Linear and quadratic equations - Higher - Simultaneous equations - Intermediate & Higher tier – WJEC - GCSE Maths Revision - WJEC

Simultaneous equations with one linear and one quadratic - Higher tier

A does not contain any powers higher than 1.

A contains a variable, the highest power of which is 2. For example:

\(y = x + 3\) is a linear equation and \(y = x^2 + 3x\) is a quadratic equation

Algebraic skills of substitution and factorising are required to solve these equations.

\(y = x + 3\)

\(y = x^2 + 3x\)

Substitute \(y = x + 3\) into the quadratic equation to create an equation which can be factorised and solved.

\(y = x^2 + 3x\)

Substitute \(y = x + 3\):

\(\mathbf{x~+~3} = x^2 + 3x\)

Rearrange the equation to get all terms on 1 side, so subtract \(x\) and \(-3\) from both sides:

\(-x - 3 - x - 3\)

\(0 = x^2 + 2x - 3\)

Factorise this equation:

\((x + 3)(x - 1) = 0\)

If the product of two numbers is zero, then 1 or both numbers must also be equal to zero. To solve, put each bracket equal to zero.

\(\begin{array}{rcl} x + 3 & = & 0 \\ -3 && -3 \\ x & = & -3 \end{array}\)

\(\begin{array}{rcl} x - 1 & = & 0 \\ +1 && +1 \\ x & = & 1 \end{array}\)

To find the values for \(y\), substitute the 2 values for \(x\) into the original linear equation.

\(y = x + 3\) when \(x = -3\)

\(y = \mathbf{-3} + 3\)

\(y = 0\)

\(y = x + 3\) when \(x = 1\)

\(y = \mathbf{1} + 3\)

\(y = 4\)

The answers are now in pairs: when \(x = -3\), \(y = 0\) and when \(x = 1\), \(y = 4\)