MISSION · Algebra
Nonlinear Systems of Equations
Where a line meets a curve, and why there can be two answers, one, or none
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The idea in one sentence
Solving a system is finding where the graphs cross, and once one of them bends, there can be two crossings, one, or none at all.
Substitution is almost always the method
Elimination relies on adding equations to cancel a variable, and that only works when the variables appear to the first power. With an \(x^2\) in play it stops helping.
So the routine is: rearrange the linear equation to get one variable alone, substitute it into the curve, and solve the single-variable equation that results.
Take \(y = x^2\) and \(y = x + 2\). Both give \(y\), so set them equal:
\[x^2 = x + 2 \quad \Rightarrow \quad x^2 - x - 2 = 0 \quad \Rightarrow \quad (x-2)(x+1) = 0\]so \(x = 2\) or \(x = -1\).
Then find the matching \(y\) for each, using the easier equation: \(y = 4\) and \(y = 1\). The solutions are the points \((2, 4)\) and \((-1, 1)\).
Two things to watch. The answers come in pairs — a bare \(x = 2\) is half an answer. And each \(x\) keeps its own \(y\); pairing them up the wrong way round gives points that lie on neither graph.
How many solutions to expect
The picture tells you before the algebra does.
A line and a parabola: 2, 1, or 0. One crossing means the line is tangent.
A line and a circle: 2, 1, or 0 — through it, touching it, or missing it entirely.
Two circles: 2, 1, or 0.
A line and a hyperbola: up to 2, one on each branch.
Once the algebra reduces to a quadratic, the discriminant answers the counting question directly:
\[b^2 - 4ac > 0 \Rightarrow 2 \text{ points} \qquad = 0 \Rightarrow \text{tangent} \qquad < 0 \Rightarrow \text{no intersection}\]That last case is a real answer, not a mistake. “No solutions” means the line misses the curve, and in coordinate geometry questions it is often exactly what is being tested.
Line meets circle
This is the standard exam pairing, and it has one extra step worth spelling out.
\(x^2 + y^2 = 25\) and \(y = x + 1\).
Substitute: \(x^2 + (x+1)^2 = 25\). Expand carefully — \((x+1)^2 = x^2 + 2x + 1\), not \(x^2 + 1\) — to get \(2x^2 + 2x - 24 = 0\), then \(x^2 + x - 12 = 0\), so \((x+4)(x-3) = 0\) and \(x = -4\) or \(x = 3\).
The points are \((-4, -3)\) and \((3, 4)\).
Squaring a bracket term by term is the error that costs most marks here, and it produces an answer that looks perfectly reasonable.
When squaring creates false answers
If solving requires squaring both sides — as it does when a square root is involved — some solutions that appear may not satisfy the original equations. Squaring turns \(-3 = 3\) into the true statement \(9 = 9\), so it can manufacture roots that were never there.
Every answer must be checked in the original system, not the rearranged one. These are the same extraneous solutions that appear with radical equations, arriving by the same route.
Systems in three variables
A different kind of harder. Three linear equations in \(x, y, z\) describe three planes, and the solution is where all three meet — usually a single point.
Elimination returns here: use two equations to eliminate \(z\), use a different pair to eliminate \(z\) again, and you are left with two equations in \(x\) and \(y\), which is a problem you already know how to do. Back-substitute for \(z\) at the end.
The degenerate cases have pictures too. Planes can meet in a line, giving infinitely many solutions, or fail to share any common point at all — like the three faces of a triangular prism, each pair meeting but all three never agreeing.
Common mix-ups
- Giving \(x\) without \(y\). Solutions are points.
- Mismatching the pairs. Each \(x\) goes with the \(y\) it produced.
- Expanding \((x+1)^2\) as \(x^2+1\).
- Treating “no real solutions” as an error. The graphs simply do not meet.
- Skipping the check after squaring.
- Substituting into the harder equation to recover the second variable, when the linear one is right there.
Check yourself
At most how many times can a line cross a circle? Solve \(y = x^2\) with \(y = x + 2\). And what does a negative discriminant mean for the number of solutions?
A line meets a circle at most twice. Setting the two equal gives \(x^2 - x - 2 = 0\), so \(x = 2\) or \(x = -1\), giving \((2,4)\) and \((-1,1)\). A negative discriminant means the quadratic has no real roots, so the graphs never meet: 0 solutions.
LEVEL 3 · REFLECT
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