Answer
$x=\dfrac{1}{a+b},y=\dfrac{1}{a+b}$
Work Step by Step
$\begin{cases}
ax+by=1\\
bx+ay=1
\end{cases}, (a^2-b^2\ne0)$
Multiplying Equation 2 by $-\frac{a}{b}$ and adding it together.
$\begin{cases}
ax+by=1\\
-ax-\frac{a^2}{b}y=-\frac{a}{b}\\
-- -- -- -- \\
(b-\frac{a^2}{b})y=1-\frac{a}{b},
\end{cases},$.
Thus,
$\left(\frac{b^2-a^2}{b}\right)y=\frac{b-a}{b}$. Multiplying both sides by $\left(\frac{b}{b^2-a^2}\right)$.
$y=\frac{b-a}{b^2-a^2}=\frac{b-a}{(b-a)(b+a)}=\frac{1}{b+a}, for - a\ne b$.
Substituting back in,
$ax+\frac{b}{b+a}=1,$
$ax=1-\frac{b}{b+a},$
$ax=\frac{b+a-b}{b+a},$
$ax=\frac{a}{b+a},$. Dividing both sides by $a$.
$x=\frac{1}{b+a}, for -a\ne b$.
