Number the equations.

$\begin{array}{rcl} x^{2} + y^{2} & = & 9 (1) \\ x + y & = & 2 (2) \end{array}$

There is one quadratic equation and one linear equation so this must be done by substitution.

Rearrange equation (2) so that the subject is either $x$ or $y$ .
In this case the subject is $y$ .

$\begin{array}{rcl} y & = & 2 - x (2) \end{array}$

Substitute this into equation (1) to eliminate $y$ .

$x^{2} + {(2 - x)}^{2} = 9$

[1]

Expand the brackets, remember that a bracket squared should be treated the same as double brackets.

$x^{2} + (2 - x) (2 - x) = 9 x^{2} + 4 - 2 x - 2 x + x^{2} = 9$

Simplify.

$\begin{array}{rcl} x^{2} + 4 - 4 x + x^{2} & = & 9 \\ 2 x^{2} - 4 x + 4 & = & 9 \end{array}$

Correct expansion [1]

Rearrange to form a quadratic equation that is equal to zero.

$\begin{array}{rcl} 2 x^{2} - 4 x + 4 - 9 & = & 0 \\ 2 x^{2} - 4 x - 5 & = & 0 \end{array}$

[1]

The question asks for answers to be given correct to 2 decimal places. This is a hint that the quadratic cannot be factorised and the quadratic formula should be used.

The quadratic formula is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ where $a x^{2} + b x + c = 0$ .
Find the values of $a$ , $b$ and $c$ .

$a = 2, b = - 4, c = - 5$

Substitute into the quadratic formula.

$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 \times 2 \times (- 5)}}{2 \times 2}$

[1]

$x = \frac{4 \pm \sqrt{16 - (- 40)}}{4} = \frac{4 \pm \sqrt{56}}{4}$

$x = 2.870828 . . ., x = - 0.870828 . . .$

[1]

Substitute the values of $x$ into one of the equations (the linear equation is easier) to find the values of $y$ .
Use more than 2 decimal places so that your answers for $y$ are accurate.

$\begin{array}{rcl} (2) y_{1} & = & 2 - x_{1} = 2 - 2.870828 . . . \\ y_{2} & = & 2 - x_{2} = 2 - (- 0.870828 . . .) \end{array}$

Round your answers to two decimal places.

$x = 2.87, y = - 0.87$
or $x = - 0.87, y = 2.87$ [1]

Pattern number	1	2	3	4
Number of grey mats	6	9	12	15
Total number of mats	6	10	15	21

Simultaneous Equations (CIE IGCSE Maths: Extended)

Exam Questions