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Here is everything you need to know about simultaneous equations for GCSE maths (Edexcel, AQA and OCR).

You’ll learn what simultaneous equations are and how to solve them algebraically. We will also discuss their relationship to graphs and how they can be solved graphically.

Look out for the simultaneous equations worksheets and exam questions at the end.

Simultaneous equations are two or more algebraic equations that share variables e.g.

For example, below are some simultaneous equations:

\[\begin{aligned}
2x+4y&=14\\
4x-4y&=4
\end{aligned} \begin{aligned}
6a+b&=18\\
4a+b&=14
\end{aligned}
\begin{aligned}
3h+2i&=8\\
2h+5i&=-2
\end{aligned}\]

When we have at least as many equations as variables we **may** be able to solve them.

We can consider each equation as a function which, when displayed graphically, may intersect at a specific point. This point of intersection gives the solution to the simultaneous equations.

E.g.

\[\begin{aligned}
x+y=6\\
-3x+y&=2
\end{aligned}\]

When we draw the graphs of these two equations, we can see that they intersect at (1,5).

So the solutions to the simultaneous equations are:

We will learn how to solve linear and quadratic simultaneous equations.

A linear equation contains terms that are raised to a power that is no higher than one.

E.g.

\[2x + 5=0\]

A quadratic equation contains terms that are raised to a power that is no higher than two.

E.g.

\[x^{2}-2x+1=0\]

Here we will learn how to solve simultaneous linear equations algebraically and link it to the graphical representation of the solution.

To solve pairs of simultaneous equations you need to:

- Use the elimination method to get rid of one of the variables.
- Find the value of one variable.
- Find the value of the remaining variables using substitution.
- Clearly state the final answer.
- Check your answer by substituting both values into either of the original equation.

See below for example solutions to the three most common forms of simultaneous equations and a worded example ‘assessment style’ question.

Within simultaneous equations, you will find lessons on quadratic simultaneous equations.

Each method of simultaneous equations is summarised below. For detailed examples, practice questions and worksheets, follow the link to the step by step guide below.

You’ll learn what quadratic simultaneous equations are and how to solve them algebraically. We will also discuss their relationship to graphs and how they can be solved graphically.

**Step by step guide: Quadratic simultaneous equations (coming soon)**

Solve:

\[\begin{aligned}
2x+4y&=14\\
4x-4y&=4
\end{aligned}\]

- Eliminate one of the variables.

*By adding the two equations together we can eliminate the variable y.*

\[\begin{aligned}
2x+4y&=14\\
4x-4y&=4\\
\hline
6x&=18
\end{aligned}\]

2Find the value of one variable.

3Find the value of the remaining variable via substitution.

*We know x = 3 so we can substitute this value into either of our original equations.*

4Clearly state the final answer.

\[x=3 \qquad\qquad y=2\]

5Check your answer by substituting both values into either of the original equation.

\[\begin{aligned}
4(3)+4(2)&=4\\
12-8&=4\\
\end{aligned}\]

*This is correct so we can be confident our answer is correct.*

When we draw the graphs of these linear equations they produce two straight lines. These two lines intersect at (1,5). So the solution to the simultaneous equations is *and*

Solve:

\[\begin{array}{l}
6 a+b=18 \\
4 a+b=14
\end{array}\]

- Eliminate one of the variables.

*By subtracting the two equations together we can eliminate the variable yb.*

\[\begin{aligned}
6a+b&=18 \\
4a+b&=14 \\
\hline
2a&=4
\end{aligned}\]

*NOTE: b − b = 0 so b is eliminated*

2Find the value of one variable.

3Find the value of the remaining variable/s via substitution.

*We know a = 2 so we can substitute this value into either of our original equations.*

\[\begin{aligned}
6 a+b &=18 \\
6(2)+b &=18 \\
12+b &=18 \\
b &=6
\end{aligned}\]

4Clearly state the final answer.

\[a=2 \qquad\qquad b=6\]

5Check your answer by substituting both values into either of the original equation.

\[\begin{aligned}
4(2)+(6) &=14 \\
8+6 &=14
\end{aligned}\]

*This is correct so we can be confident our answer is correct.*

When graphed these two equations intersect at (1,5). So the solution to the simultaneous equations is

Solve:

\[\begin{array}{l}
3 h+2 i=8 \\
2 h+5 i=-2
\end{array}\]

**Notice that adding or subtracting the equations does not eliminate either variable (see below).**

\[\begin{array}{l}
3 h+2 i=8 \\
2 h+5 i=-2 \\
\hline
5 h+7 i=6
\end{array} \begin{aligned}
3 h+2 i&=8 \\
2 h+5 i&=-2 \\
\hline
h-3 i&=10
\end{aligned}\]

This is because neither of the **coefficients** of

So our first step in eliminating one of the variables is to make either coefficients of

- Eliminate one of the variables.

*We are going to equate the variable of h.*

Multiple **every term** in the first equation by

Multiple **every term** in the first equation by

\[\begin{aligned}
3h+2 i&=8 \\
2h+5 i&=-2 \\
\\
6h+4 i&=16 \\
6h+15 i&=-6
\end{aligned}\]

*Now the coefficients of h are the same in each of these new equations, we can proceed with our steps from the first two examples. In this example, we are going to subtract the equations.*

\[\begin{aligned}
6 h+4 i&=16 \\
6 h+15 i&=-6 \\
\hline
-11 i&=22
\end{aligned}\]

*NOTE: 6h − 6h = 0 so h is eliminated*

*CAREFUL: 16 − − 6 = 22*

2Find the value of one variable.

