For the following system of equations

$$ 2^{x-y} = 32 $$ $$ 2^{x+y} = 8, $$

$y=$ ?

$-3$

$-1$

$1$

$2$

$3$

Approach 1

We can divide the top equation by the bottom one.

$$ \frac{2^{x-y}}{2^{x+y}}=\frac{32}{8} $$

Using exponent rules,

$$ 2^{(x-y)-(x+y)} = 4 $$ $$ 2^{-2y} = 4 $$ $$ 2^{-2y} = 2^2 $$ $$ -2y = 2 $$ $$ y = \boxed{-1} $$

Approach 2

Setting each side to base 2, we can obtain simpler equations:

$2^{x-y}=32$

$$ 2^{x-y} = 2^5 $$ $$ x-y =5 $$

$2^{x+y}=8$

$$ 2^{x+y} = 2^3 $$ $$ x+y = 3 $$

We can subtract (elimination) these equations to obtain:

$$ (x-y)-(x+y) = 5-3 $$ $$ -2y = 2 $$ $$ y = \boxed{-1} $$