How to Solve Simultaneous Equations Involving a Line and a Parabola
Solving simultaneous equations where a straight line meets a parabola is a key Edexcel IGCSE Maths skill. The reliable method is substitution: rearrange the line for one variable, substitute into the curve, and solve the resulting quadratic. The number of solutions tells you the geometry: the line can miss the parabola (0 points), touch it as a tangent (1 point), or cross it twice (2 points). The worked examples and auto-marked practice questions below give instant feedback.
How to Solve a Line and a Parabola Together
When a straight line \(y=mx+c\) and a parabola \(y=ax^2+bx+c\) are solved simultaneously, you are finding where they cross. The method is substitution every time.
- Equate the two expressions for \(y\): set \(mx+c = ax^2+bx+c\).
- Rearrange to a single quadratic \(ax^2+bx+c=0\) (everything on one side).
- Solve the quadratic by factorising, or with \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\).
- Find \(y\) by substituting each \(x\) back into the line \(y=mx+c\).
- Write the solutions as coordinate pairs \((x,y)\) and check in both equations.
Core ideas
A line can meet a parabola in three ways
The number of real solutions equals the number of crossing points: two, one (a tangent), or none.
Two solutions
The line cuts the curve twice.
One solution
The line is a tangent (repeated root).
No real solution
The line misses the curve.
Method 1: graphical check
Plot both equations on the same axes and read the intersection coordinates. This works exactly when the crossings sit on whole-number points.
💡 Example (graphical)
Use the graph to solve \(y=2x-4\) and \(y=x^2-4\).
Read where the line meets the curve: the crossings are at \((0,-4)\) and \((2,0)\).
So \(x=0,\ y=-4\) and \(x=2,\ y=0\).
Method 2: substitution (algebra)
💡 Example 1: factorising
Solve \(y=x+2\) and \(y=x^2+3x+2\).
So \(x=0\) or \(x=-2\). Substitute into \(y=x+2\):
\[ \begin{array}{rcl} x=0 &:& y=2 \\ x=-2 &:& y=0 \end{array} \]Solutions: \((0,2)\) and \((-2,0)\).
💡 Example 2: quadratic formula
Solve \(y=2x-1\) and \(y=x^2-4x+5\).
With \(a=1,\ b=-6,\ c=6\):
\[ \begin{array}{rcl} x &=& \dfrac{6\pm\sqrt{36-24}}{2} \\ &=& 3\pm\sqrt{3} \end{array} \]Substitute into \(y=2x-1\):
\[ \begin{array}{rcl} x=3+\sqrt{3} &:& y=5+2\sqrt{3} \\ x=3-\sqrt{3} &:& y=5-2\sqrt{3} \end{array} \]Solutions: \(\left(3+\sqrt{3},\,5+2\sqrt{3}\right)\) and \(\left(3-\sqrt{3},\,5-2\sqrt{3}\right)\).
Optional: how many solutions before you solve
After rearranging to \(ax^2+bx+c=0\), the discriminant \(\Delta=b^2-4ac\) tells you the geometry without finishing: \(\Delta>0\) two crossings; \(\Delta=0\) a tangent (one); \(\Delta<0\) no real solutions. This is exactly Room 3 in the practice below.
🔑 Key points
- Equate the line and the parabola, then rearrange to \(ax^2+bx+c=0\).
- Factorise when you can; otherwise use \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\).
- Find each \(y\) from the line \(y=mx+c\), not the parabola.
- Write solutions as ordered pairs \((x,y)\) and check both equations.
- \(\Delta=b^2-4ac\) gives the number of crossings: two, one, or none.
⚠️ Common pitfalls
- Only finding \(x\): each \(x\) needs its matching \(y\) for a full solution.
- Substituting \(y\) into the parabola (harder); use the line instead.
- Sign slips when moving terms across to form \(ax^2+bx+c=0\).
- Pairing the wrong \(x\) with the wrong \(y\): keep each pair together.