In this tutorial we’re going to look at simultaneous equations, covering the skills you need to deal with this type of question at GCSE Foundation level. It’s one of more challenging Foundation topics, so what we’re covering here is Grade 4/5 content.

There are two methods for solving simultaneous equations: for Foundation you need only the elimination method, which is covered below, and for Higher you also need the substitution method, which I’ll cover in a separate tutorial.

Solving basic simultaneous equations

What are simultaneous equations?

You can only solve an equation if there’s just ONE variable in it that you don’t know the value of. If there are TWO unknown values then you need TWO equations, with each one giving you different information.

The “different information” bit is important, because if one equation says a + b = 8 and the other says 2a + 2b = 16 then the second equation is just the first one with everything multiplied by 2; it doesn’t give you any new information.

Simple bar model example

If x + y = 10 then there are lots of possible solutions:
x = 1, y = 9
x = 2, y = 8
x = 3, y = 7
x = 4, y = 6 and so on
(and that’s ignoring negative and decimal solutions!)

BUT if we also know that x + 2y = 14 then we can compare the two equations and find a single solution – i.e. a unique pair of values for x and y that make both equations true.

Using a bar model to show this:

Simultaneous equations bar model showing x + y = 10 and x + 2y = 14

It’s easy to see that the difference between the two bars is the second “y” block, and that accounts for the difference between the totals of 14 – 10 = 4
so y = 4.

If we solve our simultaneous equations algebraically then it looks like this:
x + y = 10
x + 2y = 14
Swap them round so that the bigger one is on top (not essential but makes the subtraction easier), then take the bottom line away from the top line:
x + 2y = 14 –
x + y = 10
0 + y = 4
Notice how the xs cancel out because x – x = 0, and you’re left with just y = 4.
(You don’t have to include the 0 in your working, you can just leave it blank if you prefer.)

Then, once we know what y is, we can substitute its value back into either one of the equations to find x:
x + y = 10 becomes
x + 4 = 10
so x = 10 – 4
x = 6

Finally, you should always check your work by substituting both values back into the other equation to make sure that it’s true for those values. If it isn’t then you know you’ve made a mistake somewhere!
In this case we have
x + 2y = 6 + 2×4 = 14 ✔️

The elimination method

What you’ve just seen is a simple example of using the elimination method to solve a pair of simultaneous equations. In this case it was x that we eliminated (got rid of), when we subtracted one equation from the other and got x – x = 0.

To eliminate an unknown value, you write the two equations one above the other, with the number of either xs or ys matching, and eliminate the unknown value that matches in both equations.

Sometimes you need to rearrange the simultaneous equations so that they’re in the same order: the like terms and the equals signs need to be lined up with each other. Usually we use the form ax + by = c (where a, b and c are known values and you’re trying to find x and y), as in the example above, but it doesn’t really matter as long as they both match.

Your turn 1

Now you try it with these pairs of simultaneous equations. In question 2 you’re finding a and b rather than x and y, but it works in exactly the same way.

x + y = 8 and x + 2y = 13
2a + 3b = 25 and 2a + 5b = 39
3x + 2y = 13 and 3x + 5y = 10

Click here for the solutions

What if not everything is added?

Of course, sometimes you’ll get simultaneous equations with subtraction in them.

Example

x + y = 10 and 3x – y = 22

Start by writing them in columns as before:
x + y = 10
3x – y = 22
You can see that we have the same number of ys in both equations… BUT one is positive and the other is negative. What do you get if you subtract -y from y?
y – -y = y + y = 2y
So that doesn’t work to eliminate y; we just end up with a new equation containing both x and y.

