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Maths · Equations and inequalities

Solving linear and quadratic simultaneous equations

Solve a linear equation and a quadratic equation together by substitution, find the points where a line meets a curve or circle, and show that a line is a tangent.

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  • 6 key terms
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Teacher resources

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Student handouts

The same files the students see, to print or hand out.

Warm-up

Answer each one, then check.

  1. 1

    Expand \((x + 1)^2\).

    Show answerHide answer

    \(x^2 + 2x + 1\)

  2. 2

    Solve \(x^2 + x - 12 = 0\).

    Show answerHide answer

    \(x = 3\) or \(x = -4\)

  3. 3

    Solve \(5x^2 = 20\).

    Show answerHide answer

    \(x = \pm 2\)

  4. 4

    What is the equation of a circle with centre the origin and radius 5?

    Show answerHide answer

    \(x^2 + y^2 = 25\)

  5. 5

    What is a tangent to a curve?

    Show answerHide answer

    A line that touches the curve at one point

Learning Objectives

  1. 1Solve a linear and a quadratic equation together by substitution.
  2. 2Solve a line with a circle \(x^2 + y^2 = r^2\).
  3. 3Find the points where a line meets a curve.
  4. 4Show that a line is a tangent by getting a repeated root.

Substitution Method

Get the linear equation into \(y = \dots\) or \(x = \dots\) first.

  1. 1 Rearrange the linear equation

    So one unknown is on its own
  2. 2 Substitute into the quadratic

    You now have a quadratic in one unknown
  3. 3 Rearrange to zero and solve

    Factorise, or use the formula
  4. 4 Find the other unknown

    Substitute each solution into the LINEAR equation
  5. 5 Write pairs

    Each solution is a pair of coordinates

A Line and a Parabola

Solve simultaneously \(y = x^2 - 2x - 3\) and \(y = x + 1\).

Show the solutionHide the solution
  1. 1 Set the y-values equal \(x^2 - 2x - 3 = x + 1\)
  2. 2 Rearrange to zero \(x^2 - 3x - 4 = 0\)
  3. 3 Factorise \((x - 4)(x + 1) = 0\), so \(x = 4\) or \(x = -1\)
  4. 4 Find y from the line \(y = x + 1\) \(x = 4\): \(y = 5\); \(x = -1\): \(y = 0\)

Answer\((4, 5)\) and \((-1, 0)\)

A Line and a Circle

Solve simultaneously \(x^2 + y^2 = 25\) and \(y = x + 1\).

Show the solutionHide the solution
  1. 1 Substitute \(y = x + 1\) \(x^2 + (x + 1)^2 = 25\)
  2. 2 Expand \(2x^2 + 2x + 1 = 25\), so \(2x^2 + 2x - 24 = 0\)
  3. 3 Divide by 2 and factorise \(x^2 + x - 12 = 0\), so \((x + 4)(x - 3) = 0\)
  4. 4 Find y \(x = 3\): \(y = 4\); \(x = -4\): \(y = -3\)

Answer\((3, 4)\) and \((-4, -3)\)

Another Circle Problem

Solve simultaneously \(x^2 + y^2 = 20\) and \(y = 2x\).

Show the solutionHide the solution
  1. 1 Substitute \(x^2 + 4x^2 = 20\)
  2. 2 Simplify \(5x^2 = 20\), so \(x^2 = 4\)
  3. 3 Solve \(x = 2\) or \(x = -2\)
  4. 4 Find y \(y = 4\) or \(y = -4\)

Answer\((2, 4)\) and \((-2, -4)\)

Showing a Line Is a Tangent

Show that the line \(y = 2x - 1\) is a tangent to the curve \(y = x^2\).

Show the solutionHide the solution
  1. 1 Set equal \(x^2 = 2x - 1\)
  2. 2 Rearrange \(x^2 - 2x + 1 = 0\)
  3. 3 Factorise \((x - 1)^2 = 0\)
  4. 4 One repeated solution The line touches the curve at one point only, \((1, 1)\)

AnswerThe equation has one repeated root \(x = 1\), so the line is a tangent.

