Understand the method
If \(y=f(x)\) and \(y=g(x)\), their common points satisfy \(f(x)=g(x)\). If the linear equation is not already written for y, rearrange it first.
The number of real roots of the substituted quadratic gives the number of intersection points. A repeated root gives one point; a negative discriminant gives no real intersection. Check restrictions if a denominator is introduced.
Three worked examples
Example 1
Solve \(y=x+2\) and \(y=x^2-4\)
Equate the expressions for y: \(x+2=x^2-4\).
- \[x^2-x-6=(x-3)(x+2)=0\]
For \(x=3\), \(y=5\). For \(x=-2\), \(y=0\). The solutions are \((3,5)\) and \((-2,0)\).
Substitution checks both points in the original curve and line.
Example 2
Solve \(x+2y=5\) and \(xy=2\)
From the linear equation, \(y=(5-x)/2\).
- \[\frac{x(5-x)}2=2\]
- \[x^2-5x+4=(x-1)(x-4)=0\]
For \(x=1\), \(y=2\). For \(x=4\), \(y=1/2\). So the pairs are \((1,2)\) and \((4,1/2)\).
Do not round a fractional coordinate before checking the pair.
Example 3
Solve \(y=x^2+1\) and \(y=2x\)
- \[x^2-2x+1=(x-1)^2=0\]
There is one repeated root, \(x=1\). Back-substitution gives \(y=2\).
The only point is \((1,2)\). The line is tangent to the curve there.
A repeated root does not create two different intersection points.
Apply and explain
Recognise when there are no real intersections
Replacing the curve in Example 3 with \(y=x^2+2\) gives \(x^2-2x+2=0\), whose discriminant is -4. There is no real intersection.
Exam-style check: what makes a complete solution?
Show the substitution and resulting quadratic, find every real root, recover the matching coordinates and verify the pairs in both original equations. Listing only x-values is incomplete.
These study examples explain the reasoning; they are not an official mark scheme.
Common mistakes to catch
- Stopping after finding x
- The question asks for simultaneous pairs, so find every corresponding y.
- Mixing coordinates from different roots
- Keep a separate substitution line for each x-value.
- Losing a solution by division
- Dividing by x or y can discard a valid zero value; substitution without division avoids that loss.
Now try it yourself
Put the method into practice
Return later and solve \(y=x-2\), \(y=x^2-4\). Check each point in both equations.