# Solving simultaneous equations

## Introduction

A **simultaneous** equation is two (or more) equations which contain more than one letter term.

To solve a pair of simultaneous equations, first eliminate one of the letter terms and find the value of the remaining letter. You can then substitute this value into the original equations to find the value of the other letter term.

A common way of solving simultaneous equations is by equating the **coefficients**.
This means getting the number in front of a letter term the same by multiplying one of both of the equations.

## How to solve simultaneous equations

### Example 1 - A simple case where the coefficient of x is the same

Solve this pair of simultaneous equations: \begin{aligned} 2x + 5y & = 27\\ 2x + 2y & = 12\\ \end{aligned}

In both equations notice we have the same number of \(x\) terms.

Subtracting the second equation from the first equation gives \(3y = 15\) and so \(y = 5\).To find the value of \(x\), substitute \(y = 5\) into the first equation. This gives: \begin{aligned} 2x + 5(5) & = 27\\ 2x + 25 & = 27\\ 2x & = 2\\ x & = 1\\ \end{aligned}

### Example 2 - Solving by multiplying one of the equations

Solve this pair of simultaneous equations: \begin{aligned} 3a + 3b & = 21\\ 6a - 4b & = 22\\ \end{aligned}

To get the coefficients of \(a\) the same multiply the first equation by 2 to get:

\begin{aligned} 6a + 6b & = 42\\ 6a - 4b & = 22\\ \end{aligned}

Subtract the second equation from the first to eliminate the \(a\) terms. Always aim to subtract the largest from the smallest to avoid dealing with negative numbers where possible.

\begin{aligned} 6b - (-4b) & = 42 - 22\\ 10b & = 20\\ b & = 2\\ \end{aligned}

To find the value of \(a\), substitute \(b = 2\) into any of the original equations. For example: \begin{aligned} 3a + 3(2) & = 21\\ 3a + 6 & = 21\\ 3a & = 15\\ a & = 5\\ \end{aligned}

### Example 3 - Solving by multiplying both of the equations

Solve this pair of simultaneous equations: \begin{aligned} 6e + 2f & = -4\\ 8e + 7f & = 12\\ \end{aligned}

This time multiply the first equation by 4 and the second equation by 3. This will give us a coefficient of 24 for the \(e\) terms:

\begin{aligned} 24e + 8f & = -16\\ 24e + 21f & = 36\\ \end{aligned}

Now subtracting gives: \begin{aligned} 21f - 8f & = 36 - (-16)\\ 13f & = 52\\ f & = 4\\ \end{aligned}

Finally, substitute \(f = 4\) into any of the original equations. For example: \begin{aligned} 8e + 7(4) & = 12\\ 8e + 28 & = 12\\ 8e & = -16\\ e & = -2\\ \end{aligned}

### Example 4 - A real life example involving simultaneous equations

On a farmyard there are chickens and dogs. In total there are 46 legs and 19 heads.

How many chickens and dog are there?

Let \(c\) be the number of chickens and \(d\) be the number of dogs.

There are 19 heads in total. This gives the equation \(c + d = 19\).

Since chickens have two legs and dogs have four legs then \(2c + 4d = 46\).

So the pair of equations we need to solve are: \begin{aligned} c + d & = 19\\ 2c + 4d & = 46\\ \end{aligned}

Multiply the first equation by 2 to get:

\begin{aligned} 2c + 2d & = 38\\ 2c + 4d & = 46\\ \end{aligned}

Now subtracting gives: \begin{aligned} 4d - 2d & = 46 - 38\\ 2d & = 8\\ d & = 4\\ \end{aligned}

Since there are 4 dogs, from the first equation there must be 19 - 4 = 15 chickens.

## Worksheets to practise finding solving linear simultaneous equations

Try these worksheets to practise your skills.

- Solving simultaneous equations by equating coefficients
- Solving simultaneous equations by substitution
- Sum and product equations