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Retired Moderator Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Expert's post 00:00

Question Stats: 96% (00:16) correct 3% (01:06) wrong based on 28 sessions
If $$x \neq -y$$, then $$\frac{x^2 + 2xy + y^2}{2(x+y)^2}=$$

(A) $$1$$
(B) $$\frac{1}{2}$$
(C)$$\frac{1}{x+y}$$
(D) $$xy$$
(E) $$2xy$$
[Reveal] Spoiler: OA

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Sandy
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Manager Joined: 26 Jan 2018
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The numerator is nothing but x+y whole square. Hence on solving we get 1/2 GRE Instructor Joined: 10 Apr 2015
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Expert's post
sandy wrote:
If $$x \neq -y$$, then $$\frac{x^2 + 2xy + y^2}{2(x+y)^2}=$$

(A) $$1$$
(B) $$\frac{1}{2}$$
(C)$$\frac{1}{x+y}$$
(D) $$xy$$
(E) $$2xy$$

APPROACH #1: Factoring

Given: $$\frac{x^2 + 2xy + y^2}{2(x+y)^2}$$

Factor the numerator to get: $$\frac{(x+y)(x+y)}{2(x+y)^2}$$

In other words: $$\frac{(x+y)^2}{2(x+y)^2}$$

Divide numerator and denominator by $$(x+y)^2$$ to get: $$\frac{1}{2}$$

APPROACH #2: Plug-in values

If $$x = 0$$ and $$y = 1$$, we get: $$\frac{x^2 + 2xy + y^2}{2(x+y)^2}=\frac{0^2 + 2(0)(1) + 1^2}{2(0+1)^2}=\frac{1}{2}$$

From here we need to test answer choices C, D and E since there are variables in those answer choices.
So we'll plug in $$x = 0$$ and $$y = 1$$ to see if any of them also evaluate to be $$\frac{1}{2}$$

(C)$$\frac{1}{x+y}=\frac{1}{0+1}=1$$. No good
(D) $$xy=(0)(1)=0$$. No good
(E) $$2xy=2(0)(1)=0$$. No good

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Brent Hanneson – Creator of greenlighttestprep.com  Re: If x -y,   [#permalink] 21 Aug 2020, 07:01
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