The ratios are equivalent to each other, this can be easiest to look at by writing one above the other.

$\begin{array}{rcl} 2 x - 1 : x - 4 \\ 16 x + 1 : 2 x - 1 \end{array}$

As the ratios are equal, their scale factors must also be equal.

Find the scale factors by looking at the expressions on each side of the ratios separately.
Considering the left-hand side and letting the scale factor be a.

$\begin{array}{rcl} (2 x - 1) a & = & 16 x + 1 \\ a & = & \frac{16 x + 1}{2 x - 1} \end{array}$

Considering the right-hand side and letting the scale factor be a.

$\begin{array}{rcl} (x - 4) a & = & 2 x - 1 \\ a & = & \frac{2 x - 1}{x - 4} \end{array}$

Setting the expressions found for a equal to each other gives:

$\begin{array}{rcl} \frac{16 x + 1}{2 x - 1} & = & \frac{2 x - 1}{x - 4} \end{array}$

[1]

Remove the fractions by multiplying each side by each expression on the denominator.
Remember to use brackets when multiplying by an expression.

$\begin{array}{rcl} (x - 4) (2 x - 1) \frac{16 x + 1}{2 x - 1} & = & \frac{2 x - 1}{x - 4} (x - 4) (2 x - 1) \\ (x - 4) (16 x + 1) & = & (2 x - 1) (2 x - 1) \end{array}$

[1]

Expand the brackets on both sides.

$\begin{array}{rcl} 16 x^{2} + x - 64 x - 4 & = & 4 x^{2} - 2 x - 2 x + 1 \end{array}$

Simplify.

$\begin{array}{rcl} 16 x^{2} - 63 x - 4 & = & 4 x^{2} - 4 x + 1 \end{array}$

Rearrange to make a quadratic equation equal to zero.

$\begin{array}{rcl} 12 x^{2} - 59 x - 5 & = & 0 \end{array}$

[1]

Solve by substituting the values into the quadratic formula.
The quadratic formula is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ where $a x^{2} + b x + c = 0$ .

$a = 12, b = - 59, c = - 5$

$x = \frac{- (- 59) \pm \sqrt{{(- 59)}^{2} - 4 \times 12 \times (- 5)}}{2 \times 12}$

[1]

$x = \frac{59 \pm \sqrt{3721}}{24}$

$x = 5 or x = - \frac{1}{12}$ [1]

Solving Quadratic Equations (AQA GCSE Maths)

Exam Questions