Simultaneous Equations: Every Method and When to Use Each

Quick answer: For GCSE simultaneous equations, use elimination when both equations are in the form ax + by = c (match one variable's coefficients, then add or subtract to cancel it), use substitution when one variable is already the subject or when you have one linear and one quadratic equation, and use the graphical method only when the question tells you to draw graphs or you want a quick estimate. Two straight lines give one solution; a line-and-quadratic pair usually gives two.

I'm sitting Cambridge IGCSE and A-Level maths, and simultaneous equations look scarier than they are - right up until the quadratic ones appear and half my class freezes. You only ever need three methods, and if you read the two equations before you start, the paper basically tells you which one to use. Here's how I choose in about five seconds, plus the slips that quietly cost method marks.

Pick your method in five seconds

• Both equations look like ax + by = c (for example 2x + y = 8 and 3x - 2y = 5): elimination is usually fastest.
• One variable is already the subject (for example y = 2x + 1): substitution - just plug it straight in.
• One equation is a quadratic or a circle (it contains x^2, y^2 or xy): substitution is the only reliable route.
• The question says "by drawing graphs" or "estimate": graphical.

On a GCSE higher paper (AQA, Edexcel or OCR all do this) the linear-and-quadratic version carries the most marks and trips the most people, so I've given it its own worked example below.

Worked example: one linear pair, solved both ways

Solve 2x + y = 8 and 3x - 2y = 5.

Method A - elimination

1. Match a coefficient. Multiply the first equation by 2 so the y terms match in size: 4x + 2y = 16.
2. Add to cancel y. The y terms are now +2y and -2y (opposite signs), so add the two equations: (4x + 2y) + (3x - 2y) gives 7x = 21.
3. Solve for x. x = 3.
4. Back-substitute. Put x = 3 into 2x + y = 8: 6 + y = 8, so y = 2.
5. Check. 3(3) - 2(2) = 9 - 4 = 5. Correct. Solution: x = 3, y = 2.

Method B - substitution

1. Make a variable the subject. From the first equation, y = 8 - 2x.
2. Substitute into the other equation. 3x - 2(8 - 2x) = 5.
3. Expand and solve. 3x - 16 + 4x = 5, so 7x = 21 and x = 3.
4. Find the other variable. y = 8 - 2(3) = 2.

Same answer, x = 3 and y = 2. Notice substitution was clean here only because y had a coefficient of 1. If every coefficient had been a 2 or a 3, substitution would have dragged you into ugly fractions and elimination would have been the calmer choice.

One linear, one quadratic: substitution wins

This is the higher-tier favourite - very often a straight line meeting a circle.

Solve x^2 + y^2 = 25 and y = x + 1.

1. Substitute the linear equation into the quadratic. Replace y with x + 1: x^2 + (x + 1)^2 = 25.
2. Expand. x^2 + x^2 + 2x + 1 = 25.
3. Rearrange to equal zero. 2x^2 + 2x - 24 = 0, then divide every term by 2: x^2 + x - 12 = 0.
4. Factorise and solve. (x + 4)(x - 3) = 0, so x = -4 or x = 3.
5. Find each y from the LINEAR equation (it's easier and safer than the curve): when x = -4, y = -4 + 1 = -3; when x = 3, y = 3 + 1 = 4.
6. Write the answers as coordinate pairs. (-4, -3) and (3, 4). Check one: (-4)^2 + (-3)^2 = 16 + 9 = 25. Correct.

Here's the mark that people throw away: the scheme awards a point specifically for pairing the right y with each x. If you get two x-values and two y-values but scramble which goes with which, you lose it. Always substitute the x-values back into the line, never the curve.

The graphical method

Plot both equations on the same axes and the solution is wherever they cross - read off both coordinates. Two straight lines cross once; a line and a curve can cross twice.

Use it when the question tells you to draw graphs, or as a quick sanity check. Don't lean on it for exact answers: reading "x is about 2.9" off a graph won't get full marks when the real answer is 3. And if two lines are parallel (same gradient, different intercept) they never meet - the graph is telling you there's no solution.

Common mistakes that cost marks

• Sign slips when subtracting equations. My fix: always scale so the matching terms end up with opposite signs, then add. Adding is far harder to get wrong than subtracting.
• Giving only one answer to a quadratic pair. A line-and-curve question almost always has two solutions - if you found one, go looking for the second.
• Forgetting to find the second variable. x = 3 is half an answer; the examiner wants both x and y.
• Substituting back into the harder equation. Put your x-values into the linear equation - fewer terms, fewer errors.
• Not showing your method. On GCSE you earn method marks even when the arithmetic slips, so write every line rather than jumping to the answer.

Stuck partway through a question, drop the pair into the math solver to see every step laid out, or ask Explain anything to reteach substitution at your own pace.

Test yourself

1. Solve by elimination: 3x + 2y = 12 and x - 2y = 4.
2. Solve by substitution: y = x - 1 and x^2 + y^2 = 25.
3. You're given 5x + 3y = 1 and y = 2x - 3. Which method would you pick, and why?

Quick answers: (1) Add the equations to cancel y: 4x = 16, so x = 4; then 4 - 2y = 4 gives y = 0. Solution (4, 0). (2) x^2 + (x - 1)^2 = 25 becomes 2x^2 - 2x - 24 = 0, then x^2 - x - 12 = 0 and (x - 4)(x + 3) = 0, so x = 4 or x = -3; the coordinate pairs are (4, 3) and (-3, -4). (3) Substitution, because y is already the subject - just replace y in the first equation.

Want these marked and endless fresh ones in your board's style? Check any answer step by step on the math solver or generate a full practice set on the mock exam generator.

FAQ

Which method is best for simultaneous equations? There's no single best one - it depends on the form. If both equations are ax + by = c, use elimination. If one variable is already the subject, or one equation is a quadratic, use substitution. Graphical is for when the question asks for it or you want an estimate.

How do you solve simultaneous equations with a quadratic? Make one variable the subject of the linear equation, substitute it into the quadratic, rearrange to equal zero, then factorise or use the quadratic formula. You'll usually get two x-values; find each matching y from the linear equation and write them as coordinate pairs.

Do you still get marks if the final answer is wrong? On GCSE, yes - method marks are awarded for a correct method even with an arithmetic slip, which is why you should never skip working. Write out the scaling, the substitution and the rearranging every time.

What does it mean if there's no solution? For two straight lines it means they're parallel - same gradient, so they never cross. For a line and a curve it means the line misses the curve entirely (the quadratic you form has no real roots), so there are no crossing points to find.

In short: Read the two equations first - matching coefficients point to elimination, a subject-ready variable or a quadratic points to substitution, and "draw the graphs" points to graphical. Show every line so the method marks are yours even when the numbers wobble, and for a line-and-curve pair remember to hand over two coordinate pairs, correctly matched.