2.7 - Linear and quadratic simultaneous equations

Find every pair of values that satisfies a line and a quadratic equation together. Learn how substitution finds exact intersections, why brackets matter when squaring, and how to recognise one solution or no real solutions.

Substitution and solution pairs (Higher)

Can a straight line meet the same parabola twice, once or not at all? Start with $y=x+2$ and $y=x^2$ . At $x=2$ , both give $y=4$ . At $x=-1$ , both give $y=1$ . There are already two shared points to explain.

You already know

Substitution replaces a variable by an equal expression. To solve a factorised quadratic equal to zero, set each factor equal to zero. An ordered pair $(x,y)$ gives the two coordinates in that order.

A shared point must use the same values in both equations. This is stronger than finding one point on each graph separately.

Simultaneous solution

A pair of values satisfying both equations at the same time. For $y=x+2$ and $y=x^2$ , $(2,4)$ is a solution; $(2,1)$ is not, because it fails both equations.

When both equations are written as $y=\dots$ , equate their right-hand sides. This removes $y$ and leaves one quadratic to solve. Then each root needs its own matching $y$ -value. The substitution method is not in the Higher formulae sheet: learn it. The quadratic formula is in that sheet if factorisation is inconvenient.

Worked example 1: finding both shared points

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

Plan. Equate the expressions for $y$ , factorise, then find the matching coordinate for each root.

\begin{aligned} x^2&=x+2 &&\text{same }y\\ x^2-x-2&=0 &&\text{subtract }x+2\\ (x-2)(x+1)&=0 &&\text{factorise}\\ x&=2\quad\text{or}\quad x=-1 &&\text{zero product}\\ x=2&:\quad y=2+2=4 &&\text{substitute in the line}\\ x=-1&:\quad y=-1+2=1 &&\text{use the other root} \end{aligned}

Why equate the expressions? At a shared point, both expressions name the same height.

Answer. $(x,y)=(2,4)$ or $(-1,1)$ .

Check. $4=2+2=2^2$ and $1=-1+2=(-1)^2$ , so each pair satisfies both originals.

The diagram turns the algebraic answers into intersections. A sketch shows the possibilities; algebra establishes exact coordinates.

[DIAGRAM: asset_slug: m038-intersections; caption: The parabola y equals x squared meets three different lines twice, once and not at all; description: Three exact coordinate panels show y equals x squared with y equals x plus 2 intersecting at minus 1 comma 1 and 2 comma 4, y equals 2x minus 1 touching at 1 comma 1, and y equals 2x minus 3 with no real intersection; file: m038-intersections.png; recommended_method: matplotlib]
The parabola y equals x squared meets three different lines twice, once and not at all

Now change only the line's constant term. The parabola and the substitution decision stay the same.

Worked example 2: changing the line's height

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

Plan. Equate, factorise and pair each root with its line value.

\begin{aligned} x^2&=x+6 &&\text{equate }y\text{-values}\\ x^2-x-6&=0 &&\text{collect terms}\\ (x-3)(x+2)&=0 &&\text{factorise}\\ x&=3\quad\text{or}\quad x=-2 &&\text{solve the factors}\\ x=3&:\quad y=9 &&\text{use }y=x+6\\ x=-2&:\quad y=4 &&\text{use }y=x+6 \end{aligned}

Why calculate two heights separately? Each root identifies a different shared point.

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

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

Keep the pairing visible even when you have found all four coordinate values.

Watch out

The line “ $x=2,-1$ , $y=4,1$ , so $(2,1)$ is a solution” mismatches the coordinates. $1\ne2+2$ . The correct pairing is $(2,4)$ and $(-1,1)$ .

Try another change to the line's constant. Give complete ordered pairs.

Substituting into squared terms (Higher)

When a variable is squared, replace it by the whole linear expression inside brackets. For $y=3-x$ , the expression $y^2$ becomes $(3-x)^2$ . It does not become $9-x^2$ .

Square of a difference

(a-b)^2=a^2-2ab+b^2

The middle term comes from both cross-products in $(a-b)(a-b)$ . Not in the Higher formulae sheet: learn it.

For example, $(3-x)^2=9-6x+x^2$ . Preserve the original equations so that the final check can detect an expansion error.

Worked example 3: a squared coordinate

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

Plan. Write $y=3-x$ , substitute into the square, then solve and pair the roots.

\begin{aligned} x^2+(3-x)^2&=5 &&\text{substitute the whole expression}\\ x^2+9-6x+x^2&=5 &&\text{expand the square}\\ 2x^2-6x+4&=0 &&\text{collect terms}\\ x^2-3x+2&=0 &&\text{divide by }2\\ (x-1)(x-2)&=0 &&\text{factorise}\\ x&=1\quad\text{or}\quad x=2 &&\text{zero product} \end{aligned}

Why use brackets? The square applies to every term replacing $y$ .

Answer. From $y=3-x$ , the solutions are $(1,2)$ and $(2,1)$ .

Check. Both sums are $3$ ; both sums of squares are $1^2+2^2=5$ .

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Number

Algebra

Shape, space and measure

Handling data

2.7 - Linear and quadratic simultaneous equations

Substitution and solution pairs (Higher)

Substituting into squared terms (Higher)

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