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# If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =

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Manager
Joined: 18 Feb 2015
Posts: 84
If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

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23 Jul 2016, 13:19
6
00:00

Difficulty:

5% (low)

Question Stats:

87% (00:57) correct 13% (01:32) wrong based on 106 sessions

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If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =

(A) 0
(B) 2x
(C) y - x
(D) x - y
(E) x + y
Manager
Status: 2 months to go
Joined: 11 Oct 2015
Posts: 111
GMAT 1: 730 Q49 V40
GPA: 3.8
If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

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23 Jul 2016, 15:12
$$\left( 2a^{2\; }+\; b \right)\left( x+y \right)\; +\; \left( a\; +\; b \right)\left( x\; -\; y \right)\;$$

$$=\; \left( 2-2 \right)\left( x+y \right)\; -1\left( x-y \right)\;$$

$$=\; o\left( x+y \right)\; -\; x+y\;$$

$$=\; y-x$$

$$Answer\; \mbox{C}$$

Is it really from Gmatprep? It's really strange that they didn't put the x - 3y choice.
That choice would have probably been the most common mistake on this question.
Current Student
Joined: 12 Aug 2015
Posts: 2621
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

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29 Jul 2016, 14:14
1
Putting A=1 and B = -2 => 2A^2-B=0 ; so the first part of the term is zero
with second term => (1-2) *(x+y)=y-x => Smash C
_________________
Intern
Joined: 13 Aug 2018
Posts: 2
Re: If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

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05 Sep 2018, 07:04
DensetsuNo wrote:
$$\left( 2a^{2\; }+\; b \right)\left( x+y \right)\; +\; \left( a\; +\; b \right)\left( x\; -\; y \right)\;$$

$$=\; \left( 2-2 \right)\left( x+y \right)\; -1\left( x-y \right)\;$$

$$=\; o\left( x+y \right)\; -\; x+y\;$$

$$=\; y-x$$

$$Answer\; \mbox{C}$$

Is it really from Gmatprep? It's really strange that they didn't put the x - 3y choice.
That choice would have probably been the most common mistake on this question.

Is there a PEMDAS application?
Re: If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =   [#permalink] 05 Sep 2018, 07:04
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