Systems of Equations Question 4

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}