Find the value of $x +y$

Question

If $a=\frac{x}{x^2+y^2}$ and $b=\frac{y}{x^2+y^2}$ then find $x+y$

I find that $x+y/y=\frac{a+b}{b}$ but the ans in the form of a and B only.

Answer 1

$a^2+b^2={x^2\over{(x^2+y^2)^2}}+{y^2\over{(x^2+y^2)^2}}= {1\over{x^2+y^2}}$, $x=a(x^2+y^2), y=b(x^2+y^2)$ implies that $x+y={{a+b}\over{a^2+b^2}}$

Answer 2

Note that: $$ax+by=1 \qquad \qquad (1)\\ \frac ax=\frac by=\frac{a+b}{x+y} \qquad \qquad (2)$$ From $(2)$ we also get: $$x=\frac{a(x+y)}{a+b}; y=\frac{b(x+y)}{a+b} \qquad \qquad (3) $$ Plug $(3)$ into $(1)$: $$\frac{a^2(x+y)}{a+b}+\frac{b^2(x+y)}{a+b}=1 \Rightarrow x+y=\frac{a+b}{a^2+b^2}.$$