Simultaneous Equations (Linear) — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers solving a pair of linear equations together. By the end of this guide you should be able to solve them by elimination and by substitution, and know which method suits which pair.

You should also be able to handle equations needing multiplication before elimination, deal with negative coefficients, form a pair of equations from a worded problem, solve them graphically, and check your answer properly.

The organising idea is that two unknowns need two equations, and every method works by getting rid of one of them. A single equation such as 2x + y = 10 has endless solutions — x = 1 and y = 8, or x = 3 and y = 4, and so on indefinitely. A second equation narrows that down to one pair. Both algebraic methods do the same job by different routes: elimination removes a letter by adding or subtracting the equations, substitution removes it by replacing it. Once one letter is gone you have an ordinary linear equation, which you already know how to solve.

Key terms and definitions

Simultaneous equations — two or more equations that must be satisfied at the same time.

Linear — containing no powers above 1, so each equation draws a straight line.

Elimination — removing one unknown by adding or subtracting the equations.

Substitution — removing one unknown by replacing it with an expression from the other equation.

Coefficient — the number in front of a letter.

Solution — the pair of values satisfying both equations, and the point where the two lines cross.

Core concepts

What the solution means

Each linear equation describes a straight line, and the solution is the point where the two lines cross — the single pair of values making both true.

That geometric picture explains the unusual cases. If the two lines are parallel, they never meet and there is no solution; the algebra then produces a contradiction such as 0 = 5. If the two equations describe the same line, every point on it works and there are infinitely many solutions; the algebra produces 0 = 0.

Elimination when the coefficients already match

If a letter has the same coefficient in both equations, one operation removes it.

Same signs, subtract. Different signs, add.

For 3x + 2y = 16 and x + 2y = 8, the y terms match in both size and sign, so subtracting gives 2x = 8 and x = 4.

For 3x + 2y = 16 and 5x − 2y = 8, the y coefficients are the same size but opposite in sign, so adding gives 8x = 24 and x = 3.

The mnemonic is worth remembering, but the reasoning is simple enough to reconstruct: you want the y terms to cancel, so do whatever makes them vanish.

Elimination when they do not match

Multiply one or both equations until a pair of coefficients matches.

For 2x + 3y = 12 and 5x + 2y = 19: multiply the first by 2 and the second by 3, giving 4x + 6y = 24 and 15x + 6y = 57. The y terms now match, so subtracting gives 11x = 33 and x = 3.

Multiply every term, including the number on the right-hand side. Multiplying only the left is the commonest error in the topic, and it silently changes the equation into a different one.

Choosing which letter to eliminate is worth a moment: pick the one where the smaller multiplications are needed, or where one coefficient already divides into the other.

Finding the second unknown

Once one value is found, substitute it into either original equation to find the other.

With x = 3 in 2x + 3y = 12: 6 + 3y = 12, so 3y = 6 and y = 2.

Choose whichever equation looks simpler to substitute into — usually the one with the smaller numbers.

Substitution

Where one equation already has a letter as its subject, or easily can, substitution is faster than elimination.

For y = 2x + 1 and 3x + y = 11: replace the y in the second equation with 2x + 1, giving 3x + 2x + 1 = 11, so 5x = 10 and x = 2. Then y = 2(2) + 1 = 5.

Use brackets when substituting an expression, especially where it will be multiplied. In 3x − 2y = 4 with y = x − 3, write 3x − 2(x − 3) = 4, which expands to 3x − 2x + 6 = 4. Omitting the bracket gives 3x − 2x − 3 = 4 and the wrong answer.

Substitution is also the method that carries over to the harder case of one linear and one quadratic equation, so it is worth being comfortable with.

Choosing the method

Use substitution when one equation is already in the form y = something, or when a letter has a coefficient of 1 and so rearranges cleanly.

Use elimination when both equations are in the form ax + by = c, which is the more common presentation.

Either method gives the same answer, and no marks depend on the choice — but picking the natural one saves time and reduces the chance of error.

Negative coefficients

These cause most of the sign errors, and two habits prevent them.

Write the equations out with their signs clearly, lining up the x terms, the y terms and the constants in columns.

When subtracting, remember every term on the second line changes sign. Subtracting 5x − 2y = 8 from 3x + 2y = 16 gives 3x − 5x = −2x on the left, and +2y − (−2y) = +4y, which is why checking each column separately matters.

If subtraction looks awkward, multiplying one equation by −1 turns it into an addition, which many people find safer.