3Find the value of the remaining variable/s via substitution.

*We know i = − 2 so we can substitute this value into either of our original equations.*

4Clearly state the final answer.

\[h=4 \qquad\qquad i=-2\]

5Check your answer by substituting both values into either of the original equation.

\[\begin{aligned}
2(4)+5(-2)&=-2 \\
8-10&=-2
\end{aligned}\]

This is correct so we can be confident our answer is correct.

When graphed these two equations intersect at (1,5). So the solution to the simultaneous equations is

David buys 10 apples and 6 bananas in a shop. They cost £5 in total.

In the same shop, Ellie buys 3 apples and 1 banana. She spends £1.30 in total.

Find the cost of one apple and one banana.

We need to convert this worded example into mathematical language. We can do this by representing apples with

\[\begin{aligned}
10a+6b&=5 \\
3a+1b&=1.30
\end{aligned}\]

Notice we now have equations where we do not have equal coefficients (see example 3).

- Eliminate one of the variables.

*We are going to equate the variable of b.*

Multiple **every term** in the first equation by

Multiple **every term** in the first equation by

\[\begin{aligned}
10 a+6 b&=5 \\
3 a+1 b&=1.30 \\
\\
10 a+6 b&=5 \\
18 a+6 b&=7.80
\end{aligned}\]

*Now the coefficients of b are the same in each equation we can proceed with our steps from the previous examples. In this example, we are going to subtract the equations.*

\[\begin{aligned}
10a+6b &=5 \\
18a+6b &=7.80 \\
\hline
-8a &=-2.80
\end{aligned}\]

*NOTE: 6b − 6b = 0 so b is eliminated*

2Find the value of one variable.

*NOTE: we ÷ (− 8) not 8*

3Find the value of the remaining variable/s via substitution.

*We know xa = 0.35 so we can substitute this value into either of our original equations.*

4Clearly state the final answer.

\[a=0.35 \qquad\qquad b=0.25\]

**So**

5Check your answer by substituting both values into either of the original equation.

\[\begin{aligned}
3(0.35)+1(0.25) & =1.30 \\
1.05+0.25 & =1.30
\end{aligned}\]

*This is correct so we can be confident our answer is correct.*

When graphed these two equations intersect at (1,5). So the solution to the simultaneous equations is

**Incorrectly eliminating a variable.**

Using addition to eliminate one variable when you should subtract (and vice-versa).**Errors with negative numbers.**

Making small mistakes when+, −, ✕, ÷ with negative numbers can lead to an incorrect answer. Working out the calculation separately can help to minimise error.**Step by step guide: Negative numbers (coming soon)****Not multiplying every term in the equation.**

Mistakes when multiplying an equation. For example, forgetting to multiply**every**term by the same number.**Not checking the answer using substitution.**

Errors can quickly be spotted by substituting your solutions in the original first or second equations to check they work.

1. Solve the simultaneous equation:

\[6x + 3y = 48\]

\[6x + y = 26\]

Show answer

\[x=\frac{5}{2}=2.5, y=11\]

2. Solve the simultaneous equation:

\[x - 2y = 8\]

\[x - 3y = 3\]

Show answer

\[x = 18, y = 5\]

3. Solve the simultaneous equation:

\[4x +2y = 34\]

\[3x + y = 21\]

Show answer

\[x = 4, y = 9\]

4. Solve the simultaneous equation:

\[15x - 4y = 82\]

\[5x -9y =12\]

Show answer

\[x = 6, y = 2\]

1. Solve the simultaneous equations

\begin{array}{l} 3 y+x=-4 \\ 3 y-4 x=6 \end{array}

**(4 marks)**

Show answer

\begin{array}{l} 5x=-10 \\ x=-2 \end{array} or correct attempt to find y

**(1)**

One unknown substituted back into either equation

**(1)**

y=-\frac{2}{3} \text { oe }

**(1)**

x=-2

**(1)**

2. Solve the simultaneous equations

\begin{array}{l} x+3y=12 \\ 5x-y=4 \end{array}

**(4 marks)**

Show answer

Correct attempt to multiple either equation to equate coefficients e.g.

\begin{array}{l} 5x+15y=60 \\ 5x-y=4 \end{array}

**(1)**

Or

\begin{array}{l} x+3y=12 \\ 15x-3y=12 \end{array}

**(1)**

Correct attempt to find y or x ( 16y=56 or 16x = 24 seen)

**(1)**

One unknown substituted back into either equation

**ft (1)**

y=\frac{7}{2} \text { oe }

x=\frac{3}{2} \text { oe }

**(1)**

3. Solve the simultaneous equations

\begin{array}{l} 4x+y=25 \\ x-3y=16 \end{array}

**(4 marks)**

Show answer

Correct attempt to multiple either equation to equate coefficients e.g.

\begin{array}{l} 12x+3y=75 \\ x-3y=16 \end{array}

**(1)**

Or

\begin{array}{l} 4x+y=25 \\ 4x-12y=64 \end{array}

**(1)**

Correct attempt to find y or x ( 13x=91 or 13y=-39 seen)

**(1)**

One unknown substituted back into either equation

**ft (1)**

x=7 \text { oe }

y=-3 \text { oe }

**(1)**

Get your free simultaneous equations worksheet of 20+ questions and answers. Includes reasoning and applied questions.

- Solve two simultaneous equations with two variables (linear/linear) algebraically
- Derive two simultaneous equations, solve the equation(s) and interpret the solution

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