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# (y+3)(y-1)-(y-2)(y-1)=r(y-1) y=? 1)r=5 2)Y=r (y-1) ( 5 ) we

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CEO
Joined: 15 Aug 2003
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(y+3)(y-1)-(y-2)(y-1)=r(y-1) y=? 1)r=5 2)Y=r (y-1) ( 5 ) we [#permalink]  06 Oct 2003, 17:34
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
17.(y+3)(y-1)-(y-2)(y-1)=r(y-1) y=?
1)r=5
2)Y=r

(y-1) ( 5 )

1..clearly not sufficient

2..sufficient...

correct?

thanks
praetorian
SVP
Joined: 03 Feb 2003
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Kudos [?]: 16 [0], given: 0

(y+3)(y-1)-(y-2)(y-1)=r(y-1)
take (y-1) as common
(y-1)(y+3-y+2-r)=0
(y-1)(5-r)=0

(1) r=5; y is any
(2) y=r; (y-1)(5-y)=0; y=1 or y=5

I think we need to find a unique y, so both are not suff. I vote for E.
Intern
Joined: 12 Sep 2003
Posts: 37
Location: Peru
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Kudos [?]: 0 [0], given: 0

Why not C?

r=5
Y=r
therefore, Y=5
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