# Solving systems of equations by elimination

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.

(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 coefficients 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.

(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

(a) Arrange the equation next to each other.

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

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

(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.