Quick revision notes · Mathematics 0580 · Chapter 13 of 37
Simultaneous
Solving two equations together: elimination, substitution, word problems, and line-and-curve pairs.
4 sectionsWorked examples
Common mistakes
At a glance
Solving two equations together: elimination, substitution, word problems, and line-and-curve pairs.
1Elimination
- Make the coefficient of one letter the same in both equations (multiply one or both equations).
- Same signs → subtract; different signs → add.
- Solve for the remaining letter, then substitute back to find the other.
- Check both values in the equation you didn't substitute into.
How to do it
- Label the equations (1) and (2).
- Multiply to match one coefficient.
- Add or subtract to eliminate it.
- Solve; substitute back; check.
Worked example
Solve 3x + 2y = 19 and 2x − 5y = −19.
- (1) × 5: 15x + 10y = 95. (2) × 2: 4x − 10y = −38.
- Add: 19x = 57, so x = 3.
- 3(3) + 2y = 19 → y = 5.
- Check (2): 6 − 25 = −19 ✓
x = 3, y = 5
2Substitution
- Best when one equation already says y = … or x = …
- Replace that letter in the other equation, solve, then substitute back.
Worked example
Solve y = 3x − 4 and 2x + y = 11.
- 2x + (3x − 4) = 11.
- 5x = 15, x = 3.
- y = 3(3) − 4 = 5.
x = 3, y = 5
3Forming simultaneous equations
- Choose a letter for each unknown and say what it means.
- Write one equation for each fact in the question.
- Answer in words with units at the end.
Worked example
2 adult tickets and 3 child tickets cost $31. 1 adult and 1 child ticket cost $12. Find the price of each.
- 2a + 3c = 31 and a + c = 12.
- From the second, a = 12 − c.
- 2(12 − c) + 3c = 31 → c = 7, a = 5.
Adult $5, child $7
4One linear and one non-linear equation
- Rearrange the linear equation to y = … (or x = …) and substitute into the non-linear one.
- This gives a quadratic: solve it to get two x values.
- Find the matching y for each x using the linear equation, and pair them.
- Graphically, the solutions are where the line meets the curve (0, 1 or 2 points).
Worked example
Solve y = x + 1 and x2 + y2 = 25.
- x2 + (x + 1)2 = 25 → 2x2 + 2x − 24 = 0.
- x2 + x − 12 = 0 → (x + 4)(x − 3) = 0.
- x = 3 → y = 4; x = −4 → y = −3.
(3, 4) and (−4, −3)
✗Common mistakes
| Don’t write | Write instead |
|---|---|
| Subtracting when the signs are different | Different signs: add the equations |
| −2y − (−2y) = −4y | −2y − (−2y) = 0 |
| Giving only the x values | Find each y and pair it with its x |
| Squaring (x + 1)2 as x2 + 1 | (x + 1)2 = x2 + 2x + 1 |
★In the exam
- Write what you do to each equation: (1) × 2, (2) × 3.
- Check your pair in the original equations — a quick way to catch sign errors.
- In word problems, state the answer in words with units.