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WM88
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WM88
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WM88
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According to the MGMAT answer, The question can be factored to \(\frac{a-b}{(a-b)(a+b)}\) which can be further simplified to \(\frac{1}{(a+b)}\). Can somebody walk be through how it gets simplified to \(\frac{1}{(a+b)}\)? I am getting 1+ \(\frac{a-b}{a+b}\)?

Thanks

When you simplify a fraction, you divide the entire top and the entire bottom by the same thing.

At the beginning, the top is a-b, and the bottom is (a-b)(a+b).

Divide both the top and bottom by a-b.

When you divide a-b by a-b, you get 1. (Dividing anything by itself gives you 1.)

When you divide (a-b)(a+b) by a-b, you get a+b.

So, the whole fraction, when simplified, will come out to 1/(a+b).

In shorter form:

(a-b)/(a-b)(a+b)

= ((a-b)/(a-b))/((a-b)(a+b)/(a-b))

= 1 / (a+b)


Step 1) \(\frac{a-b}{(a+b)(a-b)}\)

Step 2) \(\frac{a-b}{a-b}\)* \(\frac{1}{a+b}\) . << What is the logic behind this step? Trying to understand the concept. Sorry, it's been years since I've had to do this type of math and am clearly rusty.

Step 3) \(\frac{1}{a+b}\)
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Explanation:

Given: a²≠b²

To find: (a-b)/(a²-b²)
(a-b) / (a-b)(a+b)
1/(a+b)

Statement 1: a+b = 8
1/8
Sufficient

Statement 2: a-b = 6
Not sufficient

IMO-A

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