Number the equations.

$\begin{array}{rcl} x^{2} + y^{2} & = & 25 (1) \\ y - 3 x & = & 13 (2) \end{array}$

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 $y$ , $y$ will be the easiest to work with here.

$\begin{array}{rcl} y & = & 13 + 3 x (2) \end{array}$

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

$x^{2} + {(13 + 3 x)}^{2} = 25$

[1]

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

$x^{2} + (13 + 3 x) (13 + 3 x) = 25 x^{2} + {(13)}^{2} + 13 (3 x) + 13 (3 x) + {(3 x)}^{2} = 25$

Simplify.

$\begin{array}{rcl} x^{2} + 169 + 39 x + 39 x + 9 x^{2} & = & 25 \\ 10 x^{2} + 78 x + 169 & = & 25 \end{array}$

Correct expansion [1]

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

$\begin{array}{rcl} 210 x^{2} + 78 x + 169 - 25 & = & 0 \\ 10 x^{2} + 78 x + 144 & = & 0 \end{array}$

[1]

This is a non-calculator paper and no degree of accuracy is asked for, so the quadratic will be possible to factorise.
Make the problem easier to deal with by dividing every term by 2.

$\begin{array}{rcl} 5 x^{2} + 39 x + 72 & = & 0 \end{array}$

We need a pair of numbers that multiply to $a c$ which in this case is $5 \times 72 = 360$ , and add to $b$ , which in this case is 39.

15 and 24 satisfy these conditions

Rewrite the middle term using 15 and 24.

$\begin{array}{rcl} 5 x^{2} + (15 + 24) x + 72 & = & 0 \\ 5 x^{2} + 15 x + 24 x + 72 & = & 0 \end{array}$

Factorise the first two terms, using $5 x$ as the highest common factor, and the second two terms using 24 as the factor.

$5 x (x + 3) + 24 (x + 3) = 0$

Note that these now have the same factor of $(x + 3)$ so this can be used as a common factor.

$(5 x + 24) (x + 3) = 0$

[1]

Once you get to this spot it may be worth expanding the brackets to check your answer.

When two numbers are multiplied and the result is zero, at least one of them must be zero, so we can say that

$5 x + 24 = 0 or x + 3 = 0$

Solving both of these equations to find the values for $x$ .

$\begin{array}{rcl} x_{1} & = & - \frac{24}{5} \\ x_{2} & = & - 3 \end{array}$

Substitute the values of $x$ into one of the equations (the linear equation is easier) to find the values of $y$ .
Leaving the value of $x_{1}$ as a fraction makes is easier to calculate with.

$y_{1} = 13 + 3 (- \frac{24}{5}) = 13 - \frac{72}{5} = 13 - 14 \frac{2}{5} = - 1 \frac{2}{5} y_{2} = 13 + 3 (- 3) = 13 - 9 = 4$

$x_{1} = - \frac{24}{5}, y_{1} = - \frac{7}{5}$
$x_{2} = - 3, y_{2} = 4$ [1]

Simultaneous Equations (Edexcel GCSE Maths)

Exam Questions