How Many Solutions?

Two solutions

  • The line crosses the curve at two points.
  • The quadratic has two different roots.
  • The discriminant \(b^2 - 4ac\) is positive.

One or no solutions

  • One repeated root: the line is a tangent.
  • No real roots: the line misses the curve.
  • The discriminant is zero or negative.

Where Do They Meet?

Find the intersection points of each pair. (a) \(y = x^2\) and \(y = 3x - 2\) (b) \(x + y = 7\) and \(x^2 + y^2 = 25\) (c) \(y = x^2 + 1\) and \(y = 3x - 1\).

1. Substitute.

2. Solve the quadratic.

3. Find y for each x.

A good answer shows: (a) \(x^2 - 3x + 2 = 0\): \((1, 1)\) and \((2, 4)\). (b) \(y = 7 - x\): \(2x^2 - 14x + 24 = 0\), \(x^2 - 7x + 12 = 0\): \((3, 4)\) and \((4, 3)\). (c) \(x^2 - 3x + 2 = 0\): \((1, 2)\) and \((2, 5)\).

Can I...?

  1. 1Rearrange the linear equation.
  2. 2Substitute into the quadratic.
  3. 3Solve the resulting quadratic.
  4. 4Find both coordinates of each point.
  5. 5Solve a line and a circle.
  6. 6Show a line is a tangent.
  7. 7Give answers as coordinate pairs.
  8. 8Check by substituting.

Summary & Exam Focus

  • Substitute the linear equation into the quadratic or the circle.
  • Solve the quadratic in one unknown.
  • Pair up each x with its y using the linear equation.
  • A repeated root means the line is a tangent.

Exam focus

Solve the simultaneous equations \(y = x^2 - 2x - 3\) and \(y = x + 1\). (5 marks) (5 marks)

Substitute the linear equation into the quadratic. When you find each x, use the LINEAR equation to find the matching y.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Tangent
A line that touches a curve at exactly one point.
Repeated root
A solution that occurs twice, such as \(x = 1\) in \((x - 1)^2 = 0\).
Intersection
A point where a line and a curve meet.
Substitution
Replacing a letter with an expression.
Circle equation
\(x^2 + y^2 = r^2\) for a circle centred on the origin.
Discriminant
\(b^2 - 4ac\); it shows the number of real roots.

Questions and answers

12 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Non-calculator 5 marks Core

Solve the simultaneous equations \(y = x^2 - 2x - 3\) and \(y = x + 1\).

Mark scheme — 5 marks available

  • Equating — M1
  • Rearranging to zero — M1
  • Factorising — M1
  • Both x values — A1
  • Both correct pairs — A1

Model answer

\(x^2 - 2x - 3 = x + 1\), so \(x^2 - 3x - 4 = 0\) and \((x - 4)(x + 1) = 0\). \(x = 4\) or \(x = -1\). The solutions are \((4, 5)\) and \((-1, 0)\).

2. Exam question Non-calculator 5 marks Core

The circle \(x^2 + y^2 = 25\) and the line \(y = x + 1\) intersect at two points. Find the coordinates of the two points.

A circle of radius 5 centred at the origin and a straight line crossing it twice.

Mark scheme — 5 marks available

  • Substituting — M1
  • Simplifying to a quadratic — M1
  • Solving — M1
  • Both x values — A1
  • Both correct pairs — A1

Model answer

\(x^2 + (x + 1)^2 = 25\), so \(2x^2 + 2x - 24 = 0\), \(x^2 + x - 12 = 0\), \((x + 4)(x - 3) = 0\). The points are \((3, 4)\) and \((-4, -3)\).

3. Exam question Non-calculator 4 marks Easier

Solve the simultaneous equations \(x^2 + y^2 = 20\) and \(y = 2x\).

