Simultaneous Equations (AQA GCSE Maths)

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26 October 2023

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Linear Simultaneous Equations

What are linear simultaneous equations?

When there are two unknowns (say x and y) in a problem, we need two equations to be able to find them both: these are called simultaneous equations
- you solve two equations to find two unknowns, x and y
  - for example, 3x + 2y = 11 and 2x - y = 5
  - the solutions are x = 3 and y = 1
If they just have x and y in them (no x² or y² or xy etc) then they are linear simultaneous equations

How do I solve linear simultaneous equations by elimination?

"Elimination" completely removes one of the variables, x or y
To eliminate the x's from 3x + 2y = 11 and 2x - y = 5
- Multiply every term in the first equation by 2
  - 6x + 4y = 22
- Multiply every term in the second equation by 3
  - 6x - 3y = 15
- Subtract the second result from the first to eliminate the 6x's, leaving 4y - (-3y) = 22 - 15, i.e. 7y = 7
- Solve to find y (y = 1) then substitute y = 1 back into either original equation to find x (x = 3)
Alternatively, to eliminate the y's from 3x + 2y = 11 and 2x - y = 5
- Multiply every term in the second equation by 2
  - 4x - 2y = 10
- Add this result to the first equation to eliminate the 2y's (as 2y + (-2y) = 0)
  - The process then continues as above
Check your final solutions satisfy both equations

How do I solve linear simultaneous equations by substitution?

"Substitution" means substituting one equation into the other
Solve 3x + 2y = 11 and 2x - y = 5 by substitution
- Rearrange one of the equation into y = ... (or x = ...)
  - For example, the second equation becomes y = 2x - 5
- Substitute this into the first equation (replace all y's with 2x - 5 in brackets)
  - 3x + 2(2x - 5) = 11
- Solve this equation to find x (x = 3), then substitute x = 3 into y = 2x - 5 to find y (y = 1)
Check your final solutions satisfy both equations

How do you use graphs to solve linear simultaneous equations?

Plot both equations on the same set of axes
- to do this, you can use a table of values or rearrange into y = mx + c if that helps
Find where the lines intersect (cross over)
- The x and y solutions to the simultaneous equations are the x and y coordinates of the point of intersection
e.g. to solve 2x - y = 3 and 3x + y = 7 simultaneously, first plot them both (see graph)
- find the point of intersection, (2, 1)
- the solution is x = 2 and y = 1

Solving Equations Graphically Notes Diagram 1, A Level & AS Level Pure Maths Revision Notes

How do I solve linear simultaneous equations from worded contexts?

EPS Notes fig2 (1), downloadable IGCSE & GCSE Maths revision notes

EPS Notes fig2 (2), downloadable IGCSE & GCSE Maths revision notes

Examiner Tip

Always check that your final solutions satisfy the original simultaneous equations - you will know immediately if you've got the right solutions or not.

Worked example

Solve the simultaneous equations

5x + 2y = 11
4x - 3y = 18

Number the equations.

$\begin{array}{rcl} 5 x + 2 y & = & 11 \\ 4 x - 3 y & = & 18 \end{array}$ $\begin{array}{rcl} (1) \\ (2) \end{array}$

Make the y terms equal by multiplying all parts of equation (1) by 3 and all parts of equation (2) by 2.
This will give two 6y terms with different signs. The question could also be done by making the x terms equal by multiplying all parts of equation (1) by 4 and all parts of equation (2) by 5, and subtracting the equations.

$\begin{array}{rcl} 15 x + 6 y & = & 33 \\ 8 x - 6 y & = & 36 \end{array}$ $\begin{array}{rcl} (3) \\ (4) \end{array}$

The 6y terms have different signs, so they can be eliminated by adding equation (4) to equation (3).

$15 x + 6 y = 33 + (8 x - 6 y = 36) 23 x = 69$

Solve the equation to find x by dividing both sides by 23.

$\begin{array}{rcl} x & = & \frac{69}{23} = 3 \end{array}$

Substitute $x = 3$ into either of the two original equations.

$(1)$ $5 (3) + 2 y = 11$

Solve this equation to find y.

$\begin{array}{rcl} 15 + 2 y & = & 11 \\ 2 y & = & 11 - 15 \\ 2 y & = & - 4 \\ y & = & \frac{- 4}{2} = - 2 \end{array}$

Substitute x = 3 and y = - 2 into the other equation to check that they are correct

$\begin{array}{rcl} (2) \end{array}$ $\begin{array}{rcl} 4 x - 3 y & = & 18 \end{array}$
$\begin{array}{rcl} 4 (3) - 3 (- 2) & = & 18 \\ 12 - (- 6) & = & 18 \\ 18 & = & 18 \end{array}$

$x = 3, y = - 2$

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Quadratic Simultaneous Equations

What are quadratic simultaneous equations?

