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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
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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] New post 06 Oct 2003, 17:34
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17.(y+3)(y-1)-(y-2)(y-1)=r(y-1) y=?
1)r=5
2)Y=r

(y-1) ( 5 )
we are asked what's y?

1..clearly not sufficient

2..sufficient...

Answer B

correct?

thanks
praetorian
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 [#permalink] New post 06 Oct 2003, 22:27
(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.
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 [#permalink] New post 07 Oct 2003, 21:50
Why not C?

r=5
Y=r
therefore, Y=5
  [#permalink] 07 Oct 2003, 21:50
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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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