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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.

  • Higher
  • 6 key terms
  • All boards
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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.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 5 marks

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

    Show answerHide answer

    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 \((-1, 0)\).

    Mark scheme

    • Equating — M1
    • Rearranging to zero — M1
    • Factorising — M1
    • Both x values — A1
    • Both correct pairs — A1
  2. Question 2 Non-calculator 5 marks

    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.
    Show answerHide answer

    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 \((-4, -3)\).

    Mark scheme

    • Substituting — M1
    • Simplifying to a quadratic — M1
    • Solving — M1
    • Both x values — A1
    • Both correct pairs — A1
  3. Question 3 Non-calculator 4 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • Substituting — M1
    • \(5x^2 = 20\) — M1
    • \(x = 2\) and \(x = -2\) — A1
    • Both pairs — A1
  4. Question 4 Show that 4 marks

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

    Show answerHide answer

    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.

    Mark scheme

    • Equating — M1
    • Rearranging to \(x^2 - 2x + 1 = 0\) — M1
    • \((x - 1)^2 = 0\) — M1
    • Conclusion — C1
  5. Question 5 Non-calculator 5 marks

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

    Show answerHide answer

    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 \((4, 3)\).

    Mark scheme

    • Rearranging and substituting — M1
    • Expanding \((7 - x)^2\) — M1
    • Solving — M1
    • Both x values — A1
    • Both correct pairs — A1
  6. Question 6 Non-calculator 5 marks

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

    Show answerHide answer

    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 \((2, 5)\).

    Mark scheme

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

Quick check

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

    1. ASubstitute one equation into the other
    2. BAdd the equations
    3. CMultiply the equations
    4. DDraw a graph only
    Show answerHide answer

    A: Substitute one equation into the other

    Substitute the linear equation into the quadratic.

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

    1. AThe quadratic
    2. BThe circle
    3. CThe linear equation
    4. DA guess
    Show answerHide answer

    C: The linear equation

    The linear equation is simpler and avoids extra pairs.

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

    1. ALinear
    2. BQuadratic
    3. CCubic
    4. DReciprocal
    Show answerHide answer

    B: Quadratic

    It simplifies to a quadratic in x.

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

    1. ANo intersection
    2. BTwo intersections
    3. CThe line is perpendicular
    4. DThe line is a tangent
    Show answerHide answer

    D: The line is a tangent

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

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

    1. AOne
    2. BTwo
    3. CThree
    4. DFour
    Show answerHide answer

    B: Two

    Two intersection points, from two roots.

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

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

    A: \((2, 4)\) and \((-2, 4)\)

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

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