Linear and quadratic simultaneous equations

Visual model

A line can meet a curve twice

two intersections means two solutions

Gold-standard guide

26 mins

What you will learn

Solve linked equations when one relationship is curved.

Use a clear step-by-step method for linear and quadratic simultaneous equations.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier algebra skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Substitute the linear equation into the quadratic to form a single quadratic equation

Step 1

Substitute y = x + 3 into the quadratic

x + 3 = x² − 1 → x² − x − 4 = 0

Step 2

Solve the resulting quadratic

x² − x − 4 = 0 does not factorise neatly; use the formula: x = (1 ± $\sqrt{1 + 16}$ ) / 2 = (1 ± $\sqrt{17}$ ) / 2

Step 3

Find corresponding y-values using the linear equation

y = x + 3, so for each x value, add 3

Watch out

Students set the two equations equal but forget to rearrange to zero before solving

Substitute

Set the line equation equal to the quadratic equation.

Solutions

Two intersections give two coordinate pairs.

Worked example

Solve y = x + 3 and y = x² − 1 simultaneously.

Substitute y = x + 3 into the quadratic: x + 3 = x² − 1 → x² − x − 4 = 0. Rearrange so one side is zero.

Solve the resulting quadratic: x² − x − 4 = 0 does not factorise neatly; use the formula: x = (1 ± $\sqrt{1 + 16}$ ) / 2 = (1 ± $\sqrt{17}$ ) / 2.

Find corresponding y-values using the linear equation: y = x + 3, so for each x value, add 3.

State both solution pairs: x = (1 + $\sqrt{17}$ )/2, y = (7 + $\sqrt{17}$ )/2 and x = (1 − $\sqrt{17}$ )/2, y = (7 − $\sqrt{17}$ )/2.

Final answer

x = (1 ± $\sqrt{17}$ )/2, y = (7 ± $\sqrt{17}$ )/2

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

Solve y = x + 3 and y = x² − 1 simultaneously.

4 marks4 minslinear-and-quadratic-simultaneous-equations-worked

Show solution

Worked solution

1.Substitute y = x + 3 into the quadratic: x + 3 = x² − 1 → x² − x − 4 = 0. Rearrange so one side is zero.
2.Solve the resulting quadratic: x² − x − 4 = 0 does not factorise neatly; use the formula: x = (1 ± $\sqrt{1 + 16}$ ) / 2 = (1 ± $\sqrt{17}$ ) / 2.
3.Find corresponding y-values using the linear equation: y = x + 3, so for each x value, add 3.
4.State both solution pairs: x = (1 + $\sqrt{17}$ )/2, y = (7 + $\sqrt{17}$ )/2 and x = (1 − $\sqrt{17}$ )/2, y = (7 − $\sqrt{17}$ )/2.

Final answer

x = (1 ± $\sqrt{17}$ )/2, y = (7 ± $\sqrt{17}$ )/2

Mark points

M1: substitute y = x + 3 into the quadratic
M1: solve the resulting quadratic
M1: find corresponding y-values using the linear equation
M1: state both solution pairs
A1: x = (1 ± $\sqrt{17}$ )/2, y = (7 ± $\sqrt{17}$ )/2

Watch out

Students set the two equations equal but forget to rearrange to zero before solving.Writing 'x² − x − 4 = 0' explicitly before factorising or using the formula prevents this error.

Diagnosticrecall

Solve y = x and y = x² − 2.

1 mark2 minslinear-and-quadratic-simultaneous-equations-q1

Show solution

Worked solution

1.Spot the skill: Substitute the linear equation into the quadratic to form a single quadratic equation.
2.Use the substitute y = x + 3 into the quadratic stage first, then solve the resulting quadratic.
3.Keep the final answer visible: x = 2, y = 2 and x = −1, y = −1.

Final answer

x = 2, y = 2 and x = −1, y = −1

Mark points

M1: use the correct substitute the linear equation into the quadratic to form a single quadratic equation.solve by factorising or the formula. each x-value gives one point — find the matching y using the linear equation.there may be 0, 1 or 2 intersection points.
A1: x = 2, y = 2 and x = −1, y = −1

Watch out

Students set the two equations equal but forget to rearrange to zero before solving.Writing 'x² − x − 4 = 0' explicitly before factorising or using the formula prevents this error.

Easyprocedure

Solve y = 2x + 1 and y = x² − 1.

2 marks3 minslinear-and-quadratic-simultaneous-equations-q2

Show solution

Worked solution

1.Spot the skill: Substitute the linear equation into the quadratic to form a single quadratic equation.
2.Use the solve the resulting quadratic stage first, then find corresponding y-values using the linear equation.
3.Keep the final answer visible: x = 2, y = 5 and x = −1, y = −1.

