What are simultaneous equations?
Simultaneous equations are two or more independent equations that
share one or more variables such as x and y.
They are called simultaneous equations because the equations are
solved at the same time with a single solution.
The number of variables in simultaneous equations must match the
number of equations for it to be solved.
An example of simultaneous equations is
2x + 4y = 14
4x − 4y = 4
Here are some more:
6a + b = 18
4a + b = 14
3h + 2i = 8
2h + 5i = −2
Each of these equations on their own could have infinite possible
solutions.
However when we have at least as many equations as variables
we may be able to solve them using methods for solving simultaneous
equations.
Representing simultaneous equations
graphically
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.
x+y=6−3x+y=2x+y=6−3x+y=2
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 in this instance are:
x = 1 and y = 5
What are simultaneous
equations?
Simultaneous equations worksheets
Get your free simultaneous equations worksheet of 20+ questions and
answers. Includes reasoning and applied questions.
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Simultaneous equations worksheets
Get your free simultaneous equations worksheet of 20+ questions and
answers. Includes reasoning and applied questions.
DOWNLOAD FREE
Solving simultaneous equations
When solving simultaneous equations you will need different methods
depending on what sort of simultaneous equations you are dealing
with.
There are two sorts of simultaneous equations you will need to solve:

linear simultaneous equations

quadratic simultaneous equations
A linear equation contains terms that are raised to a power that is no
higher than one.
E.g.
2x+5=02x+5=0
Linear simultaneous equations are usually solved by what’s called
the elimination method (although the substitution method is also an
option for you).
Solving simultaneous equations using the elimination method requires
you to first eliminate one of the variables, next find the value of one
variable, then find the value of the remaining variable via substitution.
Examples of this method are given below.
A quadratic equation contains terms that are raised to a power that is
no higher than two.
E.g.
x2−2x+1=0x2−2x+1=0
Quadratic simultaneous equations are solved by the substitution
method.
See also: 15 Simultaneous equations questions
What are linear and quadratic
simultaneous equations?
How to solve simultaneous
equations
To solve pairs of simultaneous equations you need to:
1. Use the elimination method to get rid of one of the variables.
2. Find the value of one variable.
3. Find the value of the remaining variables using substitution.
4. Clearly state the final answer.
5. Check your answer by substituting both values into either of the
original equations.
How do you solve pairs of
simultaneous equations?
See the examples below for how to solve the simultaneous linear
equations using the three most common forms of simultaneous
equations.
See also: Substitution
Quadratic simultaneous equations
Quadratic simultaneous equations have two or more equations that
share variables that are raised to powers up to 2 e.g. x2x2x2 and y2y2y2.
Solving quadratic simultaneous equations algebraically by substitution
is covered, with examples, in a separate lesson.
Step-by-step guide: Quadratic simultaneous equations
Simultaneous equations examples
For each of the simultaneous equations examples below we have
included a graphical representation.
Step-by-step guide: Solving simultaneous equations graphically
Example 1: Solving simultaneous equations by
elimination (addition)
Solve:
2x+4y=144x−4y=42x+4y=144x−4y=4
1. Eliminate one of the variables.
By adding the two equations together we can eliminate the variable y.
2x+4y=144x−4y=46x=182x+4y=144x−4y=46x=18
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=3y=2x=3y=2
5Check your answer by substituting both values into either of the
original equations.
4(3)+4(2)=412−8=44(3)+4(2)=412−8=4
This is correct so we can be confident our answer is correct.
Graphical representation of solving by
elimination (addition)
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 x = 3 and y = 2.
Example 2: Solving simultaneous equations by
elimination (subtraction)
Solve:
6a+b=184a+b=146a+b=184a+b=14
1. Eliminate one of the variables.
By subtracting the two equations we can eliminate the variable b.
6a+b=184a+b=142a=46a+b=184a+b=142a=4
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.
6a+b=186(2)+b=1812+b=18b=66a+b=186(2)+b=1812+b=18b=6
4Clearly state the final answer.
a=2b=6a=2b=6
5Check your answer by substituting both values into either of the
original equations.
4(2)+(6)=148+6=144(2)+(6)=148+6=14
This is correct so we can be confident our answer is correct.
Graphical representation of solving by
elimination (subtraction)
When graphed these two equations intersect at (1,5). So the solution
to the simultaneous equations is a = 2 and b = 6.
Example 3: Solving simultaneous equations by
elimination (different coefficients)
Solve:
3h+2i=82h+5i=−23h+2i=82h+5i=−2
Notice that adding or subtracting the equations does not
eliminate either variable (see below).
3h+2i=82h+5i=−25h+7i=63h+2i=82h+5i=−2h−3i=103h+2i=82h+5i=−25h+7i=63h+
2i=82h+5i=−2h−3i=10
This is because neither of the coefficients of h or i are the same. If
you look at the first two examples this was the case.
So our first step in eliminating one of the variables is to make either
coefficients of h or i the same.
1. Eliminate one of the variables.
We are going to equate the variable of h.
Multiply every term in the first equation by 2.
Multiply every term in the second equation by 3.
3h+2i=82h+5i=−26h+4i=166h+15i=−63h+2i=82h+5i=−26h+4i=166h+15i=−6
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.
6h+4i=166h+15i=−6−11i=226h+4i=166h+15i=−6−11i=22
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=4i=−2h=4i=−2
5Check your answer by substituting both values into either of the
original equations.
2(4)+5(−2)=−28−10=−22(4)+5(−2)=−28−10=−2
This is correct so we can be confident our answer is correct.
Graphical representation of solving by
elimination (different coefficients)
When graphed these two equations intersect at (1,5). So the solution
to the simultaneous equations is h = 4 and i = − 2.
Example 4: Worded simultaneous equation
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.
Additional step: conversion
We need to convert this worded example into mathematical language.
We can do this by representing apples with a and bananas with b.
10a+6b=53a+1b=1.3010a+6b=53a+1b=1.30
Notice we now have equations where we do not have equal
coefficients (see example 3).
1. Eliminate one of the variables.
We are going to equate the variable of b.
Multiply every term in the first equation by 1.
Multiply every term in the second equation by 6.
10a+6b=53a+1b=1.3010a+6b=518a+6b=7.8010a+6b=53a+1b=1.3010a+6b=518a
+6b=7.80
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.
10a+6b=518a+6b=7.80−8a=−2.8010a+6b=518a+6b=7.80−8a=−2.80
NOTE: 6b − 6b = 0 so b is eliminated
16 − − 6 = 22
2Find the value of one variable.
Note: we ÷ (− 8) not 8
3Find the value of the remaining variable/s via substitution.
We know a = 0.35 so we can substitute this value into either of our
original equations.
4Clearly state the final answer.
a=0.35b=0.25a=0.35b=0.25
So
1 apple costs £0.35 (or 35p) and 1 banana costs £0.25 (or 25p).
5Check your answer by substituting both values into either of the
original equations.
3(0.35)+1(0.25)=1.301.05+0.25=1.303(0.35)+1(0.25)=1.301.05+0.25=1.30
This is correct so we can be confident our answer is correct.
Graphical representation of the worded
simultaneous equatio
When graphed these two equations intersect at (1,5). So the solution
to the simultaneous equations is a = 0.35 and b = 0.25.
Common misconceptions

