Solve the following pair of linear equations:
$ax + by = c$
$bx + ay = 1 + c$

(N/A) $ax + by = c \dots(1)$
$bx + ay = 1 + c \dots(2)$
Multiplying equation $(1)$ by $a$ and equation $(2)$ by $b$,we obtain:
$a^2x + aby = ac \dots(3)$
$b^2x + aby = b + bc \dots(4)$
Subtracting equation $(4)$ from equation $(3)$:
$(a^2 - b^2)x = ac - bc - b$
$x = \frac{c(a - b) - b}{a^2 - b^2}$
Substituting the value of $x$ in equation $(1)$:
$a \left[ \frac{c(a - b) - b}{a^2 - b^2} \right] + by = c$
$by = c - \frac{ac(a - b) - ab}{a^2 - b^2}$
$by = \frac{c(a^2 - b^2) - (a^2c - abc - ab)}{a^2 - b^2}$
$by = \frac{a^2c - b^2c - a^2c + abc + ab}{a^2 - b^2}$
$by = \frac{abc - b^2c + ab}{a^2 - b^2}$
$by = \frac{bc(a - b) + ab}{a^2 - b^2}$
$y = \frac{c(a - b) + a}{a^2 - b^2}$