Solve the following pair of linear equations:
$px + qy = p - q$
$qx - py = p + q$

(N/A) Given equations are:
$px + qy = p - q \quad \dots(1)$
$qx - py = p + q \quad \dots(2)$
To eliminate $y$,multiply equation $(1)$ by $p$ and equation $(2)$ by $q$:
$p(px + qy) = p(p - q) \implies p^2x + pqy = p^2 - pq \quad \dots(3)$
$q(qx - py) = q(p + q) \implies q^2x - pqy = pq + q^2 \quad \dots(4)$
Adding equations $(3)$ and $(4)$:
$(p^2x + pqy) + (q^2x - pqy) = (p^2 - pq) + (pq + q^2)$
$p^2x + q^2x = p^2 + q^2$
$x(p^2 + q^2) = p^2 + q^2$
$x = \frac{p^2 + q^2}{p^2 + q^2} = 1$
Substituting $x = 1$ in equation $(1)$:
$p(1) + qy = p - q$
$p + qy = p - q$
$qy = -q$
$y = -1$
Thus,the solution is $x = 1$ and $y = -1$.