Question

Please give a worked explanation! Thanks :)

Answer 1

When you add any fractions, you need to find a common denominator.  In this case they have no common factors so you just multiply them together.  First multiply the first fraction by $\frac{x - 2}{x - 2}$ which gets you $\frac{3x - 6}{ x^{2} - 2x}$.  Then multiply the second fraction by $\frac{x}{x}$ to get $\frac{5x}{ x^{2} - 2x}$.  Now you can add them together to get $\frac{-2x - 6}{ x^{2} - 2x}$.

(b) $\sqrt{18} = \sqrt{9 * 2} = 3 \sqrt{2}$
$\sqrt{72} = \sqrt{36 * 2} = 6 \sqrt{2}$
now you can pull out root 2 and get $\sqrt{2}(3 - 1 + 6)$ which equals $8 \sqrt{2}$

Answer 2

a) We need a common denominator to be able to subtract the fractions, so by multiplying both the numerator and the denominator of each fraction by the denominator of the other.
$\frac{x + 2}{x + 2} ( \frac{3}{x}) - \frac{x}{x} ( \frac{5}{x + 2} )$
Then simplify the fraction:
$\frac{6 + 3x}{x^{2} + 2x} - \frac{5x}{x^{2} + 2x}$
Then you can just subtract the nominators from each other, but leave the denominator as it is
$\frac{(6 + 3x) - 5x}{x^{2} + 2x}$
to get: 
$\frac{6 - 2x}{x^{2} + 2x}$
and then you can simplify to get the final answer:
$\frac{2(3 - x)}{x(x + 2)}$

b) you need to have the number inside the root the same to add or subtract:
$\sqrt{2(9)} - \sqrt{2} + \sqrt{2(36)}$
Then you can 'take out' the numbers that square root easily:
$3\sqrt{2} - \sqrt{2} + 6\sqrt{2}$
now you can just add and subtract using the whole numbers outside the root of 2 ( since 2 is a surd):
$8\sqrt{2}$