Sometimes tested MM1.4

Linear and Quadratic Simultaneous Equations

This topic covers finding the common solutions to a pair of equations, specifically a linear and a quadratic one. This is equivalent to finding the coordinate(s) where the graphs of a straight line and a parabola intersect.

Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.

Key points

  • To solve a system of one linear and one quadratic equation, use the substitution method.
  • First, rearrange the linear equation to make either x or y the subject.
  • Substitute this expression into the quadratic equation. This will result in a new quadratic equation in a single variable.
  • Solve the resulting quadratic to find the value(s) for one variable (e.g., x).
  • Substitute these value(s) back into the rearranged linear equation to find the corresponding values of the other variable (e.g., y).
  • The number of solutions (0, 1, or 2) corresponds to the number of times the line and the parabola intersect.

Diagram

GraphGraph with axes x and y. solutionsolutionxy
A line and a curve meet where their equations are both satisfied: each intersection point is a solution of the simultaneous equations.

Formulae

b2 - 4ac

On the quadratic equation formed after substitution (ax2+bx+c=0) to determine the number of intersection points. If > 0, there are two distinct points. If = 0, the line is tangent (one point). If < 0, there are no real points of intersection.

Definitions

Simultaneous Equations
A set of two or more equations containing two or more variables, where the goal is to find the variable values that satisfy all equations in the set at the same time.
Substitution Method
An algebraic technique for solving simultaneous equations by rearranging one equation to isolate a variable and then substituting that expression into the other equation.

Worked example

Find the coordinates of the points of intersection between the line y - x = 2 and the curve y = x2 - 4.

  1. 1

    Rearrange the linear equation to make y the subject:

    y = x + 2
  2. 2

    Substitute this expression for y into the quadratic equation:

    x + 2 = x2 - 4
  3. 3

    Rearrange the new equation into the standard quadratic form ax2 + bx + c = 0:

    x2 - x - 6 = 0
  4. 4

    Solve the quadratic for x by factorising:

    (x - 3)(x + 2) = 0.

    This gives x = 3 and x = -2.

  5. 5

    Substitute each x-value back into the linear equation y = x + 2 to find the corresponding y-values.

  6. 6
    For x = 3, y = 3 + 2 = 5

    The first point is (3, 5).

  7. 7
    For x = -2, y = -2 + 2 = 0

    The second point is (-2, 0).

Answer: The points of intersection are (3, 5) and (-2, 0).

Common mistakes

  • ×Making sign errors when rearranging the equation after substitution. Be especially careful when moving terms across the equals sign.
  • ×Forgetting to find the corresponding y-values after solving for x. A solution consists of a coordinate pair (x, y).
  • ×Arithmetic mistakes when substituting the x-values back to find y, especially with negative numbers.

No-calculator tips

  • ✓When solving the quadratic, always check for simple factorisation before attempting to use the quadratic formula. ESAT questions often have integer solutions.
  • ✓Substitute your x-values back into the original linear equation to find y, as it involves simpler arithmetic than the quadratic equation.
  • ✓If you find an x-value, you can quickly plug it and the corresponding y-value back into the original quadratic equation as a mental check to see if it holds true.

Read this topic in the official UAT-UK ESAT guide →

All Mathematics 2 topics