If a > 0, is [2/(a+b)] +[2/(a-b)] = 1?

[TamL732]

If a > 0, is [2/(a+b)] +[2/(a-b)] = 1 ?

1.) b = 0
2.) (a^2)-(b^2) = 4a

The correct answer is B.

I thought the answer was A.
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The question simplified: 4a = (a+b) (a-b)

1.) If b = 0, then a has to equal to 4, then we can plug in a and b to to figure if [2/(a+b)] +[2/(a-b)] = 1 ?

2.) I understand when you simplify the question, you get the same statement as statement two. However, I am having a hard time interpreting how this helps determine if statement is equal to 1. How can I figure if 4a = (a+b) (a-b) = (a^2)-(b^2) is equal to 1?

Am I missing something or assuming something extra?

[Sage Pearce-Higgins]

Is this question taken from a Manhattan Prep CAT? If so, please can you give me the question name? (I couldn't locate it on our system.)

You seem to be adding something extra in your solution. Taking statement 1 on it's own, how do you calculate that a = 4? What I think you are doing is forgetting that the question is a question and not a fact. We don't know that the initial equation is true! Sure, we can rephrase it to make 4a = a^2 - b^2, but it's still a question.

Let me know if the Algebra of the rephrase is where you're getting stuck.