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WAEC Mathematics 2020 Paper

Question : 7

Total: 48

Given that

\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}

Solution:

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = x + y $\sqrt{15}$

cross multiply to have: $\sqrt{3}$ + $\sqrt{5}$ = x $\sqrt{5}$ + 5y $\sqrt{3}$

Collect like roots : x $\sqrt{5}$ = $\sqrt{5}$ → x = 1

5y $\sqrt{3}$ = $\sqrt{3}$ → y = $\frac{1}{5}$

∴ ( x + y ) = 1 + $\frac{1}{5}$

= 1 $\frac{1}{5}$

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