Forming equations from a problem

Worded problems supply two pieces of information, each becoming one equation.

Define the letters explicitly first — "let x be the cost of an adult ticket and y the cost of a child ticket" — since that definition is usually worth a mark and prevents confusion later.

Then translate each sentence. "Two adults and three children cost £34" becomes 2x + 3y = 34, and "one adult and four children cost £27" becomes x + 4y = 27.

Solve as usual, then answer the question in words with units: the adult ticket costs £11 and the child ticket £4.

Read carefully what is being asked — sometimes the question wants a total or a difference rather than the values themselves.

Solving graphically

Plotting both lines and reading the crossing point is a valid method, and some questions require it.

Draw each line accurately, ideally using a small table of values, and read the coordinates where they meet.

Graphical answers are estimates, limited by the accuracy of the drawing and the scale, so questions asking for exact values expect the algebraic method instead.

Checking

Substitute both values into both original equations. If either fails, something has gone wrong.

Checking against only one equation is not enough, since an error in the second stage can still satisfy the first.

Worked examples

Example 1: Elimination with multiplication

Solve 3x + 4y = 18 and 5x − 2y = 4.

The y coefficients are 4 and −2. Multiplying the second equation by 2 gives 10x − 4y = 8, so the y terms are now equal in size and opposite in sign.

Add the equations: 3x + 10x = 13x, and 4y + (−4y) cancels, and 18 + 8 = 26.

So 13x = 26 and x = 2.

Substitute into the first equation: 6 + 4y = 18, so 4y = 12 and y = 3.

Check in the second: 5(2) − 2(3) = 10 − 6 = 4. ✓ Both equations are satisfied.

Note that every term of the second equation was multiplied by 2, including the 4 on the right.

Example 2: Substitution with a bracket

Solve y = 3x − 5 and 2x + 3y = 7.

The first equation already gives y, so substitute it into the second.

2x + 3(3x − 5) = 7. The bracket is essential here, since the 3 multiplies the whole expression.

Expand: 2x + 9x − 15 = 7, so 11x = 22 and x = 2.

Then y = 3(2) − 5 = 1.

Check in the second equation: 2(2) + 3(1) = 4 + 3 = 7. ✓

Example 3: Forming equations from a problem

Three pens and two notebooks cost £13. Five pens and four notebooks cost £23. Find the cost of each.

Let p be the cost of a pen and n the cost of a notebook.

The two equations are 3p + 2n = 13 and 5p + 4n = 23.

Multiply the first by 2 to match the n terms: 6p + 4n = 26.

Subtract the second from it: 6p − 5p = p, the n terms cancel, and 26 − 23 = 3. So p = 3.

Substitute into the first: 9 + 2n = 13, so 2n = 4 and n = 2.

A pen costs £3 and a notebook costs £2. Check in the second equation: 5(3) + 4(2) = 15 + 8 = 23. ✓

Common mistakes and how to avoid them

Multiplying only the left-hand side. Every term must be multiplied, including the constant.

Adding when you should subtract. Same signs subtract, different signs add.

Sign errors when subtracting. Every term on the second line changes sign; check each column.

Omitting brackets when substituting. Write 3(3x − 5), not 3 × 3x − 5.

Giving only one value. The solution is a pair.

Checking against one equation only. Substitute into both.

Not answering the worded question. Return to the context and give the answer with units.

Exam technique for "Simultaneous Equations"

Line the equations up in columns — x terms, y terms, constants — before doing anything. Most sign errors come from misaligned working.

Label your equations (1) and (2), and label any you create, so the examiner can follow which was multiplied by what.

Write the multiplied equation out in full as a new line rather than adjusting in your head.

Substitute back into the simpler equation to find the second unknown, then check in the other one.

For worded problems, define the letters first and give the final answer as a sentence with units.

If a graphical method is asked for, use a table of values and a ruler, and describe the answer as an estimate.

Quick revision summary

Two unknowns need two equations, and every method works by getting rid of one unknown.

The solution is the point where the two lines cross. Parallel lines give no solution; identical lines give infinitely many.

Elimination: match a pair of coefficients, then same signs subtract, different signs add. Multiply every term when scaling an equation.

Substitution: replace one letter using the other equation, with brackets around any expression substituted.

Choose substitution when a letter is already the subject, and elimination when both equations are in the form ax + by = c.

Find the second unknown by substituting into the simpler equation, then check both values in both equations.

For worded problems, define the letters, form one equation per statement, and answer in a sentence with units.