What the question is really asking
Solving two equations simultaneously means finding the values of \(x\) and \(y\) that satisfy both at the same time. Graphically, that is exactly where the two graphs intersect.
When both equations are straight lines, they meet at most once. But when one is linear and the other is a quadratic (a parabola), a straight line can slice through the curve twice, so you should usually expect a pair of solutions.
- Non-linear system
- A set of equations where at least one is not a straight line, for example a line paired with \(y = x^2 - 1\). The curve is why two intersection points are possible.
Where this is assessed
The algebra sits in Criterion A (selecting and applying substitution) and Criterion C (setting out clear, ordered working). Explaining what the two points mean on the graph, and pairing each \(x\) with its correct \(y\), is Criterion D.
The substitution method
Elimination (adding or subtracting equations) works for two lines, but it fails here because of the \(x^2\) term. Instead you substitute. The recipe is reliable:
- Make sure the linear equation is written as \(y = \ldots\) (or \(x = \ldots\)).
- Substitute that expression into the quadratic so only one variable is left.
- Rearrange into the form \(ax^2 + bx + c = 0\) and solve it (factorise or use the formula).
- Put each \(x\) back into the linear equation to get its matching \(y\).
Solve simultaneously: \(y = x + 1\) and \(y = x^2 - 1\).
Pair the coordinates correctly
Each \(x\) has its own \(y\). Do not mix them: \(x = 2\) goes with \(y = 3\), not with \(y = 0\). Always substitute back into the simpler linear equation, never the quadratic, to avoid extra work and a sign slip.
Interpreting the two intersection points
The two answers are not just numbers, they are the coordinates where the line \(y = x + 1\) crosses the parabola \(y = x^2 - 1\). You can and should check them in both equations:
- \((2, 3)\): line gives \(2 + 1 = 3\) ✓, curve gives \(2^2 - 1 = 3\) ✓.
- \((-1, 0)\): line gives \(-1 + 1 = 0\) ✓, curve gives \((-1)^2 - 1 = 0\) ✓.
Both points satisfy both equations, so the line genuinely passes through the curve at these two places. That checking sentence is worth writing out: it confirms the answer and shows you understand what "simultaneous" means.
Sketch to sanity-check
A quick sketch of the U-shaped parabola and the line makes it obvious whether you should expect zero, one, or two crossings. If your algebra gives two solutions but a sketch shows the line missing the curve, you know to go back and recheck.
How many solutions?
After substituting, you always land on a single quadratic. The number of real solutions of that quadratic is the number of intersection points, and the discriminant \(b^2 - 4ac\) tells you which case you are in.
| Discriminant | Solutions | Geometry |
|---|---|---|
| \(b^2 - 4ac > 0\) | Two | Line cuts the curve at two points |
| \(b^2 - 4ac = 0\) | One (repeated) | Line is a tangent, just touching |
| \(b^2 - 4ac < 0\) | None (no real) | Line misses the curve entirely |
In the worked example the quadratic was \(x^2 - x - 2 = 0\), with discriminant \((-1)^2 - 4(1)(-2) = 1 + 8 = 9 > 0\), confirming the two crossings we found.
Check yourself
Solve each fully, then check your coordinates in both equations.
1. Solve \(y = x + 3\) and \(y = x^2 + 1\) simultaneously. +
Set equal: \(x + 3 = x^2 + 1\), so \(x^2 - x - 2 = 0\), which factorises to \((x - 2)(x + 1) = 0\). Thus \(x = 2\) or \(x = -1\). Using \(y = x + 3\): \(x = 2 \Rightarrow y = 5\); \(x = -1 \Rightarrow y = 2\). Solutions: \((2, 5)\) and \((-1, 2)\).
2. Solve \(y = 2x\) and \(y = x^2 - 3\) simultaneously. +
Set equal: \(2x = x^2 - 3\), so \(x^2 - 2x - 3 = 0\), which factorises to \((x - 3)(x + 1) = 0\). Thus \(x = 3\) or \(x = -1\). Using \(y = 2x\): \(x = 3 \Rightarrow y = 6\); \(x = -1 \Rightarrow y = -2\). Solutions: \((3, 6)\) and \((-1, -2)\).
3. Without solving fully, how many times does the line \(y = x - 5\) meet \(y = x^2\)? +
Set equal: \(x - 5 = x^2\), so \(x^2 - x + 5 = 0\). Discriminant \(= (-1)^2 - 4(1)(5) = 1 - 20 = -19 < 0\). A negative discriminant means no real solutions, so the line and curve meet zero times (the line misses the parabola).
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