An algebra problem by Dan Andrei

$X^2-Y^2=5=(X+Y)(X-Y) \\ 5=5(X-Y)\Rightarrow 1=X-Y$

$x - y = 5$ $\implies$ $x = 5 + y$

$x^2 - y^2 = 5$ $\implies$ $(5 - y)^2 - y^2 = 5$ $\implies$ $25 + 10y + y^2 - y^2 = 5$ $\implies$ $\color{#D61F06}y = -2$

$x = 5 + y$ $\implies$ $x = 5 - 2$ $\implies$ $\color{#D61F06}x = 3$

$\large\therefore$ $x + y = 3 - 1 =$ $\boxed{\color{#20A900}1}$

Dividing the second equation by the first gives:
$\frac{X^2-Y^2}{X-Y} = \frac{5}{5} \Rightarrow \frac{(X+Y)(X-Y)}{(X-Y)} = 1 \Rightarrow X+Y = 1$

(x^2 - y^2) =(x - y) (x + y) = 1 5. Ed Gray