If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?

[mithunsam]

Can we just use x=1 and y=1 and plug in for x+y and x-y? This results in 16 and I just want to make sure I'm not making a mistake.

that is a very effective way to solve the problem.

Since the question doesn't state that x and y are integer, it is better to avoid substitution. We may not know what other possibilities are possible. The question is pretty simple.

[2^(x+y)^2]/[2^(x-y)^2] = 2^([x+y]^2 - [x-y]^2) = 2^4xy = 2^4*1 (xy=1 from question) = 2^4 = 16

[RonPurewal]

Since the question doesn't state that x and y are integer, it is better to avoid substitution. We may not know what other possibilities are possible.

hmm?

the question is multiple-choice, and the answers are different numbers.
therefore, if you plug in anything -- whether integers or not -- and get 16, then the answer is 16.

[zeshan.dhanani]

For those who didn't know Ron's epic perfect square trinomial subtraction shortcut, factoring out is just as easy, although slightly longer:
2^(x+y)^2 / 2^(x-y)^2 = 2^(x^2 + 2xy + y^2) / 2^(x^2 - 2xy + y^2).
Plugging in 1 for xy gives you 2^(x^2 + y^2 +2) / 2^(x^2 + y^2 - 2) which can then be written as 2^(x^2 + y^2) x 2^2 / 2^(x^2 + y^2) x 2^-2
The 2^(x^2 +y^2) cancel out, leaving you with 2^2 / 2^-2, which is just 4 / (1/4), which is 16

[jlucero]

Nice work. After the fact, I think there's huge value to being able to break down an equation. But I agree with Ron that plugging in numbers (x = 1, y = 1), would probably be faster on test day. Moral of the story: learn the algebra, but also learn when you can avoid algebra.