Solving simultaneous equations by elimination and substitution, including one linear and one quadratic.
Imagine you walk into a shop and buy 3 coffees and 2 teas for £11.50. Your friend buys 1 coffee and 4 teas for £10.00. Can you work out the price of each drink?
You have two unknowns and two pieces of information — that’s exactly when simultaneous equations come in. At GCSE you solved pairs of linear equations. At A-Level, the new challenge is solving a linear equation alongside a quadratic — and understanding what the solutions mean graphically.
You already know two methods for a pair of linear equations:
Both methods work for linear pairs. At A-Level, you’ll occasionally need these as a first step inside a bigger problem, so keep them sharp.
Solve simultaneously: 3x+2y=12 and x−2y=4
Add the two equations to eliminate y:
3x+2y+x−2y=12+4
4x=16⟹x=4
Substitute into the second equation:
4−2y=4⟹y=0
Solution: (x,y)=(4,0)
Answer at least 3 of 4 correctly to complete this section.
When one equation is quadratic (e.g. x2+y2=25, or y=x2−3x+1), elimination won’t work. You must use substitution:
Why substitution? Elimination relies on matching coefficients of a variable across two equations. When one equation has x2 and the other has x, there is no way to eliminate — the powers don’t match.
Use this interactive tool to see the substitution method applied to different types of problem. Click through each step:
Write down the two equations
Solve simultaneously: y=x+3 and y=x2+1
Since both equations are written as y=…, set them equal:
x+3=x2+1
Rearrange to 0:
x2−x−2=0
Factorise:
(x−2)(x+1)=0
x=2orx=−1
Substitute each x-value into the linear equation y=x+3:
Solutions: (x,y)=(2,5) or (−1,2)
Note: Always substitute back into the linear equation, not the quadratic. It’s quicker and you’re less likely to make errors.
Find the points of intersection of x+y=5 and x2+y2=25
Rearrange the linear equation: y=5−x
Substitute into the circle equation:
x2+(5−x)2=25
x2+25−10x+x2=25
2x2−10x=0
2x(x−5)=0
x=0orx=5
Substitute back into y=5−x:
Solutions: (x,y)=(0,5) or (5,0)
Graphically, the line x+y=5 meets the circle x2+y2=25 at these two points on the axes.
Answer at least 3 of 4 correctly to complete this section.
Solving simultaneous equations is the same as finding intersection points of two graphs. A line can intersect a quadratic curve in 0, 1, or 2 points:
After substituting the linear equation into the quadratic, you get a single quadratic in one variable. The discriminant b2−4ac tells you which case you’re in:
b2−4ac>0⟹2 solutionsb2−4ac=0⟹1 solution (tangent)b2−4ac<0⟹no solutions
This connects directly to the quadratics and discriminant work you’ve already done.
Determine the number of points of intersection of y=2x−1 and y=x2+x+3.
Set the equations equal:
2x−1=x2+x+3
Rearrange to zero:
x2−x+4=0
Find the discriminant:
b2−4ac=(−1)2−4(1)(4)=1−16=−15
Since b2−4ac<0, the quadratic has no real roots. The line and curve do not intersect.
Drag the sliders to move the line and the parabola. Watch the discriminant change and see the intersection points appear and disappear:
Answer at least 3 of 4 correctly to complete this section.
Exam questions on this topic often ask you to use the discriminant to prove something about the number of solutions. Here are two classic types.
Show that the line y=x+6 does not intersect the curve y=x2+3x+8.
Set the equations equal:
x+6=x2+3x+8
Rearrange:
x2+2x+2=0
Find the discriminant:
b2−4ac=(2)2−4(1)(2)=4−8=−4
Since b2−4ac<0, the quadratic has no real roots, so the line and curve do not intersect.
The line y=2x+k is tangent to the curve y=x2+4x+1. Find the value of k.
Set equal:
2x+k=x2+4x+1
Rearrange:
x2+2x+(1−k)=0
For tangency (exactly one solution), set the discriminant equal to zero:
b2−4ac=0
(2)2−4(1)(1−k)=0
4−4+4k=0
4k=0⟹k=0
So the tangent line is y=2x. You can check: substituting gives x2+2x+1=0, i.e. (x+1)2=0, which has exactly one solution x=−1, confirming tangency.
Answer at least 3 of 3 correctly to complete this section.
| Equation type | Method | Possible number of solutions |
|---|---|---|
| Linear + Linear | Elimination or substitution | 0 (parallel lines) or 1 |
| Linear + Quadratic | Substitution only | 0, 1 (tangent), or 2 |
The discriminant of the resulting quadratic tells you how many solutions to expect — and exam questions regularly ask you to use it to prove a line misses or is tangent to a curve.
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.