System of equations that I'm having trouble with

Question

$a/(x+y) - b/(x-y) = 1$

$b/(x+y) + a/(x-y) = (b^2-a^2)/2ab$

The answer to the values of $x$ and $y$ are given as $x=a-b, y=a+b$. How is that achieved?

Answer 1

Here is a straightforward solution without using any trick, but you can follow it step by step. First, we multiply the first equation by $x^2-y^2$ to obtain a polynomial equation $$ ax - ay - bx - by - x^2 + y^2=0 $$ Similarly, we multiply the second equation by $2ab(x^2-y^2)$ to obtain a polynomial equation. Then we compute $(a^2-b^2)$ times the first equation minus the second one. This gives a third equation, namely $$ (a^2 + b^2)(ax - ay + bx + by)=0. $$ We can dispose of $a^2+b^2=0$, because we assume $ab\neq 0$ from the beginning. Hence we obtain $x(a+b)+y(b-a)=0$. Substituting $x$ from this in the first equation gives $(a+b-y)aby=0$ hence $y=a+b$. Then $x=a-b$.

Answer 2

One way to solve that is to let $u=x+y$, $v=x-y$, and solve for $u$ and $v$. Then use

$$x=\frac{u+v}2,\quad y=\frac{u-v}2$$

That is probably easier than solving for $x$ and $y$ directly.