Simultaneous equations, explained with examples

By The STEM-Tutor team · Updated October 2026

Simultaneous equations are two equations that share the same unknowns and are true at the same time. Solving them means finding the values that make both equations work.

What “simultaneous” means

The equation x + y = 10 has endless solutions: 1 and 9, 2 and 8, 4.5 and 5.5, and so on. Add a second equation, x − y = 2, and only one pair works for both: x = 6, y = 4. That pair is the solution of the simultaneous equations. On a graph, each linear equation is a straight line, and the solution is the point where the lines cross.

Method 1: elimination

Elimination works by adding or subtracting the equations so that one unknown disappears.

1. Line up the equations so x is under x, y is under y and the numbers are on the right.
2. If needed, multiply one or both equations so that one unknown has the same coefficient in both.
3. If the signs are the same, subtract. If they are different, add. A memory aid: “same sign subtract”.
4. Solve the resulting equation, then substitute back to find the other unknown.
5. Check both values in the equation you did not use for substitution.

Example 1: adding to eliminate

Solve 3x + 2y = 16 and 5x − 2y = 0.

• The y terms are +2y and −2y, different signs, so add: 8x = 16, so x = 2.
• Substitute into the first: 6 + 2y = 16, so y = 5.
• Check in the second: 10 − 10 = 0. x = 2, y = 5.

Example 2: subtracting to eliminate

Solve 4x + 3y = 25 and 4x + y = 15.

• The x terms are both +4x, same sign, so subtract: 2y = 10, so y = 5.
• 4x + 5 = 15, so x = 2.5. x = 2.5, y = 5.

Example 3: multiplying one equation

Solve 2x + 3y = 13 and x + y = 5.

• Multiply the second by 2: 2x + 2y = 10.
• Subtract from the first: y = 3. Then x = 2. x = 2, y = 3.

Example 4: multiplying both equations

Solve 3x + 4y = 18 and 2x − 3y = −5.

• To eliminate x, multiply the first by 2 and the second by 3: 6x + 8y = 36 and 6x − 9y = −15.
• Subtract: 17y = 51, so y = 3. Then 3x + 12 = 18, so x = 2.
• x = 2, y = 3.

Method 2: substitution

Substitution works best when one equation already gives one unknown on its own, such as y = 2x + 1. Replace that unknown in the other equation, solve, then substitute back.

Example 5: substitution

Solve y = 2x + 1 and 3x + y = 11.

• Substitute: 3x + (2x + 1) = 11, so 5x = 10 and x = 2.
• y = 2(2) + 1 = 5. x = 2, y = 5.

One linear and one quadratic (Higher)

When one equation contains x² or y², elimination does not work. Use substitution: rearrange the linear equation for one unknown and substitute it into the quadratic. You will get a quadratic equation, usually with two solutions, so there are usually two pairs of answers. On a graph, this is a straight line crossing a curve or circle at two points.

Example 6: a line and a curve

Solve y = x + 1 and y = x² − 1.

• Set them equal: x² − 1 = x + 1, so x² − x − 2 = 0.
• (x − 2)(x + 1) = 0, so x = 2 or x = −1.
• Pair each with its y: x = 2, y = 3 and x = −1, y = 0.

Example 7: a line and a circle

Solve x² + y² = 25 and y = x + 1.

• Substitute: x² + (x + 1)² = 25, so 2x² + 2x + 1 = 25 and x² + x − 12 = 0.
• (x + 4)(x − 3) = 0, so x = 3 or x = −4.
• x = 3, y = 4 and x = −4, y = −3. Always expand (x + 1)² fully: it is x² + 2x + 1, not x² + 1.

Forming equations from a word problem

Many exam questions describe a real situation and expect you to write the equations yourself. Choose letters for the unknowns, write one equation for each piece of information, then solve.

Example 8: a shopping problem

3 pens and 2 rulers cost £4.10. 1 pen and 3 rulers cost £3.00. Find the cost of each.

• Let p = cost of a pen and r = cost of a ruler, in pence: 3p + 2r = 410 and p + 3r = 300.
• Multiply the second by 3: 3p + 9r = 900. Subtract the first: 7r = 490, so r = 70.
• Then p + 210 = 300, so p = 90. A pen costs 90p and a ruler 70p.
• Check: 3 × 90 + 2 × 70 = 410. Working in pence avoids decimals, and exam answers are usually neat, so an awkward value is a sign to recheck your equations.

Solving graphically

You can also solve simultaneous equations by drawing both graphs and reading off where they cross. This gives approximate answers, so exam questions usually say “use the graph to estimate”. If the lines are parallel, they never meet, and there is no solution. If the two equations describe the same line, every point on it is a solution.

Common mistakes

• Subtracting when the signs are different, or adding when they are the same.
• Multiplying only one side of an equation, or only some terms.
• Forgetting to find the second unknown after solving for the first.
• In linear and quadratic pairs, pairing an x with the wrong y.
• Not checking the answer in the original equations.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Are simultaneous equations on the Foundation paper?

Linear simultaneous equations can appear on Foundation papers, usually as a higher-mark question. Linear and quadratic pairs are Higher tier only.

Which method is best?

Elimination is usually quickest for two linear equations. Substitution is best when one equation is already in the form y = …, and it is the only method for a linear and a quadratic.

Are simultaneous equations in A level Maths?

Yes. A level builds on GCSE with harder linear and quadratic pairs, and uses them in coordinate geometry, for example to find where a line meets a circle.

Key takeaways

• The solution satisfies both equations at once.
• Elimination: match coefficients, then same sign subtract, different sign add.
• Substitution: replace one unknown using the other equation.
• Linear and quadratic pairs usually have two solution pairs.
• Always check your answers in the original equations.

Related guides

• The quadratic formula
• Factorising quadratics
• GCSE Maths topics