The two fractions are being multiplied and involve three quadratic expressions.
Factorise these expressions as this will make them easier to multiply.

$\begin{array}{rcl} 25 x^{2} - 64 & = & (5 x - 8) (5 x + 8) \\ 5 x^{2} - 13 x - 6 & = & (5 x + 2) (x - 3) \\ x^{2} - 8 x + 15 & = & (x - 5) (x - 3) \end{array}$

One quadratic correctly factorised [1]
Two quadratics correctly factorised [1]

Rewrite the product of the fractions.

$\frac{25 x^{2} - 64}{5 x^{2} - 13 x - 6} \times \frac{x^{2} - 8 x + 15}{5 x + 8} = \frac{(5 x - 8) (5 x + 8)}{(5 x + 2) (x - 3)} \times \frac{(x - 5) (x - 3)}{(5 x + 8)}$

Look to cancel any factors before multiplying.

$\frac{(5 x - 8) (5 x + 8)}{(5 x + 2) (x - 3)} \times \frac{(x - 5) (x - 3)}{(5 x + 8)} = \frac{(5 x - 8) (x - 5)}{(5 x + 2)}$

Now we need to include the third term, $- (x - 7)$ , and make a common denominator.

$\frac{(5 x - 8) (x - 5)}{(5 x + 2)} - (x - 7) = \frac{(5 x - 8) (x - 5) - (x - 7) (5 x + 2)}{(5 x + 2)}$

[1]

Expand and simplify the numerator.

$\begin{array}{rcl} \frac{(5 x - 8) (x - 5) - (x - 7) (5 x + 2)}{(5 x + 2)} & = & \frac{5 x^{2} - 33 x + 40 - (5 x^{2} - 33 x - 14)}{(5 x + 2)} \\ = & \frac{5 x^{2} - 33 x + 40 - 5 x^{2} + 33 x + 14}{(5 x + 2)} \\ = & \frac{54}{(5 x + 2)} \end{array}$

Nothing else can be simplified or cancelled.

$\frac{25 x^{2} - 64}{5 x^{2} - 13 x - 6} \times \frac{x^{2} - 8 x + 15}{5 x + 8} = \frac{54}{5 x + 2}$ [1]

	Distance (miles)	Speed (mph)	Time (hours)
1st stage	731	$x$	$\frac{731}{x}$
2nd stage	287	$x - 24$	$\frac{287}{x - 24}$

Algebraic Fractions (Edexcel GCSE Maths)

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