MrWallSt wrote:

If x / (x+y) = n, and x / (x-y) = m, then x/y=? (|x| does not equal |y|, xy does not equal 0)

A. 3mn/2

B. (3m)/(2n)

C. (n * (m+2))/2

D. 2nm / (m-n)

E. (n^2 - m^2) / nm

You can do it either algebraically or by assuming numbers:

Algebra:

Note that we are happier with (x+y)/x rather than x/(x+y) since in the former case we can make manipulations easily. So let's take the inverse of both n and m

\(\frac{1}{n} = \frac{(x+y)}{x} = 1 + \frac{y}{x}\) ....(I)

\(\frac{1}{m} = \frac{(x-y)}{x} = 1 - \frac{y}{x}\) .....(II)

Since we have both m and n in our answer, lets subtract II from I to get

\(\frac{1}{n} - \frac{1}{m} = \frac{2y}{x}\)

\(\frac{(m-n)}{2mn} = \frac{y}{x}\)

\(\frac{x}{y} = \frac{2mn}{(m-n)}\)

Answer (D)

Or Plug in numbers: x = 2, y = 1

n = x/(x+y) = 2/3

m = x/(x-y) = 2

x/y = 2

Now put n = 2/3 and m = 2 in the options. Only option (D) gives you x/y = 2.

Hence answer (D)

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