When there are two unknowns (say x and y) in a problem, we need two equations to be able to find them both: these are called simultaneous equations
If there is an x² or y² or xy in one of the equations then they are quadratic (or non-linear) simultaneous equations

How do I solve quadratic simultaneous equations?

Use the method of substitution
- Substitute the linear equation, y = ... (or x = ...), into the quadratic equation
  - Do not try to substitute the quadratic equation into the linear equation
Solve x² + y² = 25 and y - 2x = 5
- Rearrange the linear equation into y = 2x + 5
- Substitute this into the quadratic equation, replacing all y's with (2x + 5) in brackets
  - x² + (2x + 5)² = 25
- Expand and solve this quadratic equation (x = 0 and x = -4)
- Substitute each value of x into the linear equation, y = 2x + 5, to get their value of y
- Present your solutions in a way that makes it obvious which x belongs to which y
  - x = 0, y = 5 or x = -4, y = -3
Check your final solutions satisfy both equations

How do you use graphs to solve quadratic simultaneous equations?

Plot both equations on the same set of axes
- to do this, you can use a table of values (or, for straight lines, rearrange into y = mx + c if it helps)
Find where the lines intersect (cross over)
- The x and y solutions to the simultaneous equations are the x and y coordinates of the point of intersection
e.g. to solve y = x² + 3x + 1 and y = 2x + 1 simultaneously, first plot them both (see graph)
- find the points of intersection, (-1, -1) and (0, 1)
- the solutions are x = -1 and y = -1 or x = 0 and y = 1

Solving Equations Graphically - Notes Diagram 4, A Level & AS Level Pure Maths Revision Notes

Examiner Tip

If the resulting quadratic has a repeated root then the line is a tangent to the curve. If the resulting quadratic has no roots then the line does not intersect with the curve – or you have made a mistake!
When giving your final answer, make sure you indicate which x and y values go together. If you don’t make this clear you can lose marks for an otherwise correct answer.
Don't make the common mistake of thinking each squared term in x² + y² = 25 can be square-rooted to give x + y = 5 (they can't, the most you can do is $\sqrt{x^{2} + y^{2}} = \pm 5$ , but you shouldn't be making x or y the subject of this anyway!)

Worked example

Solve the equations

x² + y² = 36
x = 2y + 6

Number the equations.

$x^{2} + y^{2} = 36 x = 2 y + 6$ $(1) (2)$

There is one quadratic equation and one linear equation so this must be done by substitution.

Equation (2) is equal to $x$ so this can be eliminated by substituting it into the $x$ part for equation (1).
Substitute $x = 2 y + 6$ into equation (1).

${(2 y + 6)}^{2} + y^{2} = 36$

[1]

Expand the brackets, remember that a bracket squared should be treated the same as double brackets.

$(2 y + 6) (2 y + 6) + y^{2} = 36 4 y^{2} + 6 (2 y) + 6 (2 y) + 6^{2} + y^{2} = 36$

[1]

Simplify.

$\begin{array}{rcl} 4 y^{2} + 12 y + 12 y + 36 + y^{2} & = & 36 \\ 5 y^{2} + 24 y + 36 & = & 36 \end{array}$

[1]

Rearrange to form a quadratic equation that is equal to zero.

$\begin{array}{rcl} 5 y^{2} + 24 y + 36 - 36 & = & 0 \\ 5 y^{2} + 24 y & = & 0 \end{array}$

The question does not give a specified degree of accuracy, so this can be factorised.
Take out the common factor of $\begin{array}{rcl} y \end{array}$ .

$\begin{array}{rcl} y (5 y + 24) & = & 0 \end{array}$

Solve to find the values of $y$ .
Let each factor be equal to 0 and solve.

$\begin{array}{rcl} y_{1} & = & 0 \\ 5 y_{2} + 24 & = & 0 \Rightarrow y_{2} = - \frac{24}{5} = - 4.8 \end{array}$

[1]

Substitute the values of $y$ into one of the equations (the linear equation is easier) to find the values of $x$ .

$x_{1} = 2 (0) + 6 = 6 x_{2} = 2 (- \frac{24}{5}) + 6 = - 9.6 + 6$

$x_{1} = 6, y_{1} = 0$
$x_{2} = - 3.6, y_{2} = - 4.8$ [1]

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