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.
Practice simultaneous equations questions
1. Solve the Simultaneous Equation
6x+3y=48
6x+y=266x+y=266x+y=26
6x+3y=486x+3y=48
x=52=2.5,y=11x=52=2.5,y=11
x=25=2.5,y=11
x=11,y=52=2.5x=11,y=52=2.5
x=11,y=25=2.5
x=6,y=1x=6,y=1
x=6,y=1
x=3,y=6x=3,y=6
x=3,y=6
2. Solve the Simultaneous Equation
x−2y=8
x−3y=3x−3y=3x−3y=3
x−2y=8x−2y=8
x=1,y=2x=1,y=2
x=1,y=2
x=1,y=3x=1,y=3
x=1,y=3
x=18,y=5x=18,y=5
x=8,y=3x=8,y=3
x=18,y=5
x=8,y=3
3. Solve the Simultaneous Equation
4x+2y=34
3x+y=213x+y=213x+y=21
4x+2y=344x+2y=34
x=4,y=2x=4,y=2
x=4,y=2
x=4,y=9x=4,y=9
x=4,y=9
x=3,y=1x=3,y=1
x=3,y=1
x=3,y=2x=3,y=2
x=3,y=2
4. Solve the Simultaneous Equation:
15x−4y=82
5x−9y=125x−9y=125x−9y=12
15x−4y=8215x−4y=82
x=6,y=2x=6,y=2
x=6,y=2
x=15,y=4x=15,y=4
x=5,y=9x=5,y=9
x=15,y=4
x=5,y=9
x=−6,y=−2x=−6,y=−2
x=−6,y=−2
Simultaneous equations GCSE questions
1. Solve the simultaneous equations
3y+x=−43y−4x=63y+x=−43y−4x=6
3y+x=−43y−4x=6
(4 marks)
Show answer
2. Solve the simultaneous equations
x+3y=125x−y=4x+3y=125x−y=4
x+3y=125x−y=4
(4 marks)
Show answer
3. Solve the simultaneous equations
4x+y=25x−3y=164x+y=25x−3y=16
4x+y=25x−3y=16
(4 marks)
Show answer
Learning checklist

Solve two simultaneous equations with two variables (linear/linear)
algebraically

Derive two simultaneous equations, solve the equation(s) and
interpret the solution