How to solve simultaneous equations (GCSE and A-Level)

Simultaneous equations means finding values that satisfy two (or more) equations at the same time. At GCSE this usually means two linear equations, or one linear and one quadratic. At A-Level it extends to systems involving three variables and more complex non-linear pairs. The method depends entirely on what type of equations you're given, so the first job is always to identify the type before choosing a technique.

Linear simultaneous equations

When both equations are linear (no squared terms, no products of variables), there are two standard methods: elimination and substitution. Both give the same answer; elimination is usually faster when the coefficients line up conveniently.

Elimination method

Solve: 3x + 2y = 16 and 5x − 2y = 8.

1. The y-coefficients are already +2 and −2, so adding the equations eliminates y directly: (3x + 2y) + (5x − 2y) = 16 + 8, giving 8x = 24, so x = 3.
2. Substitute x = 3 into either original equation: 3(3) + 2y = 16, so 9 + 2y = 16, so y = 3.5.
3. Check both original equations with x = 3 and y = 3.5 to confirm.

If the coefficients don't match directly, multiply one or both equations by a constant first so that a pair of coefficients becomes equal (or opposite). For example, to solve 2x + 3y = 13 and 3x + 4y = 18, multiply the first by 3 and the second by 2 to make the x-coefficients both 6, then subtract.

Substitution method

Substitution is often quicker when one equation is already arranged as y = ... or x = ..., or can easily be rearranged that way.

1. Rearrange one equation to make one variable the subject.
2. Substitute that expression into the other equation, replacing every occurrence of that variable.
3. Solve the resulting equation in one variable.
4. Substitute back to find the second variable.

Linear and quadratic simultaneous equations

This is the standard GCSE Higher-tier version, appearing on Edexcel, AQA and OCR papers. One equation is linear, the other quadratic (often a circle equation or y = expression with an x² term). Substitution is the only method that works here.

Worked example

Solve the simultaneous equations y = x + 1 and x² + y² = 25.

1. Substitute y = x + 1 into the quadratic equation: x² + (x + 1)² = 25.
2. Expand: x² + x² + 2x + 1 = 25, giving 2x² + 2x − 24 = 0.
3. Divide through by 2: x² + x − 12 = 0.
4. Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3.
5. Find the corresponding y for each: when x = −4, y = −3; when x = 3, y = 4.
6. Give both full pairs of solutions: (−4, −3) and (3, 4).

Geometrically, this pair of equations represents a straight line intersecting a circle. Two solutions mean the line crosses the circle at two points; one repeated solution (discriminant = 0) means the line is a tangent; no real solutions mean the line misses the circle entirely. Recognising this connects the algebra to the graph, which is often asked about directly.

Simultaneous equations at A-Level

At A-Level the same substitution method is used, but the equations are typically more demanding — one might involve a quadratic in both x and y, or the system might come embedded inside a larger question about intersections of curves, or need the discriminant to determine the number of intersection points without fully solving.

Using the discriminant to avoid solving

A common A-Level question style asks you to find the value(s) of a constant k for which a line is tangent to a curve, without asking for the point of intersection itself. Substitute to form a quadratic, then set the discriminant b² − 4ac = 0 (tangent, one repeated root) rather than solving the quadratic directly — this is faster and is exactly what the mark scheme expects.

Three-variable systems

A-Level (and some further maths courses) also include systems of three linear equations in three unknowns. The method extends elimination: use one pair of equations to eliminate a variable, then a different pair to eliminate the same variable again, leaving two equations in two unknowns that can be solved as usual, before substituting back through both stages.

Common mistakes

• Substituting into the same equation the expression came from, rather than the other equation — this produces a true but useless identity.
• Sign errors when expanding a bracket like (x + 1)² — always expand fully rather than trying to do it in one step.
• Losing one of the two solutions in a linear-quadratic pair by only finding one value of x.
• Mismatching x and y values when writing final coordinate pairs.
• Forgetting to check answers in both original equations, which would catch most of the above.

A general checklist

1. Identify whether both equations are linear, or one is quadratic.
2. For two linear equations, choose elimination if coefficients align easily, otherwise substitution.
3. For linear and quadratic, always substitute the linear equation into the quadratic one.
4. Solve the resulting single-variable equation fully — expect two solutions from a quadratic.
5. Find the paired second value for each solution, and present full (x, y) pairs.
6. Check every solution pair in both original equations.

Simultaneous equations reward a calm, systematic approach far more than cleverness. Identify the type, pick the matching method, and check your answers at the end — that routine covers the vast majority of questions from Year 9 through to A-Level.

Worked example: a trickier non-linear pair

Solve the simultaneous equations y = 2x − 3 and y² = 4x + 1.

1. Substitute y = 2x − 3 into the second equation: (2x − 3)² = 4x + 1.
2. Expand fully: 4x² − 12x + 9 = 4x + 1.
3. Rearrange to standard quadratic form: 4x² − 16x + 8 = 0, then divide by 4: x² − 4x + 2 = 0.
4. This does not factorise, so use the quadratic formula: x = [4 ± √(16 − 8)] / 2 = 2 ± √2.
5. Find the paired y-values from y = 2x − 3 for each x-value.

Simultaneous equations with fractions

Some GCSE and A-Level questions present simultaneous equations with fractional coefficients or terms on both sides. The extra step is to clear fractions or expand brackets fully before applying elimination or substitution, so that the underlying method is unaffected by the presentation.

Worked example

Solve: (x/2) + y = 7 and x − (y/3) = 4.

1. Multiply the first equation by 2 to clear the fraction: x + 2y = 14.
2. Multiply the second equation by 3 to clear the fraction: 3x − y = 12.
3. From the second equation, y = 3x − 12. Substitute into the first: x + 2(3x − 12) = 14.
4. Solve: x + 6x − 24 = 14, so 7x = 38, so x = 38/7. Substitute back to find y.

This example deliberately does not give whole-number answers, which is realistic for many exam questions — do not assume you have made an error simply because the solution is a fraction. Check your working rather than second-guessing a correct but untidy answer.

Simultaneous equations in context questions

GCSE papers often embed a simultaneous equations question inside a real-world scenario — for example, finding the cost of two types of ticket from two different combinations bought by two different groups. The algebra is identical to the abstract version; the extra skill is translating the wording into two equations correctly before solving.

Worked example

Two adult tickets and three child tickets cost £29. Three adult tickets and one child ticket cost £27. Find the cost of one adult ticket and one child ticket.

1. Let a = cost of an adult ticket, c = cost of a child ticket.
2. Form the equations: 2a + 3c = 29 and 3a + c = 27.
3. Multiply the second equation by 3: 9a + 3c = 81.
4. Subtract the first equation: 7a = 52, so a = 52/7 — check this against the original wording; if the numbers don't give a sensible answer, recheck the equations were set up correctly, since ticket prices should usually be tidy amounts in a well-posed question.

How simultaneous equations connect to graphs

Every simultaneous equations question has a graphical meaning: the solution(s) are the coordinates where the graphs of the two equations intersect. Two linear equations meet at exactly one point unless they are parallel (no solution) or identical (infinitely many solutions). A line and a curve can meet at zero, one or two points, which is why linear-quadratic pairs often produce two solutions rather than one.