unit > Algebra > Equations

# Linear Equation with Examples and Methods to Solve

## Introduction

Before we take a deep dive into linear equations, let us shed some light on the general term equations used in mathematics.

### What is an equation ?

An equation is a condition in which two algebraic expressions are equal.
When two algebraic expressions involve sign of equality, it is called statement of equality.
The expression written on left side of the sign of equality (=) is called left hand side (LHS) and expression written on the right of equality (=) is called right hand side (RHS).
The equality sign shows value of left hand side is equal to value of right hand side.
In addition to LHS expression, RHS expression and the equality sign, at least one of the two expressions always contain minimum one variable.

Example

Example 1
x + 8 = 14
x + 8 = 14 is an equation in which variable x has the highest power of 1.
Here, LHS expression is x + 8
and RHS expression is 14.

Example 2
x2 + 4x + 4 = 0
x2 + 4x + 4 = 0 is an equation in which variable x has the highest power of 2.
Here, LHS expression is x2 + 4x + 4
and RHS expression is 0.

## What is linear equation?

An equation is called a linear equation in which the highest power of a variable is 1.

Example

Example 1
y + 7 = 8
y + 7 = 8 is linear equation because variable y has the highest power of 1.

Example 2
x + 5 = 8
x + 5 = 8 ia also a linear equation because it has one variable x and its the highest power is 1.

## Methods to solve linear equation

### 1.Trail and error method

In this method, the value of an unknown variable in an equation is calculated by putting different values of the variable, one at a time, in the equation.
By putting values of variable in LHS and in RHS, both sides are checked, if they become equal.
If LHS and RHS becomes equal for a value then that value of variable is called root of equation or solution of equation.

Example

Solve equation 2x + 5 = 15.
In 2x + 5 = 15 equation, LHS is 2x + 5 and RHS is 15.
LHS has variable x.
To solve the above equation, different values of variable x will be used to check which makes LHS 2x + 5 equal to RHS 15.
Value of x are put in LHS until both LHS and RHS become equal.

Table: Solving of equation 2x + 5 = 15 for different values of x

Value of xLHS of equationRHS of equationIs LHS = RHS ?
12(1) + 5 = 715No
22(2) + 5 = 915No
32(3) + 5 = 1115No
42(4) + 5 = 1315No
52(5) + 5 = 1515Yes

Hence, LHS = RHS when for the value of x = 5
Therefore, x = 5 is solution of equation 2x + 5 = 15.

### 2. Systematic method

The above trial and error method to find solution of equation is a bit time consuming for cases when many values of a variable have to be used until LHS becomes equal to RHS.
Systematic method is considered a better method than trial and error method to find value of variable because it can solve the equation with lesser number of steps than trial and error method.
Solving and equation using systematic method involves following the minimum steps to be taken:

1. A same number can be added to both sides of an equation without changing the equality.
2. A same number can be subtracted from both sides of an equation without changing the equality.
3. A same number (a non zero number) can be multiplied on both sides of an equation without changing the equality.
4. Both sides of equation can be divided by a same number (a non zero number) without changing the equality .
Example

Solve equation 2x + 5 = 15.
2x + 5 – 5 = 15 – 5       (subtract 5 on both sides)
2x = 10
$\frac{2x}{2}=\frac{10}{2}$      (Now divide both sides by 2)
x = 5
x = 5 is solution of equation.

### 3. Transposition method

In this method, variables are shifted to LHS and constants are shifted to RHS of an equation. By transposition of terms means a term is shifted to another side alongwith its sign changed.
It means sign of addition is changed to sign of subtraction and sign of subtraction is changed after shifting the term on the other side.

Example

Solve equation 2x + 5 = 15.
Shift variables on LHS and constants on RHS of the equation
∴ 2x = 15 – 5
2x = 10
$x=\frac{10}{2}$
x = 5
∴ x = 5 is solution of linear equation.

Let’s take another example
Solve equation 6x + 5 = 3x + 20
Shift variables on LHS and constants on RHS of the equation
∴ 6x – 3x = 20 – 5
3x = 15
$x=\frac{15}{3}$
x = 5
∴ x = 5 is solution of linear equation.

