WAEC 2020 · Paper 1 · Q8

Given that $\dfrac{\sqrt3 + \sqrt5}{\sqrt5} = x + y\sqrt{15}$ , find the value of $(x + y)$ .

Worked solution (try it first)

Split the fraction: $\dfrac{\sqrt3 + \sqrt5}{\sqrt5} = \dfrac{\sqrt3}{\sqrt5} + 1$ .
Rationalise: $\dfrac{\sqrt3}{\sqrt5} = \dfrac{\sqrt3 \times \sqrt5}{5}$ , which is $\frac15\sqrt{15}$ .
So $x = 1$ and $y = \frac15$ , and $x + y = 1\frac15$ , option C.