Simultaneous equations: linear-linear and linear-quadratic [Higher tier]
Two equations, two unknowns. The solution is the pair (x, y) that satisfies both at once. Geometrically, it's where the two graphs cross.
Linear-linear: elimination
The most reliable method. Make the coefficients of one variable equal (in absolute value), then add or subtract to eliminate it.
Example: solve 2x + 3y = 12 and 5x + 3y = 21 simultaneously.
The y-coefficients already match. Subtract: (5x + 3y) - (2x + 3y) = 21 - 12 → 3x = 9 → x = 3.
Substitute back into the first: 6 + 3y = 12 → y = 2. Solution: (3, 2).
Elimination with sign-matching
3x + 2y = 14 and 4x - 2y = 6. y-coefficients are +2 and -2. Add to eliminate y:
7x = 20 → x = 20/7. Substitute: messy — usually the question chooses nicer numbers, so always check whether to add or subtract.
Elimination with multiplication
2x + 3y = 13 and 5x - 2y = 4. To eliminate y, multiply first by 2 and second by 3:
4x + 6y = 26 and 15x - 6y = 12. Add: 19x = 38 → x = 2. Substitute: 4 + 3y = 13 → y = 3.
Linear-quadratic: substitution
When one equation is quadratic, substitute the linear one into the quadratic.
Example: y = x + 1 and y = x² - 5. Substitute: x + 1 = x² - 5 ⇒ x² - x - 6 = 0 ⇒ (x - 3)(x + 2) = 0 ⇒ x = 3 or x = -2.
Find y for each: x = 3 → y = 4; x = -2 → y = -1. Solutions: (3, 4) and (-2, -1).
A linear-quadratic system typically has two solutions (the line meets the curve twice), but can have one (tangent) or zero (no intersection).
✦Worked example— Worked example — circle and line
x² + y² = 25 and y = x + 1. Substitute: x² + (x + 1)² = 25 → x² + x² + 2x + 1 = 25 → 2x² + 2x - 24 = 0 → x² + x - 12 = 0 → (x + 4)(x - 3) = 0.
x = -4 → y = -3 and x = 3 → y = 4. Two intersection points: (-4, -3) and (3, 4).
⚠Common mistakes— Common mistakes (examiner traps)
- Wrong sign when adding/subtracting equations. Watch the signs of every term.
- Forgetting to find the second variable. A solution is a pair (x, y).
- Substituting back into the wrong equation. Either works — but pick the simpler one.
- Linear-quadratic: substituting the quadratic into the linear. Always go the other way (linear into quadratic) so you don't have a y² term to square-root.
- Reporting only one of two solutions in a linear-quadratic system. There are usually two intersection points.
➜Try this— Quick check
Solve y = 2x + 1 and y = x² + 1. Sub: 2x + 1 = x² + 1 ⇒ x² - 2x = 0 ⇒ x(x - 2) = 0. So x = 0 or x = 2; y = 1 or y = 5. Points: (0, 1) and (2, 5).
AI-generated · claude-opus-4-7 · v3-deep-algebra