How do you solve a linear and quadratic simultaneous equation?

Make one variable the subject of the linear equation, substitute into the quadratic, solve, then substitute back.

How do you know if a line is a tangent to a curve?

Substitute the line into the curve. A tangent gives a quadratic with a repeated root, so its discriminant is zero.

Do I always need to solve for x and y?

No. Sometimes a symmetric expression can be found directly from the given sums and products.

TMUA Simultaneous Equations

Simultaneous equations ask for values that satisfy two equations at once, usually one linear and one quadratic. Geometrically you are finding where a line meets a curve.

Paper 1 and Paper 2 topic. Specification codes: MM1.4, M4.15, MM8.7. Last checked 12 November 2026.

Substitution

Rearrange the linear equation to make one variable the subject, substitute into the other equation, then solve the resulting quadratic. Substitute back to pair up each $x$ with its $y$ .

How many solutions?

After substitution, the discriminant of the quadratic tells you how many intersections there are: two, one (the line is a tangent) or none.

Look for shortcuts

Symmetric expressions such as $x^2+y^2=(x-y)^2+2xy$ can give an answer without solving for $x$ and $y$ at all.

Setting up from words

Some questions describe a situation and leave you to create the equations. Define your variables, write one equation per fact, then solve. Check each pair in both original equations and reject any that break a condition, such as a length that must be positive.

Worked examples

Example 1

Solve $2x+y=7$ and $xy=6$ .

Show the worked solution

Put $y=7-2x$ into $xy=6$ : $x(7-2x)=6$ , so $2x^2-7x+6=0$ , which factorises as $(2x-3)(x-2)=0$ . Then $x=\tfrac32,\ y=4$ or $x=2,\ y=3$ .

Answer: $(\tfrac32,4)$ and $(2,3)$

Example 2

The line $y=2x+c$ is a tangent to $y=x^2-4x+10$ . Find $c$ .

Show the worked solution

Set equal: $x^2-6x+(10-c)=0$ . A tangent means a repeated root, so $36-4(10-c)=0$ , giving $4c=4$ and $c=1$ .

Answer: $c=1$

Example 3

If $x-y=1$ and $xy=12$ , find $x^2+y^2$ .

Show the worked solution

$x^2+y^2=(x-y)^2+2xy=1+24=25$ . No need to solve for $x$ and $y$ .

Answer: 25

Common mistakes

Finding the $x$ -values but forgetting to find the matching $y$ -values.
Pairing the wrong $x$ and $y$ by substituting into the quadratic instead of the linear equation.
Dividing by a variable and losing a solution.
Treating a tangent condition as 'discriminant greater than zero'.

Test yourself on this topic with our free practice questions.

Quick answers

Practise it under timed conditions

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TMUA Simultaneous Equations

Substitution

How many solutions?

Look for shortcuts

Setting up from words

Worked examples

Common mistakes

Quick answers

Practise it under timed conditions

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