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If x-y= R and xy=S, then (x-2)(y+2)=

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If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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Question Stats:

87% (01:01) correct 13% (01:16) wrong based on 165 sessions

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Re: If x-y= R and xy=S, then (x-2)(y-2)= [#permalink]

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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 23 Jul 2017, 10:38
D it should be
xy-2y+2x -4
Xy-2(y-x)-4
S-2(-R)-4
S+2R-4
Hence d

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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 23 Jul 2017, 11:33
xy + 2x - 2y -4
= xy + 2 (x-y) -4
= S + 2R - 4 . Ans D

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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 24 Jul 2017, 21:47
carcass wrote:
If x-y= R and xy=S, then (x-2)(y+2)=

A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S

(x-2)(y+2) = xy +2x - 2y -4
=xy +2(x-y) -4
=S + 2R - 4

Answer D
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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 24 Jul 2017, 22:19
(x-2) (y+2) = xy - 2y + 2x - 4

x-y = R, xy = S

xy - 2y + 2x - 4 = S + 2(x-y) - 4 = S + 2R - 4. Ans - D.
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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 24 Jul 2017, 22:58
carcass wrote:
If x-y= R and xy=S, then (x-2)(y+2)=

A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S


x-y = R
xy = S
(x-2)(y+2) = xy - 2y + 2x -4 = xy + 2(x-y) - 4
= S + 2R - 4

Answer D
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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 26 Jul 2017, 16:16
carcass wrote:
If x-y= R and xy=S, then (x-2)(y+2)=

A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S


(x-2)(y+2)

= xy + 2x - 2y - 4

= xy + 2(x - y) - 4

= S + 2R - 4

Answer: D
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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 28 Aug 2017, 00:14
I made a frivulous mistake in hurry and picked B.

I solved xy+2(x-y)-4
But in hurry picked R+2S-4 Instead of S+2R-4.

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Re: If x-y= R and xy=S, then (x-2)(y+2)= [#permalink]

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New post 05 Sep 2017, 06:05
1. (x-2)(y+2)
2. xy + 2x -2y - 4 (order of operations/multiplication/parenthesis)

3. xy = S; 2x - 2y = 2R, for R = x - y; and - 4 stays the same
4. xy + 2x - 2y - 4 ----> S + 2R - 4

Ans = (D) 2R + S - 4

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Re: If x-y= R and xy=S, then (x-2)(y+2)=   [#permalink] 05 Sep 2017, 06:05
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