Final answer

x = 2, y = 5 and x = −1, y = −1

Mark points

M1: use the correct substitute the linear equation into the quadratic to form a single quadratic equation.solve by factorising or the formula. each x-value gives one point — find the matching y using the linear equation.there may be 0, 1 or 2 intersection points.
A1: x = 2, y = 5 and x = −1, y = −1

Watch out

Students set the two equations equal but forget to rearrange to zero before solving.Writing 'x² − x − 4 = 0' explicitly before factorising or using the formula prevents this error.

Mediumreasoning

Show that y = x − 5 and y = x² do not intersect.

3 marks4 minslinear-and-quadratic-simultaneous-equations-q3

Show solution

Worked solution

1.Spot the skill: Substitute the linear equation into the quadratic to form a single quadratic equation.
2.Use the find corresponding y-values using the linear equation stage first, then state both solution pairs.
3.Keep the final answer visible: Discriminant = 1 − 20 = −19 < 0.

Final answer

Discriminant = 1 − 20 = −19 < 0

Mark points

M1: use the correct substitute the linear equation into the quadratic to form a single quadratic equation.solve by factorising or the formula. each x-value gives one point — find the matching y using the linear equation.there may be 0, 1 or 2 intersection points.
A1: Discriminant = 1 − 20 = −19 < 0

Watch out

Students set the two equations equal but forget to rearrange to zero before solving.Writing 'x² − x − 4 = 0' explicitly before factorising or using the formula prevents this error.

Hardproblem solving

Solve y = 3 and y = x² − 6.

3 marks5 minslinear-and-quadratic-simultaneous-equations-q4

Show solution

Worked solution

1.Spot the skill: Substitute the linear equation into the quadratic to form a single quadratic equation.
2.Use the state both solution pairs stage first, then substitute y = x + 3 into the quadratic.
3.Keep the final answer visible: x = 3 or x = −3, y = 3.

Final answer

x = 3 or x = −3, y = 3

Mark points

M1: use the correct substitute the linear equation into the quadratic to form a single quadratic equation.solve by factorising or the formula. each x-value gives one point — find the matching y using the linear equation.there may be 0, 1 or 2 intersection points.
A1: x = 3 or x = −3, y = 3

Watch out

Students set the two equations equal but forget to rearrange to zero before solving.Writing 'x² − x − 4 = 0' explicitly before factorising or using the formula prevents this error.

Exam-stylemulti-step

Solve x² + y² = 25 and y = x + 1 (circle and line).

4 marks6 minslinear-and-quadratic-simultaneous-equations-q5

Show solution

Worked solution

1.Spot the skill: Substitute the linear equation into the quadratic to form a single quadratic equation.
2.Use the substitute y = x + 3 into the quadratic stage first, then solve the resulting quadratic.
3.Keep the final answer visible: x = 3, y = 4 and x = −4, y = −3.

Final answer

x = 3, y = 4 and x = −4, y = −3

Mark points

M1: use the correct substitute the linear equation into the quadratic to form a single quadratic equation.solve by factorising or the formula. each x-value gives one point — find the matching y using the linear equation.there may be 0, 1 or 2 intersection points.
A1: x = 3, y = 4 and x = −4, y = −3

Watch out

Students set the two equations equal but forget to rearrange to zero before solving.Writing 'x² − x − 4 = 0' explicitly before factorising or using the formula prevents this error.

Grade 9 stretchproblem solving

The line y = x + 1 intersects the circle x² + y² = 25. Find the coordinates of both intersection points.

4 marks7 minssimultaneous-g9

Show solution

Worked solution

1.Substitute y = x + 1 into the circle equation.
2.Expand and simplify the resulting quadratic.
3.Solve for both x-values, then find each matching y-value.

Final answer

(3, 4) and (-4, -3)

Mark points

M1: form x² + (x + 1)² = 25
M1: simplify to x² + x - 12 = 0
A1: find x = 3 and x = -4
A1: state both matching coordinate pairs

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

16 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Linear and quadratic simultaneous equations - 2 marksSolve y = x and y = x² − 2.Mark answer

Answer

x = 2, y = 2 and x = −1, y = −1

2Algebraic notation and substitution - 2 marksFind t² + 4t when t = −2Mark answer

Answer

−4

3Simplifying expressions - 2 marksSimplify 2xy + 3x²y − xy + x²yMark answer

Answer

3x²y + xy

4Expanding and factorising - 3 marksFactorise fully 6x² + 9xMark answer

Answer

3x(2x + 3)

Mastery check

I can explain the method for linear and quadratic simultaneous equations.
I can show clear working without skipping key steps.
I can avoid this mistake: Students set the two equations equal but forget to rearrange to zero before solving.Writing 'x² − x − 4 = 0' explicitly before factorising or using the formula prevents this error.

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A line can meet a curve twice

What you will learn

Know the rule, then use it

Method

Substitute y = x + 3 into the quadratic

Solve the resulting quadratic

Find corresponding y-values using the linear equation

Watch out

Solve y = x + 3 and y = x2 − 1 simultaneously.

Build up to the hardest questions

Switch between skills

Get help with anything that still feels tricky.

Solve y = x + 3 and y = x² − 1 simultaneously.