What are simultaneous equations?
Simultaneous means "at the same time". Simultaneous equations are equations that must all be true together.
Example: x + y = 7 and x − y = 1. Many pairs fit the first one (like 3 and 4, or 5 and 2). Only x = 4, y = 3 fits both. That pair is the solution of the system.
A group of equations solved together is also called a system of equations.
Two linear equations: substitution and elimination
Substitution: make one letter the subject of one equation, then put it into the other.
x + y = 7 → y = 7 − x. Put this into x − y = 1: x − (7 − x) = 1 → 2x = 8 → x = 4, so y = 3.
Elimination: add or subtract the equations so one letter disappears. (x + y) + (x − y) = 7 + 1 → 2x = 8 → x = 4.
Two straight lines can cross once (one solution), be parallel (no solution) or be the same line (infinitely many solutions).
One linear and one quadratic equation
When one equation is a straight line and the other has a square (like y = x² − 1 or x² + y² = 25), use substitution. Elimination usually does not work here.
- Make y (or x) the subject of the linear equation.
- Put it into the quadratic equation.
- Tidy up to get ax² + bx + c = 0 and solve it.
- Put each x back into the linear equation to get its y.
Answers come in pairs: each x has its own y. Write them as points, like (−1, 0) and (2, 3).
How many solutions? Use the discriminant
After substitution you get a quadratic ax² + bx + c = 0. Its discriminant is D = b² − 4ac.
- D > 0: the line cuts the curve at 2 points → 2 solutions.
- D = 0: the line just touches the curve (it is a tangent) → 1 solution.
- D < 0: the line misses the curve → no real solution.
This is how exam questions ask you to "find k so that the line is a tangent": set D = 0 and solve for k.
Two quadratic equations and symmetric systems
Some systems use x + y and xy. Example: x + y = 5, xy = 6. Then x and y are the two roots of t² − 5t + 6 = 0, so t = 2 or 3. The solutions are (2, 3) and (3, 2).
For two quadratics like x² + y² = 13 and xy = 6, use (x + y)² = x² + y² + 2xy = 25, so x + y = ±5, and then solve as above.
Three linear equations in three unknowns
With three letters you need three equations. Remove one letter twice to get two equations in two letters, solve those, then go back.
x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
(1) + (3): 2x + 3y = 8. (2) + (3): 3x + y = 5. From the second, y = 5 − 3x. Then 2x + 15 − 9x = 8 → x = 1, y = 2, and z = 6 − 1 − 2 = 3.
Each linear equation in three letters is a flat plane in 3D. The solution is the point where all three planes meet. Later this is done with matrices.
Try it: the meeting game
In the 3D graph, set m = 2 and slide c. Watch the two red points move together and join when c = −2 (the line becomes a tangent). Predict first: at what value of c will the line stop touching the curve? Then check with the slider and with D = b² − 4ac.
At home: draw y = x² on squared paper. Lay a ruler on it in different positions. Count how many times the ruler edge meets the curve: 2, 1 or 0.
Key formulas and definitions
- Substitution: y = mx + c into the other equation
- Discriminant: D = b² − 4ac (D > 0: 2, D = 0: 1, D < 0: none)
- If x + y = s and xy = p, then x, y are roots of t² − st + p = 0
- (x + y)² = x² + y² + 2xy
Worked examples
1. Solve x + y = 7 and x − y = 1.
Add the equations: 2x = 8, so x = 4. Then y = 7 − 4 = 3. Solution (4, 3).
2. Solve y = x + 1 and y = x² − 1.
Substitute: x² − 1 = x + 1 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0 → x = 2 or x = −1. Then y = 3 or y = 0. Solutions (2, 3) and (−1, 0).
3. Solve x + y = 7 and xy = 12.
y = 7 − x, so x(7 − x) = 12 → x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0. x = 3, y = 4 or x = 4, y = 3.
4. Solve x² + y² = 25 and y = x + 1.
x² + (x + 1)² = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0 → (x + 4)(x − 3) = 0. x = 3, y = 4 or x = −4, y = −3.
5. Find k so that the line y = 2x + k touches the curve y = x².
x² = 2x + k → x² − 2x − k = 0. For one touching point, D = 0: 4 + 4k = 0 → k = −1. It touches at x = 1, y = 1.
6. A rectangle has perimeter 20 m and area 24 m². Find its sides.
2(x + y) = 20 → x + y = 10, and xy = 24. y = 10 − x → x(10 − x) = 24 → x² − 10x + 24 = 0 → x = 4 or 6. Sides 4 m and 6 m.
7. Solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
(1)+(3): 2x + 3y = 8. (2)+(3): 3x + y = 5 → y = 5 − 3x. Then 2x + 15 − 9x = 8 → x = 1, y = 2, z = 3. Check: 2 − 2 + 3 = 3 ✓.
Common mistakes
- Finding x and stopping. Each x needs its own y; give answers as pairs.
- Putting x back into the quadratic instead of the linear equation. This can give extra wrong y values.
- Forgetting brackets when substituting: x − (7 − x), not x − 7 − x.
- Saying "no solution" when D < 0 without saying "no real solution". The line simply misses the curve.