To write two separate fractions as one single fraction, they both must be over the same lowest common denominator
Find the lowest common denominator by multiply $(x - 2)$ by 2

$(x - 2) \times 2 = 2 (x - 2)$

[1]

Write each of the separate fractions over this lowest common denominator (by multiplying top and bottom by the missing terms needed)

$\frac{5 \times 2}{2 (x - 2)} - \frac{(x - 5) \times (x - 2)}{2 (x - 2)} \frac{10}{2 (x - 2)} - \frac{(x - 5) (x - 2)}{2 (x - 2)}$

Combine into one single fraction over the lowest common denominator (by subtracting the numerators)

$\frac{10 - (x - 5) (x - 2)}{2 (x - 2)}$

correct numerator before expanding [1]

Expand $(x - 5) (x - 2)$ by multiplying everything in the first bracket by everything in the second bracket

$\begin{array}{rcl} (x - 5) (x - 2) & = & x^{2} - 5 x - 2 x + 10 \\ = & x^{2} - 7 x + 10 \end{array}$

Substitute this expression back into the numerator using brackets around the three terms

$10 - (x - 5) (x - 2) = 10 - (x^{2} - 7 x + 10)$

Expand the right-hand side and collect like terms, it may help to think of it as $10 - 1 (x^{2} - 7 x + 10)$

$\begin{array}{rcl} 10 - (x^{2} - 7 x + 10) & = & 10 - x^{2} + 7 x - 10 \\ = & - x^{2} + 7 x \end{array}$

Write the final answer as an algebraic fraction

$\frac{- x^{2} + 7 x}{2 (x - 2)}$ [1]
The following are also accepted: $\frac{7 x - x^{2}}{2 (x - 2)}, \frac{7 x - x^{2}}{2 x - 4}, \frac{- x^{2} + 7 x}{2 x + 4}$

Algebraic Fractions (Cambridge O Level Maths)

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