## Algebra 1: Common Core (15th Edition)

$x=7$ $y=4$
$\frac{a^x}{a^y}=a^3$ $a^{x-y}=a^3$ Comparing the exponents of a on both sides: $x-y=3$ (1) $\frac{a^x}{a^3y}=a^{-5}$ $a^{x-3y}=a^{-5}$ Comparing the exponents of a on both sides: $x-3y=-5$ (2) From (1) and (2) we get: $x-y=3$ $x-3y=-5$ Multiply both sides of the second equation with $-1$: $x-y=3$ $-x+3y=5$ Using substitution: $2y=8$ Solve for $y$: $y=4$ Solve for $x$: $x-4=3$ $x=7$