Solving Simultaneous Equations (Linear & Quadratic)
How do you find the exact point where a straight road crosses a curved river on a map? In maths, we find where a linear and a quadratic equation intersect. This guide covers the substitution method for solving these higher-tier problems.
What are the Possible Solutions?
When a straight line meets a parabola, there are three possible scenarios. The number of solutions equals the number of intersection points.
Two Solutions
The line cuts through the curve at two distinct points.
One Solution
The line is a tangent, touching the curve at a single point.
No Real Solutions
The line and the curve never meet — no real solutions exist.
The Substitution Method: Step by Step
The goal is to combine both equations into one new quadratic equation with a single variable.
- Rearrange the linear equation to make one variable the subject (e.g., $y = \ldots$).
- Substitute this expression into the quadratic equation in place of that variable.
- Rearrange into the standard form $ax^2 + bx + c = 0$.
- Solve the quadratic by factorising, the quadratic formula, or completing the square.
- Substitute each $x$ value back into the linear equation to find the corresponding $y$ values.
- Write your answers as coordinate pairs $(x, y)$.
Worked Example
Solve simultaneously: $y = x + 1$ and $y = x^2 – 3x + 4$.
- The linear equation is already rearranged: $y = x + 1$.
- Substitute $(x+1)$ for $y$ in the quadratic:
$x + 1 = x^2 – 3x + 4$ - Rearrange to $= 0$:
$0 = x^2 – 4x + 3$ - Factorise:
$(x-1)(x-3) = 0$
So $x = 1$ or $x = 3$. - Find $y$ values using $y = x + 1$:
If $x=1$: $y = 2$. If $x=3$: $y = 4$.
Solutions: $(1,\ 2)$ and $(3,\ 4)$. These are the coordinates where the two graphs intersect.
Tutor Insights
🤔 Common Misunderstandings
- Stopping after finding $x$. The solution is a coordinate pair — you must substitute each $x$ value back into the linear equation to find the corresponding $y$.
- Expansion errors. Mistakes when expanding brackets after substituting, especially with negative signs or terms like $(x+3)^2$.
📝 Common Exam Mistakes
- Sign errors when rearranging to $ax^2 + bx + c = 0$.
- Errors in solving the quadratic — double-check your factorisation by expanding.
- Only giving one solution pair when there are two.
- Substituting back into the quadratic instead of the linear equation — using the linear is quicker and less error-prone.
Practice Questions
- Solve simultaneously: $y = x + 2$ and $y = x^2 – 4$.
- Solve simultaneously: $x + y = 7$ and $y = x^2 – 2x + 7$.
- Solve simultaneously: $x – y = 5$ and $x^2 + y^2 = 25$.
Show Answers
- Working: $x+2=x^2-4 \implies x^2-x-6=0 \implies (x-3)(x+2)=0$.
$x=3 \Rightarrow y=5$; $x=-2 \Rightarrow y=0$.
Answers: $(3,\ 5)$ and $(-2,\ 0)$. - Working: $y=7-x$. Substitute: $7-x=x^2-2x+7 \implies x^2-x=0 \implies x(x-1)=0$.
$x=0 \Rightarrow y=7$; $x=1 \Rightarrow y=6$.
Answers: $(0,\ 7)$ and $(1,\ 6)$. - Working: $x=y+5$. Substitute: $(y+5)^2+y^2=25 \implies y^2+10y+25+y^2=25 \implies 2y^2+10y=0 \implies 2y(y+5)=0$.
$y=0 \Rightarrow x=5$; $y=-5 \Rightarrow x=0$.
Answers: $(5,\ 0)$ and $(0,\ -5)$.
Need Help with Simultaneous Equations?
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