So what do you need to do to get the y terms to cancel out?
Answer: ADD the two equations instead!
x + y = 10 +
3x – y = 22
4x + 0 = 32
x = 32/4 = 8
As before, sub back into one of the equations to find y:
8 + y = 10
y = 10 – 8 = 2
So the solution is x = 8, y = 2
Check using the other equation:
3x – y = 3×8 – 2 = 22 ✔️

If BOTH of the matching terms are negative, for example with the equation pair x – y = 2 and 3x – y = 10, then we’re back to having to subtract to get them to cancel out, because
-y – -y = -y + y = 0

Remember this rule:
OPPOSITE SIGNS, ADD
SAME SIGN, SUBTRACT
When you’re deciding what to do, only look at the matched term – but then once you’ve decided, you either add or subtract the whole equation.

Your turn 2

Try these simultaneous equations, with a mixture of adding and subtracting. Don’t forget that in some cases you’ll have to rearrange them first to make them match!
Also don’t worry if your answers aren’t all integers.

5p + q = 20 and 2p – q = 1
3y = 2x + 8 and 4x + 3y = 20
3y – 2x = 19 and 5y – 2x = 25
2x + 3y = -3 and 4x = 3y + 12

Click here for the answers

Adding more challenge: Multiplying up to get a match

Sometimes your simultaneous equations won’t have a matching term. If that’s the case then you’ll need to multiply up one or both of the equations to get something that matches.

(You won’t often see a question this hard on a Foundation paper – we’re well into Grade 5 territory here!)

Example 1

x + 2y = 12 and 3x – y = 22
Start by writing them in columns as before. As you can see, neither the xs nor the ys match.
However, if we multiply the whole of the first equation by 3 then both equations will contain 3x and that will allow us to eliminate the x by subtraction:

Sub back in to find x:
x + 2×2 = 12
x = 12 – 4 = 8
and use the other equation to check:
3x – y = 3×8 – 2 = 22 ✔️

Alternatively, we could have done this one by multiplying the second equation by 2 to make the ys match and then eliminating y. Try it that way and check that you get the same solution.

Example 2

3x + 2y = 8 and 2x – 3y = 14
In this case we need to multiply BOTH equations up.
Either top x 2 and bottom x 3 to make the xs match
or top x 3 and bottom x 2 to make the ys match.
Let’s do the second option. We get 6y and -6y; these are opposite signs, so we add:

Then sub x = 4 back into one of the original equations to get y = -2, and use the other original equation to check your solution.

Your turn 3

Here’s your final set of simultaneous equations in this lesson. Enjoy the challenge!

2x + 3y = 19 and 4x + y = 13
3x – 5y = -21 and y = 2x + 7
7a – 2b = 26 and 5a + 3b = 7.5

Click here for the answers

That covers everything you need to know about solving simultaneous equations by the elimination method. For Higher GCSE you also need to be familiar with the substitution method, which I’ll cover in another tutorial.

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Answers

Your turn 1 answers:

x = 5, y = 3
a = 2, b = 7
x = 5, y = -1
Full solutions below.

1.
2x + y = 13 –
x + y = 8
x + 0 = 5 (ys match so y is eliminated)
Sub back into one of the equations to find y:
5 + y = 8
y = 8 – 5 = 3
(Check with the other equation: 2x + y = 2×5 + 3 = 13 ✔️)

2.
2a + 5b = 39 –
2a + 3b = 25
0 + 2b = 14 (a is eliminated since we had 2 in each equation, and be careful: 5b – 3b = 2b, not just b)
b = 14/2 = 7
Sub back in to find a:
2a + 3×7 = 25
2a = 25 – 21 = 4
a = 4/2 = 2
(Check: 2a + 5b = 2×2 + 5×7 = 4 + 35 = 39 ✔️)

3.
3x + 5y = 10 –
3x + 2y = 13
0 + 3y = -3 (eliminate x)
y = -3/3 = -1
Sub back in:
3x + 2×-1 = 13
3x – 2 = 13
3x = 13 + 2 = 15
x = 15/3 = 5
(Check: 3x + 5y = 3×5 + 5×-1 = 15 – 5 = 10 ✔️)

Click here to return to questions

Your turn 2 answers:

x = 2, y = 5
x = -2, y = 3
a = 3, b = -2.5

Full solutions:

Click here to return to questions