The *solving simultaneous equations introduction* page explained how we can solve 2 simultaneous linear equations with 2 variables together.

Along with the fact that a combined solution will graphically be a point of intersection of the 2 straight lines that the equations represent.

This page will show the process of solving two simultaneous equations, with the methods of solving equations by substitution and solving by elimination.

Both are common methods used for solving when dealing with simultaneous equations in Math.

This page will focus on the approach of solving by substitution.

## Solving Equations by Substitution Approach

When dealing with 2 equations involving 2 variables, such as.

x + 2y = 4

3x \space {\text{--}} \space y = 5

When solving by substitution we want to put one of the equations into the form “x =” or “y =“.

Then ‘substitute’ said equation into the other equation in place of the relevant variable, then solve to obtain a value for the other variable.

This value obtained can then be used in the other equation to solve fully for both variables in the equations.

So if we follow this approach with the equations above, we can start by labelling them with numbers.

x + 2y = 4 [1]

3x \space {\text{--}} \space y = 5 [2]

Subtracting 2y from both sides of [1] will give us an equation in the form “x =“.

x + 2y \space {\text{--}} \space 2y = 4 \space {\text{--}} \space 2y => x = 4 \space {\text{--}} \space 2y

Now substituting this into [2] we have.

3(4 \space {\text{--}} \space 2y) \space {\text{--}} \space y = 5 = 12 \space {\text{--}} \space 6y \space {\text{--}} \space y = 5

12 \space {\text{--}} \space 7y = 5 , {\text{-}}7y = {\text{-}}7 => y = 1

Now substituting this y value into [1] can give us x.

x + 2(1) = 4 => x + 2 = 4 , x = 2

__Example__

*(1.1)*Solve the simultaneous equations:

\space x + y = 6

2x \space {\text{--}} \space 3y = 7

by solving by substitution.

*Solution*If we begin by labelling each equation.

\space x + y = 6 [1]

2x \space {\text{--}} \space 3y = 7 [2]

We can see that subtracting y from both sides of [1] will get the equation in terms of just x.

x + y \space {\text{--}} \space y = 6 \space {\text{--}} \space y => x = 6 \space {\text{--}} \space y

Substituting this into [2] gives.

2(6 \space {\text{--}} \space y) \space {\text{--}} \space 3y = 7 = 12 \space {\text{--}} \space 2y \space {\text{--}} \space 3y = 7

12 \space {\text{--}} \space 5y = 7 , {\text{-}}5y = {\text{-}}5 => y = 1

Substituting this y value into [1] to obtain x.

x + 1 = 6 , x = 5

It’s generally good practice to also check that the values for the variables also work for the other equation, to be sure they are correct.

So here we can input the values in to equation [2].

2(5) \space {\text{--}} \space 3(1) = 7 => 10 \space {\text{--}} \space 3 = 7

**( CORRECT )**

**Return to TOP of page**