Mark scheme — 4 marks available

  • Substituting — M1
  • \(5x^2 = 20\) — M1
  • \(x = 2\) and \(x = -2\) — A1
  • Both pairs — A1

Model answer

\(x^2 + 4x^2 = 20\), so \(x^2 = 4\), \(x = \pm 2\). The solutions are \((2, 4)\) and \((-2, -4)\).

4. Exam question Show that 4 marks Easier

Show that the line \(y = 2x - 1\) is a tangent to the curve \(y = x^2\).

Mark scheme — 4 marks available

  • Equating — M1
  • Rearranging to \(x^2 - 2x + 1 = 0\) — M1
  • \((x - 1)^2 = 0\) — M1
  • Conclusion — C1

Model answer

\(x^2 = 2x - 1\), so \(x^2 - 2x + 1 = 0\) and \((x - 1)^2 = 0\). There is only one solution, \(x = 1\), so the line touches the curve at one point and is a tangent.

5. Exam question Non-calculator 5 marks Core

Solve the simultaneous equations \(x + y = 7\) and \(x^2 + y^2 = 25\).

Mark scheme — 5 marks available

  • Rearranging and substituting — M1
  • Expanding \((7 - x)^2\) — M1
  • Solving — M1
  • Both x values — A1
  • Both correct pairs — A1

Model answer

\(y = 7 - x\), so \(x^2 + (7 - x)^2 = 25\), \(2x^2 - 14x + 24 = 0\), \(x^2 - 7x + 12 = 0\), \((x - 3)(x - 4) = 0\). The solutions are \((3, 4)\) and \((4, 3)\).

6. Exam question Non-calculator 5 marks Core

Solve the simultaneous equations \(y = x^2 + 1\) and \(y = 3x - 1\).

Mark scheme — 5 marks available

  • Equating — M1
  • Rearranging to zero — M1
  • Factorising — M1
  • Both x values — A1
  • Both correct pairs — A1

Model answer

\(x^2 + 1 = 3x - 1\), so \(x^2 - 3x + 2 = 0\) and \((x - 1)(x - 2) = 0\). The solutions are \((1, 2)\) and \((2, 5)\).

7. Multiple choice 1 mark Easier

To solve a line and a quadratic together, first...

  1. A Substitute one equation into the other Correct
  2. B Add the equations
  3. C Multiply the equations
  4. D Draw a graph only

Why: Substitute the linear equation into the quadratic.

8. Multiple choice 1 mark Core

After finding x for a line and curve, you find y using...

  1. A The quadratic
  2. B The circle
  3. C The linear equation Correct
  4. D A guess

Why: The linear equation is simpler and avoids extra pairs.

9. Multiple choice 1 mark Core

The line meets a circle where \(x^2 + (x+1)^2 = 25\). What type of equation is this?

  1. A Linear
  2. B Quadratic Correct
  3. C Cubic
  4. D Reciprocal

Why: It simplifies to a quadratic in x.

10. Multiple choice 1 mark Core

A line and a curve meet where \((x - 3)^2 = 0\). What does this show?

  1. A No intersection
  2. B Two intersections
  3. C The line is perpendicular
  4. D The line is a tangent Correct

Why: A repeated root means one point of contact: the line is a tangent.

11. Multiple choice 1 mark Core

How many pairs of coordinates does a line and a circle usually give?

  1. A One
  2. B Two Correct
  3. C Three
  4. D Four

Why: Two intersection points, from two roots.

12. Multiple choice 1 mark Stretch

Solve \(y = x^2\) and \(y = 4\).

  1. A \((2, 4)\) and \((-2, 4)\) Correct
  2. B \((4, 4)\) only
  3. C \((2, 4)\) only
  4. D \((16, 4)\) and \((-16, 4)\)

Why: \(x^2 = 4\), so \(x = \pm 2\): the points \((2, 4)\) and \((-2, 4)\).