A word problem gives you information in sentences. To use simultaneous equations, you identify two unknown quantities, write one equation for each piece of information, then solve the pair. The key skill is translating the English sentence by sentence into algebra before choosing your method.
Why do word problems need simultaneous equations?
When a problem has two unknown quantities and provides two separate facts about them, you have exactly the information needed to form two equations with two unknowns. Neither equation alone pins down the values — but together they do.
If you had only one equation with two unknowns (e.g. 2x + y = 10), infinitely many pairs (x, y) would satisfy it. The second equation narrows the answer to one solution pair.
Step-by-step process
1. Define your variables. Name each unknown with a letter and write what it represents. Be specific: "Let a = the cost of one adult ticket in pounds."
2. Write two equations. Translate each piece of information from the problem into an equation. Each sentence or condition typically gives one equation.
3. Solve. Use elimination or substitution.
4. Answer the question. State what x and y represent and include units. Check your answers make sense in the original context.
Worked example 1: ticket prices
At a cinema, 3 adult tickets and 2 child tickets cost £28.50. 1 adult ticket and 4 child tickets cost £22.50. Find the cost of each type of ticket.
Step 1 — Define variables. Let a = cost of one adult ticket (£), c = cost of one child ticket (£).
Step 2 — Write the equations. 3a + 2c = 28.50 ... (1) a + 4c = 22.50 ... (2)
Step 3 — Solve by elimination. Multiply equation (2) by 3: 3a + 12c = 67.50 ... (3) Subtract (1) from (3): 10c = 39.00, so c = 3.90 Substitute c = 3.90 into (2): a + 4(3.90) = 22.50 a + 15.60 = 22.50 a = 6.90
Step 4 — Answer. An adult ticket costs £6.90 and a child ticket costs £3.90.
Check: 3(6.90) + 2(3.90) = 20.70 + 7.80 = 28.50 ✓ 1(6.90) + 4(3.90) = 6.90 + 15.60 = 22.50 ✓
Worked example 2: ages
Amir is twice as old as his sister Leila. In 5 years' time, the sum of their ages will be 37. Find their current ages.
Step 1 — Define variables. Let a = Amir's current age, l = Leila's current age.
Step 2 — Write the equations. Amir is twice Leila's age: a = 2l ... (1) In 5 years, sum of ages = 37: (a + 5) + (l + 5) = 37, so a + l = 27 ... (2)
Step 3 — Solve by substitution. Substitute (1) into (2): 2l + l = 27, so 3l = 27, l = 9 Then a = 2(9) = 18
Step 4 — Answer. Amir is currently 18 years old and Leila is 9 years old.
Check: 18 = 2 × 9 ✓. In 5 years: Amir 23, Leila 14. Sum = 37 ✓.
How to spot which method to use
| Situation | Best method |
|---|---|
| One equation already gives x in terms of y (e.g. y = 3x − 1) | Substitution |
| Both equations are in the form ax + by = c | Elimination |
| Coefficients of one variable are already equal or easily made equal | Elimination |
| One equation is quadratic | Substitution |
Common translation phrases
| English phrase | Algebraic meaning |
|---|---|
| "costs twice as much as" | a = 2b |
| "the sum of x and y is 15" | x + y = 15 |
| "5 more than three times x" | 3x + 5 |
| "together they have 40" | x + y = 40 |
| "the difference between their ages is 6" | x − y = 6 (or y − x = 6) |
| "the total of p and q" | p + q |
A three-step shortcut for checking
After solving, always verify both equations:
- Substitute both values into equation (1) — does it balance?
- Substitute both values into equation (2) — does it balance?
- Re-read the question — do the values make sense? (Ages can't be negative; ticket prices should be positive.)
Frequently asked questions
What if I write the equations the wrong way round?
If your equation says 2a + 3c = 28.50 when it should be 3a + 2c = 28.50, your solution will be wrong. Always re-read each sentence carefully and check the coefficient matches the number of items described in that sentence.
Can simultaneous equations have more than two unknowns?
At GCSE you will only encounter two unknowns and two equations. Three unknowns require three equations and are studied at A-level. If a question has three unknowns, look carefully — one of them is probably given an explicit value in the text.
What if the two equations are inconsistent?
If the equations contradict each other (for example, parallel lines in the graphical interpretation), there is no solution. This means the problem as stated is impossible — in a real exam this would indicate a reading error. Re-check your equations against the original problem.
Is it better to use substitution or elimination?
Neither is always better — it depends on the equations. If one equation gives you x directly (like x = 3y − 2), substitution is quick. If both equations have matching coefficients for one variable, elimination avoids extra algebra. With practice you will recognise which is cleaner for each pair.
Translate word problems into simultaneous equations with Professor Pi's Socratic guidance at aitutors.me.