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Nonlinear Systems

A free Algebra II lesson from the “Equations, Rational Functions, and Conics” unit, with a worked example and practice problems including step-by-step solutions.

A nonlinear system can include a line with a quadratic, circle, or other curve. Solutions are intersection points that satisfy both equations.

What you'll learn

Why it matters: Where a parabola meets a line, where a circle crosses a curve, and where two growth models intersect are all nonlinear systems. Substitution and graphing still find the intersection — there just may be zero, one, two, or more solutions instead of always exactly one.

Worked example

Problem. Solve y = x + 2 and y = x^2 for x-values.

  1. Set the expressions equal: x + 2 = x^2.
  2. Rearrange: x^2 - x - 2 = 0.
  3. Factor: (x - 2)(x + 1) = 0, so x = 2 or -1.

Answer: -1 and 2

Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. Solutions to a system are graphically...

Choices: Intersection points · Only y-intercepts · Only vertices · End behavior

Show solution
  1. Both equations are true at intersections.

Answer: Intersection points

2. Solve y = x and y = 2x - 3. Enter x.

Show solution
  1. Set x = 2x - 3.

Answer: 3

3. For y = x^2 and y = 4, enter the positive x-value.

Show solution
  1. x^2 = 4, so x = +/-2.

Answer: 2

4. A line and parabola can intersect in how many points?

Choices: 0, 1, or 2 · Exactly 3 · Only 0 · Always 4

Show solution
  1. Depending on the line position.

Answer: 0, 1, or 2

5. Substitute y = 2x into y = x^2. Enter the nonzero x-value.

Show solution
  1. Set x^2 = 2x.
  2. Move all terms to get x^2 - 2x = 0.
  3. Factor x(x - 2) = 0, so the nonzero solution is 2.

Answer: 2

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