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# If (y+3)(y-1) (y-2)(y-1) = r(y-1), what is the value of y?

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If (y+3)(y-1) (y-2)(y-1) = r(y-1), what is the value of y? [#permalink]  14 Jul 2006, 10:10
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If (y+3)(y-1) â€“ (y-2)(y-1) = r(y-1), what is the value of y?
(1) r2 = 25
(2) r = 5
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-B-

(y+3)(y-1) â€“ (y-2)(y-1) = r(y-1)

Solving for y --> y=(5-r)/(5-r)

St 1 --> INSUFF
r could be 5 or -5. So y could be 0 or 1

ST 2 --> SUFF
with r=5, y=0.
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With r=5, y=1 is still possible. Hence, (2) is not sufficient.

E.
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E?

factor stem to:
5(y-1) = r(y-1)

(1) if r=5, y could be anything. if r=-5, y must be 1
(2) same as above.
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Reviewing my approach I'm getting that y=(5-r)/(5-r)

So with St1 I get that when r=5 y=0/0 which is an undefined result, but when r=-5 I got that y=10/10=1.

With St 2 y=undefined result.

So I guess that it should be A.

Need some advice from the smartest people in the room...
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E

St1: r is either 5 or -5. If r = 5 then y could be anything. If r is -5 then y = 1: INSUFF

St2: If r = 5 then y could be anything: INSUFF

Combined: r = 5. Then y could be anything : INSUFF
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Silly mistake.. ups. Agree it should be E
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This one is quite easy one, but i made silly mistake on this.

Watch out for those mistakes folks.!!

OA is E
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