When one equation is linear and the other has a squared term or a product xy, use substitution. Make one variable the subject of the linear equation, substitute it, and solve the quadratic that results.
This lesson is part of systems of equations. It needs quadratic factorising, so revisit that if it feels slow.
What is the routine?
- Rearrange the linear equation so that y (or x) stands alone.
- Substitute into the nonlinear equation, using brackets.
- Expand and collect terms into a quadratic equal to zero.
- Solve for the first variable, usually two values.
- Find the partner of each value from the linear equation, then check both equations.
Worked example 1: a circle and a line
Solve x + y = 5 and x² + y² = 13.
From the linear equation, y = 5 − x. Substitute: x² + (5 − x)² = 13.
Expand: x² + 25 − 10x + x² = 13, so 2x² − 10x + 12 = 0, which simplifies to x² − 5x + 6 = 0 and factorises as (x − 2)(x − 3) = 0.
So x = 2 or x = 3. The linear equation gives y = 3 when x = 2 and y = 2 when x = 3. The answers are (2, 3) and (3, 2). Check (3, 2): 3 + 2 = 5 and 9 + 4 = 13.
Worked example 2: a product and a fractional root
Solve y = 2x − 1 and xy = 15.
Substitute: x(2x − 1) = 15, so 2x² − x − 15 = 0, which factorises as (2x + 5)(x − 3) = 0.
So x = 3 or x = −5/2. The partners are y = 2(3) − 1 = 5 and y = 2(−5/2) − 1 = −6. The answers are (3, 5) and (−2.5, −6).
Check the second pair: xy = (−2.5)(−6) = 15, and 2(−2.5) − (−6) = 1. The negative pair is valid, even though it looks unusual.
The mistake that costs marks
The common slip is to expand (5 − x)² as 25 − x² or as 25 + x², losing the middle term −10x. The quadratic then has the wrong coefficients and the roots are wrong.
| Step | Wrong | Right |
|---|---|---|
| Expand (5 − x)² | 25 − x² | 25 − 10x + x² |
| Collect terms | x² + 25 − x² = 13, no x² left | 2x² − 10x + 12 = 0 |
| Result | No quadratic to solve | x = 2 or 3 |
If the x² terms cancel completely, stop. A genuine quadratic should remain, so recheck the expansion.
Check yourself
Solve x − y = 2 and x² + y² = 10.
Answer
From the first equation, x = y + 2, so (y + 2)² + y² = 10, which gives 2y² + 4y − 6 = 0, or y² + 2y − 3 = 0. Then (y + 3)(y − 1) = 0, so y = 1 or y = −3.
The partners are x = 3 and x = −1. The answers are (3, 1) and (−1, −3). Check (−1, −3): −1 − (−3) = 2 and 1 + 9 = 10.
What to study next
Next, learn how to pick the quickest method in choosing substitution or elimination. Then see when answers must be rejected in rejecting invalid solutions in contextual systems.
If you want a teacher to work through these with you, see online one-to-one Additional Mathematics tuition.