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MathidiaMathematics Academy
Proof and advanced algebra
Lesson 4 of 75
Challenge

Simultaneous algebraic models

Objective. Solve linked linear and nonlinear equations and interpret valid roots.

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

Learn: the key idea

Simultaneous equations link two unknowns through two conditions, and when one equation is linear and the other quadratic, substituting the linear equation into the quadratic reduces the system to a single-variable quadratic. Solve that quadratic, then back-substitute each root to find matching pairs, and discard any pair that fails the original equations or a context restriction such as a negative length. For y=x+1 and y=x²−1, substituting gives x+1=x²−1, so x²−x−2=0, giving x=2 or x=−1, with corresponding y=3 or y=0. Always check both equations are satisfied by each final pair.

Worked example

Solve the simultaneous equations y=2x−1 and y=x²−4 to find the valid pairs (x,y).

2 … 1
Coordinate grid with the line 2 … 1
  1. 1Set 2x−1 = x²−4
  2. 2Rearrange to x²−2x−3=0
  3. 3Factorise as (x−3)(x+1)=0, so x=3 or x=−1
  4. 4Substitute back: (3,5) and (−1,−3)

Answer: (3, 5) and (−1, −3)

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Common mistake. Students substitute the quadratic equation into the linear one instead of the reverse, creating unnecessary complexity; always isolate a variable in the linear equation first and substitute into the quadratic one.

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