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MathidiaMathematics Academy
Simultaneous equations and inequalities
Lesson 26 of 75
Introduction

Nonlinear simultaneous equations

Objective. Solve systems containing a line and a circle or quadratic.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Nonlinear simultaneous equations pair a straight line with a curve such as a circle or quadratic, giving up to two intersection points. Rearrange the linear equation to make one variable the subject, then substitute it into the nonlinear equation to get a single equation in one variable. Solve that equation (often a quadratic) and back-substitute to find matching pairs. For example, with y = x + 1 and x² + y² = 25, substitution gives x² + (x+1)² = 25, so 2x² + 2x − 24 = 0, giving x = 3 or x = −4.

Worked example

Solve the simultaneous equations y = 2x − 1 and x² + y² = 10.

2 … 1
Coordinate grid with the line 2 … 1
  1. 1Substitute y = 2x − 1 into x² + y²: x² + (2x−1)² = 10
  2. 2Expand: x² + 4x² − 4x + 1 = 10, so 5x² − 4x − 9 = 0
  3. 3Factorise: (5x − 9)(x + 1) = 0, giving x = 9/5 or x = −1
  4. 4Find y: x = 9/5 gives y = 13/5; x = −1 gives y = −3

Answer: x = 9/5, y = 13/5 or x = −1, y = −3

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Common mistake. A common error is substituting the curve equation into the line instead of the other way round, or forgetting to back-substitute both roots, losing one solution pair. Always substitute into the equation that keeps the algebra linear in one step, and find y for each x value found.

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