A19 - Linear and quadratic simultaneous equations

A line and a curve can share two points, one point or no points. You will use substitution to find their common coordinates, keep each solution pair together and connect the algebra with a graph.

Substitution into a quadratic relationship (Higher)

Can two numbers have sum 7 and product 12? The numbers 3 and 4 work, but their order gives two pairs: $(3,4)$ and $(4,3)$ . Simultaneous equations help us find every pair systematically.

You already know

Substitution replaces a variable by an equal expression. A factorised equation such as $(x-2)(x+1)=0$ has a solution when either factor is zero.

Start with the linear equation. Make one variable the subject, then replace it everywhere in the quadratic relationship.

Simultaneous solution

A simultaneous solution is a pair $(x,y)$ satisfying both equations at the same time. Satisfying just one equation is not enough.

When both equations give $y$ , their right-hand sides must be equal at a common point.

Worked example 1: equating two expressions

Question. Solve $y=x+2$ and $y=x^2$ .

Plan. Equate the expressions for $y$ , solve the quadratic, then recover $y$ .

\begin{aligned} x^2&=x+2 &&\text{same value of }y\\ x^2-x-2&=0 &&\text{collect on one side}\\ (x-2)(x+1)&=0 &&\text{factorise}\\ x&=2\text{ or }-1 &&\text{zero-product rule}\\ y&=4\text{ or }1 &&\text{use }y=x+2 \end{aligned}

Why equate the expressions? At an intersection, they describe the same height.

Answer. $(2,4)$ and $(-1,1)$ .

Check. $2^2=4=2+2$ and $(-1)^2=1=-1+2$ .

Change only the line's constant from 2 to 6. The method stays the same; the roots move.

Worked example 2: moving the line

Question. Solve $y=x+6$ and $y=x^2$ .

Plan. Use the same substitution and factorisation route.

\begin{aligned} x^2-x-6&=0 &&\text{equate and rearrange}\\ (x-3)(x+2)&=0 &&\text{factorise}\\ x&=3\text{ or }-2 &&\text{solve each factor}\\ y&=9\text{ or }4 &&\text{add 6 to each root} \end{aligned}

Why find two $y$ -values? Each root can give a different common point.

Answer. $(3,9)$ and $(-2,4)$ .

Check. $3^2=9=3+6$ and $(-2)^2=4=-2+6$ .

The quadratic relationship need not already give $y$ . The opening puzzle has $x+y=7$ and $xy=12$ . Rearrange the line first.

Worked example 3: a product relationship

Question. Solve $x+y=7$ and $xy=12$ .

Plan. Replace $y$ by $7-x$ in the product.

\begin{aligned} y&=7-x &&\text{rearrange the line}\\ x(7-x)&=12 &&\text{substitute}\\ x^2-7x+12&=0 &&\text{expand and rearrange}\\ (x-3)(x-4)&=0 &&\text{factorise}\\ x&=3\text{ or }4 &&\text{solve} \end{aligned}

Why use the line first? It gives an expression that removes one unknown.

Answer. $(3,4)$ and $(4,3)$ .

Check. Both sums are 7 and both products are 12.

A squared replacement needs brackets around the entire expression. This also lets us solve line-and-circle systems.

Worked example 4: squaring the whole replacement

Question. Solve $x+y=3$ and $x^2+y^2=5$ .

Plan. Put $y=3-x$ into the squared equation.

\begin{aligned} x^2+(3-x)^2&=5 &&\text{replace the whole }y\\ 2x^2-6x+4&=0 &&\text{expand and collect}\\ (x-1)(x-2)&=0 &&\text{divide by 2 and factorise}\\ x&=1\text{ or }2 &&\text{solve} \end{aligned}

Why keep the brackets? The square applies to all of $3-x$ .

Answer. $(1,2)$ and $(2,1)$ .

Check. $1+2=3$ and $1^2+2^2=5$ , in either order.

Keep the replacement intact until you expand it.

Watch out

$x(7-x)=7x-x^2$ , not $7x-x$ . Also, $(3-x)^2=9-6x+x^2$ ; squaring each term alone misses the middle term.

Try the line-and-parabola method again, changing only the line's constant from Worked example 2.

Matching and checking every pair (Higher)

A root is only half a coordinate pair. Find its matching value from the original line, then check both original equations.

For $y=2x+1$ and $y=x^2-2$ , the roots are 3 and $-1$ . They give $(3,7)$ and $(-1,-1)$ . Mixing them into $(3,-1)$ fails $y=2x+1$ . Keep one row of working for each root.

When factorisation is awkward, use completing the square or the quadratic formula from quadratic equations.

Quadratic formula

For $ax^2+bx+c=0$ , where $a\ne0$ ,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

On the formula sheet (2026 version; the 2027 sheet will have the same content).

Exact roots may contain surds. Use the upper signs together and the lower signs together.

Worked example 5: pairing exact roots

Question. Solve $y=x+1$ and $y=x^2$ exactly.

Plan. Solve the reduced quadratic, then add 1 to each root.

\begin{aligned} x^2-x-1&=0 &&\text{equate the two }y\text{-values}\\ x&=\frac{1\pm\sqrt5}{2} &&\text{quadratic formula}\\ y&=\frac{3\pm\sqrt5}{2} &&\text{add 1 with matching signs} \end{aligned}

Why keep the signs together? Each $y$ -value comes from one particular $x$ .

Answer. $\left(\frac{1+\sqrt5}{2},\frac{3+\sqrt5}{2}\right)$ and $\left(\frac{1-\sqrt5}{2},\frac{3-\sqrt5}{2}\right)$ .

Check. $\left(\frac{1\pm\sqrt5}{2}\right)^2=\frac{6\pm2\sqrt5}{4}=\frac{3\pm\sqrt5}{2}=y$ . The linear equation was used to form each pair.

Change only the line's constant from 1 to 3. This changes the surd, but not the pairing rule.

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Papers 1, 2 and 3

A19 - Linear and quadratic simultaneous equations

Substitution into a quadratic relationship (Higher)

Matching and checking every pair (Higher)

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