## Solved Examples

### 1) Solve the equation -2y + 2 = -10.

-2y + 2 = -10
-2y = -10 - 2
-2y = -12
y = $\frac{-12}{-2}$
y = 6

2x - 3x = -9 + 7
-x = -2
x = 2

### 3) Solve the linear equation $\frac{4x -3}{2}=\frac{9 + x}{3}$

Cross multiply
3(4x - 3) = 2(9 + x)
12x - 9 = 18 + 2x
12x - 2x = 18 + 9
10x = 27
x = $\frac{27}{10}$
x = 2.7

### 4) Solve $\frac{x}{2}+2=\frac{x}{3}+4$

$\frac{x}{2}-\frac{x}{3}=4-2$
$\frac{3x - 2x}{6}=2$
$\frac{x}{6}=2$
x = 2 × 6
x = 12

### 5) 8(2x - 3) - 5(4x - 3) = 1

16x - 24 - 20x + 15 = 1
-4x - 9 = 1
-4x = 1 + 9
-4x = 10
x = - $\frac{10}{4}$
x = -2.5

### 6) $\frac{\mathrm{4x - 1}}{2}+1$ = $\frac{\mathrm{x - 1}}{5}+16$

$\frac{\mathrm{4x - 1}}{2}+1$ = $\frac{\mathrm{x - 1}}{5}+16$
$\frac{\mathrm{4x - 1}}{2}-\frac{\mathrm{\left(x - 1\right)}}{5}$ = 16 - 1
$\frac{\mathrm{5\left(4x - 1\right) - 2\left(x -1\right)}}{10}$ = 15
20x - 5 - 2x + 2 = 15
18x - 3 = 15
18x = 15 + 3
18x = 18
x = $\frac{18}{18}$
x = 1

### 7) $\frac{x}{3}$ + $\frac{\mathrm{x - 1}}{2}$ + $\frac{\mathrm{x - 1}}{4}$ = 2

$\frac{\mathrm{4x + 6\left(x - 2\right) + 3\left(x - 1\right)}}{12}$ = 2
$\frac{\mathrm{4x + 6x - 12\right) + 3x - 3}}{12}$ = 2
13x - 15 = 24
13x = 24 + 15
13x = 39
x = $\frac{39}{13}$
x = 3

### 8) $\frac{\mathrm{x - 1}}{2}$ + $\frac{\mathrm{x - 2}}{4}$ = 2

$\frac{\mathrm{2\left(x - 1\right) + \left(x - 2\right)}}{4}$ = 2
$\frac{\mathrm{2x - 2 + x - 2}}{4}$ = 2
$\frac{\mathrm{3x - 4}}{4}$ = 2
3x - 4 = 2 × 4
3x - 4 = 8
3x = 8 + 4
3x = 12
x = $\frac{12}{3}$
x = 4

### 9) 0.6x + 7.2 = 0.2x + 9.6

0.6x - 0.2x = 9.6 - 7.2
0.4x = 2.4
x = $\frac{2.4}{0.4}$
x = 6

### 10) 6.3x + 18 = 2.3x + 30

6.3x - 2.3x = 30 - 18
4.0x = 12
4x = 12
x = $\frac{12}{4}$
x = 3

Help box
-6
6
9
15
-2
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1
21
32
8

## Worksheet 3

### Multiple choice questions

1. 4
2. -4
3. -1
4. -3

1. 2
2. -5
3. 10
4. -10

#### 3) What is yhe value of x in 15x = -1?

1. -1
2. 15
3. - $\frac{1}{15}$
4. $\frac{1}{15}$

1. 7
2. -35
3. -5
4. 5

1. 28
2. -28
3. 0
4. 1

1. 25
2. -25
3. -50
4. 50

1. 2
2. -12
3. -2
4. 12

#### 8) Which value of x does satisfy $\frac{x}{2}=\frac{5}{12}$?

1. $\frac{2}{5}$
2. $\frac{5}{2}$
3. $\frac{6}{5}$
4. $\frac{5}{6}$

1. 1
2. -1
3. 1.5
4. 2.5

#### 10) The value of y for 3(y + 15) = 2y - 10

1. 35
2. -45
3. -55
4. 55
Last updated on: 06-08-2024