Nonlinear Systems of Equations

By the end of this lesson, you will be able to solve systems involving quadratics, circles, and other nonlinear equations using substitution and elimination, and interpret solutions graphically.

The Big Idea

Nonlinear systems arise when two curves (not just two lines) intersect. The intersection points are the solutions. Unlike linear systems (which have at most one solution or infinitely many), nonlinear systems can have 0, 1, 2, or more solutions — the exact count depends on the geometry of the curves.

Solving by Substitution

Best when one equation is linear (or easily solved for one variable): solve for one variable, substitute into the other.

Example 1 — Line and Parabola

Solve: $y = x + 2$ and $y = x^2 - 4$

Substitute the first into the second:

$x + 2 = x^2 - 4 \implies x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0$

$x = 3 \implies y = 5$; $x = -2 \implies y = 0$

Solutions: $(3, 5)$ and $(-2, 0)$

Geometric: line intersects parabola at two points.

Example 2 — Circle and Line

Solve: $x^2 + y^2 = 25$ and $y = x + 1$

Substitute: $x^2 + (x+1)^2 = 25$

$x^2 + x^2 + 2x + 1 = 25 \implies 2x^2 + 2x - 24 = 0 \implies x^2 + x - 12 = 0$

$(x+4)(x-3) = 0 \implies x = -4$ or $x = 3$

$y = -3$ and $y = 4$

Solutions: $(-4, -3)$ and $(3, 4)$

Solving by Elimination

Best when the equations can be combined to eliminate one variable (common for circle/parabola systems).

Example 3 — Two Circles (Elimination)

Solve: $x^2 + y^2 = 5$ and $x^2 + y^2 - 4x = 1$

Subtract first from second: $-4x = -4 \implies x = 1$

Substitute: $1 + y^2 = 5 \implies y^2 = 4 \implies y = \pm 2$

Solutions: $(1, 2)$ and $(1, -2)$

Example 4 — Parabola and Circle

Solve: $y = x^2$ and $x^2 + y^2 = 6$

Substitute $x^2 = y$: $y + y^2 = 6 \implies y^2 + y - 6 = 0 \implies (y+3)(y-2) = 0$

$y = 2 \implies x^2 = 2 \implies x = \pm\sqrt{2}$

$y = -3 \implies x^2 = -3$ — no real solutions (reject)

Solutions: $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$

Example 5 — No Intersection