Question 9
Two real numbers $$a$$ and $$b$$, where $$a>b$$, are given such that their sum is equal to $$4$$ times their difference. The value of $$\frac{2ab}{3(a^2-b^2)}$$ is
Solution
From $$a+b=4(a-b)$$, we get $$3a=5b$$, so we may write $$a=5k$$ and $$b=3k$$. Substitution gives $$\frac{2ab}{3(a^2-b^2)}=\frac{2 \times 5k \times 3k}{3(25k^2-9k^2)}=\frac{30}{48}=\frac{5}{8}$$.
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