Solving Linear-Nonlinear Systems

Q: Why does substitution work here?

Both equations are solved for $y$, so their right sides are equal at any intersection. Setting them equal gives one equation to solve.

Q: Do I need both coordinates?

Yes. Solve for $x$, then substitute back to find $y$, and report each solution as an ordered pair $(x, y)$.

A linear-nonlinear system pairs a line with a curve, like $y = x^2$ and $y = x + 2$ . The solutions are the points where they intersect — and there can be zero, one, or two.

Substitution is the go-to method: since both equal $y$ , set them equal and solve the resulting quadratic.

Substitute and solve the quadratic

When both equations give $y$ , set the right sides equal. That produces a single quadratic in $x$ .

Solve it by factoring or the formula. Each x-solution is the x-coordinate of an intersection point.

Geometrically, the solutions are where the line crosses the curve. In the figure below the line meets the parabola at two points — a line can also just touch it (one solution) or miss it entirely (none).

Back-substitute for full points

For each x you find, plug back into either original equation to get $y$ . Report each solution as an ordered pair.

Two x-values give two intersection points; one gives a single point of tangency; none means the graphs never meet.

Worked examples

Example 1: line meets parabola

Solve $y = x^2$ and $y = x + 2$ by substitution.

Set equal

x^2 = x + 2

Move to one side and factor

x^2 - x - 2 = (x-2)(x+1) = 0

Solve, then back-substitute

x = 2 \to (2,4), ; x = -1 \to (-1,1)

Answer: $(2, 4)$ and $(-1, 1)$

Example 2: counting intersections

If the quadratic from a system has no real roots, how many intersection points are there?

No real x-solutions

\text{discriminant} < 0

So the graphs never meet

0 \text{ points}

Answer: None

Example 3: a line that just touches the parabola

Solve $y = x^2$ and $y = 2x - 1$ by substitution.

Set equal

x^2 = 2x - 1

Move to one side and factor

x^2 - 2x + 1 = (x-1)^2 = 0

One repeated x, then back-substitute

x = 1 \to (1,1)

Answer: $(1, 1)$ , a single point of tangency

Try one yourself

Common questions

Why does substitution work here?

Both equations are solved for $y$ , so their right sides are equal at any intersection. Setting them equal gives one equation to solve.

How many solutions can there be?

A line and a parabola can meet at 0, 1, or 2 points — matching the number of real roots of the resulting quadratic.

Do I need both coordinates?

Yes. Solve for $x$ , then substitute back to find $y$ , and report each solution as an ordered pair $(x, y)$ .

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Solving Linear-Nonlinear Systems

Substitute and solve the quadratic

Back-substitute for full points

Worked examples

Example 1: line meets parabola

Example 2: counting intersections

Example 3: a line that just touches the parabola

Try one yourself

Common questions

Keep learning

Solving Quadratic Equations by Factoring

Solving Two-Variable Systems

Graphing Quadratic Functions