To solve a linear and quadratic pair, substitute the linear equation into the quadratic one and rearrange to make it equal zero. Solve it, then find each matching y from the linear equation. The answers are coordinates, so they come in pairs.
Check the current Cambridge syllabus page for your exam year to confirm whether this sits in your route. The substitution habit comes from choosing substitution when one variable is isolated, and it applies directly here.
Why does the answer have two points?
A straight line meets a parabola in zero, one or two points. Setting the two y-expressions equal finds the x-values where the heights match.
Because the equation you get is quadratic, it may have two roots. Each root is an x-coordinate of a crossing, and the line tells you the height there.
The method, step by step
- Make both equations “y = …” or rearrange one so it is.
- Set the expressions equal to each other.
- Move everything to one side so the equation reads ax² + bx + c = 0.
- Solve by factorising, or by the formula if factorising fails.
- Substitute each x into the linear equation to find y.
- Write coordinate pairs and check each in the quadratic.
Worked example
Solve the pair: y = x² − 4x + 3 and y = x − 1
Step 1, set equal: x² − 4x + 3 = x − 1.
Step 2, rearrange: subtract x and add 1 to both sides: x² − 5x + 4 = 0.
Step 3, factorise: (x − 1)(x − 4) = 0, so x = 1 or x = 4.
Step 4, find y using the line: when x = 1, y = 1 − 1 = 0. When x = 4, y = 4 − 1 = 3.
Step 5, check in the curve: at x = 1, 1 − 4 + 3 = 0, which matches. At x = 4, 16 − 16 + 3 = 3, which matches.
Answer: (1, 0) and (4, 3).
The mistake to watch for
The common slip is pairing values by order instead of by substitution, or reporting only the x-values.
Mistaken answer: “x = 1 and x = 4, so the points are (1, 3) and (4, 0).”
The student paired each x with the wrong y. The point (1, 3) is not on the line, since 1 − 1 = 0.
The correction is always to substitute each x separately into the linear equation and write the pair straight away. A second slip is making a sign error when moving terms across the equals sign, so it helps to write each move beside the line, for example “−x, +1”.
Check yourself
Try these on paper, then open each answer.
1. Solve y = 2x and y = x² − 3.
Show answer
Set equal: x² − 3 = 2x, so x² − 2x − 3 = 0. Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1. Using y = 2x gives y = 6 and y = −2. Check (3, 6): 9 − 3 = 6, which matches. Check (−1, −2): 1 − 3 = −2, which matches.
(3, 6) and (−1, −2)
2. Solve y = x + 6 and y = x².
Show answer
Set equal: x² = x + 6, so x² − x − 6 = 0. Factorise: (x − 3)(x + 2) = 0, so x = 3 or x = −2. Using y = x + 6 gives y = 9 and y = 4. Check (−2, 4): (−2)² = 4, which matches.
(3, 9) and (−2, 4)
3. Solve y = 2x − 1 and y = x². What does the result tell you about the line and the curve?
Show answer
Set equal: x² = 2x − 1, so x² − 2x + 1 = 0. This factorises as (x − 1)² = 0, so x = 1 is a repeated root. Then y = 2(1) − 1 = 1. Check: 1² = 1, which matches.
One solution, (1, 1). The line just touches the curve at one point, which means it is a tangent.
Where this leads next
After this, detecting inconsistent conditions shows what happens when the equations have no shared solution. The quadratic structure explorer lets you see a parabola and a line together. Try the simultaneous relationships practice set to mix all the cases.
For many students the algebra is fine but marks slip on signs and on pairing each x with its y. In online one-to-one Mathematics tuition, a teacher can watch your written working and catch those slips as they happen.