How Do I Solve Quadratic Simultaneous Equations?
Learn how to solve quadratic simultaneous equations using substitution, factorising and checking both solutions, with a clear GCSE Higher worked example.
📋 Spec coverage: GCSE AQA Mathematics (8300) This article covers:
AQA A19(Higher, Algebra): solving simultaneous equations where one equation is linear and the other is quadratic.Completeness check: Algebraic substitution, both possible solutions and checking are covered.
Solve quadratic simultaneous equations by substituting the linear equation into the quadratic equation. Solve the resulting quadratic, then substitute each value back to find the corresponding second value.
What are quadratic simultaneous equations?
Simultaneous equations are equations that must be true at the same time. In this GCSE topic, one equation is linear and the other is quadratic, meaning it contains a squared term such as x2x^2x2.
There are often two pairs of solutions because a straight line can cross a quadratic graph at two points. The coordinates of each intersection satisfy both equations, so every answer is an ordered pair.
Why does substitution work?
Substitution replaces one variable with an equal expression. This leaves an equation containing only xxx, which can be solved. If the quadratic does not factorise, use another suitable method, such as the quadratic formula. Keep exact values until the end to avoid rounding errors.
A reliable method
First, rearrange the linear equation if necessary so one variable is isolated. Next, substitute that expression into the quadratic equation and rearrange it to equal zero. Solve the quadratic carefully, then use the linear equation to find each partner value. Finally, check every ordered pair in both original equations.
Worked example
Solve:
y=x+1y=x2−5\begin{aligned} y &= x+1 \\ y &= x^2-5 \end{aligned}yy=x+1=x2−5Both expressions equal yyy, so set them equal:
x+1=x2−5x+1=x^2-5x+1=x2−5Rearrange and factorise:
x2−x−6=0x^2-x-6=0x2−x−6=0 (x−3)(x+2)=0(x-3)(x+2)=0(x−3)(x+2)=0Therefore:
x=3orx=−2x=3 \quad \text{or} \quad x=-2x=3orx=−2Substitute into y=x+1y=x+1y=x+1:
x=3⇒y=4x=3 \Rightarrow y=4x=3⇒y=4 x=−2⇒y=−1x=-2 \Rightarrow y=-1x=−2⇒y=−1The solutions are:
(x,y)=(3,4)and(−2,−1)(x,y)=(3,4) \quad \text{and} \quad (-2,-1)(x,y)=(3,4)and(−2,−1)Check both equations. For (3,4)(3,4)(3,4), 4=3+14=3+14=3+1 and 4=32−54=3^2-54=32−5, so the pair is correct. Check (−2,−1)(-2,-1)(−2,−1) similarly.
Common exam mistake
Do not stop after finding xxx. Each xxx-value has a matching yyy-value, so write two coordinate pairs. Take care not to match an xxx-value with the wrong yyy-value.
