In this chapter we will learn to solve linear equation with two variables using elimination method.

## Solving systems of equations using elimination

The elimination method will work **when two linear equations are given with two variables**.

To understand the process, let us imagine that we have been provided with linear equation with variables x and y.

**Follow the below steps;**

(a) **Arrange the equations next to each other **such that same variables are vertically aligned.

(b) **Multiply/Divide the equation to make coefficient of one variable same**.

(c) **Eliminate the variable with same coef**ficients using subtraction and find the exact value of other variable.

Using this 3 steps you can solve the given system of equations.

Let us understand the above process using solved examples.

## Elimination method – Solved examples

**Example 01**

Solve the below system of equations using elimination method.

x + y = 5

2x – 3y = 4

**Solution**

Here two equations are provided with two variables x & y. So we can solve the equation using elimination method.

Follow the below steps;

(a)** Arrange the equation next to each other such that same variables are in same vertical line**.

(b) **Multiply / Divide one equation to make one of the variable same**.

Multiply first equation by 2, so that we have variable x with coefficient 2 in both the equation.

After multiplying the equation we get;

**(c) Subtract both the equation**.

This can be done multiplying negative sign on second equation.

During subtraction, the variable x will get cancelled out and we get the exact value of y variable.

Solving the equation further, we get;

5y = 6**y = 6/5**

**Here we got exact value of variable y.**

Not put this value in any of the given question to get the value of x.

**Taking first equation.**

x + y = 5

\mathtt{x+\ \frac{6}{5} =5}\\\ \\ \mathtt{x\ =\ 5\ -\ \frac{6}{5}}\\\ \\ \mathtt{x\ =\ \frac{25\ -\ 6}{5}}\\\ \\ \mathtt{x\ =\frac{21}{5}}

Hence, **(21/5 , 6/5) is the solution of given equation.**

**Example 02**

Solve the below system of equation using elimination method.

3x – y = 12

2x + y = 13

**Solution**

Follow the below steps;

**(a) Arrange the equation next to each other**.

Note that if you vertically add the numbers, the y variable will cancel out.

**(b) Add the equations vertically.**

Further solving the expressions.

3x = 25

x = 25 / 3

Here we got exact value of variable x. Now put this value in any of the given equation to get value of y.

Taking first equation.

\mathtt{3x\ -\ y\ =\ 12}\\\ \\ \mathtt{3\ \left(\frac{25}{3}\right) -\ y=\ 12}\\\ \\ \mathtt{25\ -\ y\ =\ 12}\\\ \\ \mathtt{y\ =\ 25\ -\ 12}\\\ \\ \mathtt{y\ =\ 13}

Hence, **(25/3, 13) is the solution of given equation**.

**Example 03**

Solve the system of equation using elimination method.

3x – 5y – 4 = 0

9x = 2y + 7

**Solution**

First arrange both the linear equation properly.

3x – 5y = 4

9x – 2y = 7

(a)** Arrange the equations next to each other**.

(b)** Multiply the first equation by 3.**

After multiplication, we get;

(c) **Subtract the equation**.

To subtract the equation, we have to first multiply second equation with negative sign.

Now subtracting the equation, we get;

-13y = 5**y = -5/13**

Here we got the exact value of y. Put this value in any of the given equation to get the value of x.

Taking first equation.

\mathtt{3x\ -\ 5y\ =\ 4}\\\ \\ \mathtt{3x\ -5\left(\frac{-5}{13}\right) =\ 4}\\\ \\ \mathtt{3x\ +\frac{25}{13} =\ 4}\\\ \\ \mathtt{3x\ =\ 4\ -\ \frac{25}{13}}\\\ \\ \mathtt{3x\ =\ \frac{52-25}{13}}\\\ \\ \mathtt{3x\ =\ \frac{27}{13}}\\\ \\ \mathtt{x\ =\ \frac{9}{13}}

Hence, **(9/13, -5/13) is the solution.**

**Example 04**

Solve the linear equation using elimination method.

9x + 2y = 6

3x – y = 7

**Solution**

Follow the below steps;

(a)** Arrange the equations next to each other.**

(b) **Multiply the second equation by 2 so that the coefficients of y in both the equations are equal**.

After multiplication, we get;

(c) **Adding both the equations vertically.**

Further solving the equations.

15 x = 20

x = 20/15

x = 4 / 3

Here we have got exact value of x. Put this value in any of the given equation to get the value of y.

Taking first equation;

\mathtt{9x\ +\ 2y\ =\ 6}\\\ \\ \mathtt{9\left(\frac{4}{3}\right) \ +2y\ =\ 6}\\\ \\ \mathtt{12\ +\ 2y\ =\ 6}\\\ \\ \mathtt{2y\ =\ 6-12}\\\ \\ \mathtt{2y\ =\ -6}\\\ \\ \mathtt{y\ =\ -3}

Hence,** (4/3, -3) is the solution of give equation.**

**Example 05**

Solve below linear equation using elimination method.

\mathtt{\frac{1}{3} x\ -\ \frac{2}{3} y=\ 3}\\\ \\ \mathtt{\frac{1}{3} x-\frac{1}{2} y=\ \frac{8}{3}}

**Solution**

(a)** Arrange the equation next to each other.**

Note that the coefficient of x are same in both equations, so we don’t need to do any kind of multiplication or division.

(b) **Subtract the expressions**.

For subtraction, first multiply the negative sign on second equation.

After multiplying the negative sign, we get;

Further solving the equation we get;

\mathtt{\frac{-2y}{3} +\frac{1}{2} y=\ 3-\frac{8}{3}}\\\ \\ \mathtt{\frac{-4y+3y}{6} =\frac{9-8}{3}}\\\ \\ \mathtt{\frac{-y}{6} =\frac{1}{3}}\\\ \\ \mathtt{y=-2}

Here we have got exact value of y. Put this value in any given equation to get the value of x.

**Taking first equation.**

\mathtt{\frac{1}{3} x\ -\ \frac{2}{3} y=\ 3}\\\ \\ \mathtt{\frac{1}{3} x\ -\ \frac{2}{3}( -2) =\ 3}\\\ \\ \mathtt{\frac{1}{3} x+\frac{4}{3} =\ 3}\\\ \\ \mathtt{\frac{1}{3} x=\ 3-\frac{4}{3}}\\\ \\ \mathtt{\frac{1}{3} x=\ \frac{9-4}{3}}\\\ \\ \mathtt{x\ =\ 5}

Hence, (5, -2) is the solution of given equations.