Simultaneous Equations: One Linear, One Quadratic (Grade 11)

Solving a linear equation together with a quadratic one by substitution, why there are usually two solution pairs, and what those pairs mean on a graph.

In Grade 10 both equations were linear and there was one answer. In Grade 11 one of them is quadratic, so there are usually two pairs of answers — the two points where a line cuts a curve.

1The method: always substitute

Make x or y the subject of the LINEAR equation — never the quadratic one.
Substitute that into the quadratic equation.
Solve the resulting quadratic (it will have up to two answers).
Substitute each answer back into the linear equation to find its partner.
Write the answers as pairs.

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Rearranging the linear equation is always easier and never produces a square root. If you rearrange the quadratic instead you make far more work for yourself.

2Worked example

Worked ExampleSolve simultaneously: y = x + 1 and y = x² − 3x + 4

1
The linear equation already has y as the subject: y = x + 1.
Step 1 done.
2
Substitute: x + 1 = x² − 3x + 4.
Replace y in the quadratic.
3
0 = x² − 4x + 3.
Move everything to one side.
4
(x − 1)(x − 3) = 0, so x = 1 or x = 3.
Factorise.
5
If x = 1: y = 1 + 1 = 2. If x = 3: y = 3 + 1 = 4.
Substitute into the LINEAR equation — it is quicker and cannot go wrong.
6
(1 ; 2) and (3 ; 4).
Check (3 ; 4) in the quadratic: 9 − 9 + 4 = 4. ✓

⚠️

Do not stop at x = 1 and x = 3. Each x must be paired with its own y. Answers left as 'x = 1 or 3' throw away half the marks.

3When the equations are not already arranged

Worked ExampleSolve simultaneously: x + y = 5 and x² + y² = 17

1
From the linear: y = 5 − x.
Make y the subject.
2
x² + (5 − x)² = 17.
Substitute — keep the bracket.
3
x² + 25 − 10x + x² = 17.
(5 − x)² = 25 − 10x + x², not 25 + x².
4
2x² − 10x + 8 = 0 → x² − 5x + 4 = 0.
Divide through by 2.
5
(x − 1)(x − 4) = 0, so x = 1 or x = 4.
6
(1 ; 4) and (4 ; 1).
Both check: 1 + 16 = 17 and 16 + 1 = 17. ✓

4What the answers mean

The solution pairs are the points of intersection of the two graphs. Two pairs means the line cuts the curve twice; one pair means it is a tangent; no real solution means they never meet.

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If the quadratic you end up with has a negative discriminant, say so in words: 'no real solutions, so the graphs do not intersect'. That sentence earns the mark.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Solve simultaneously: y = x + 1 and y = x² − 3x + 4
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x² − 4x + 3 = 0 → x = 1 or 3 → (1 ; 2) and (3 ; 4)
2
Solve simultaneously: y = 2x and y = x² − 3
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x² − 2x − 3 = 0 → (x−3)(x+1) = 0 → (3 ; 6) and (−1 ; −2)
3
Solve simultaneously: x + y = 5 and x² + y² = 17
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(1 ; 4) and (4 ; 1)
4
Solve simultaneously: y − x = 2 and xy = 8
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x(x+2) = 8 → x² + 2x − 8 = 0 → (x+4)(x−2) = 0 → (2 ; 4) and (−4 ; −2)
5
Which equation should you rearrange first, and why?
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The linear one — making a variable the subject of a linear equation is simple and never introduces a square root.
6
Solve simultaneously: y = x − 1 and y = x² − 5x + 7
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x² − 6x + 8 = 0 → (x−2)(x−4) = 0 → (2 ; 1) and (4 ; 3)
7
The solutions of a simultaneous system are the … of the two graphs.
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points of intersection
8
What does it mean if the resulting quadratic has a negative discriminant?
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There are no real solutions, so the two graphs do not intersect.