Question

If $P=\frac{x}{x+y},Q=\frac{y}{x+y}$, then find $\frac{1}{P-Q}-\frac{2Q}{P^2-Q^2}$.

Answer 1

We are given that $P=\frac{x}{x+y}$ and $Q=\frac{y}{x+y}$. So, we find $\frac{1}{P-Q}-\frac{2Q}{P^2-Q^2}$ as follows:

$\frac{1}{P-Q}-\frac{2Q}{P^2-Q^2}=\frac{1}{\frac{x}{x+y}-\frac{y}{x+y}}-\frac{2\left(\frac{y}{x+y}\right)}{{\left(\frac{x}{x+y}\right)}^2-{\left(\frac{y}{x+y}\right)}^2}=\frac{1}{\frac{x-y}{x+y}}-\frac{\frac{2y}{x+y}}{\frac{x^2}{(x+y{)}^2}-\frac{y^2}{(x+y{)}^2}}$

$=\frac{x+y}{x-y}-\frac{\frac{2y}{x+y}}{\frac{x^2-y^2}{(x+y{)}^2}}=\frac{x+y}{x-y}-\frac{\frac{2y}{x+y}}{\frac{(x-y)(x+y)}{(x+y{)}^2}}(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)$

$=\frac{x+y}{x-y}-(\frac{2y}{x+y}\times \frac{(x+y{)}^2}{(x-y)(x+y)})=\frac{x+y}{x-y}-\frac{2y}{x-y}=\frac{x+y-2y}{x-y}=\frac{x-y}{x-y}=1$

Hence, $\frac{1}{P-Q}-\frac{2Q}{P^2-Q^2}=1$.