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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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Manager
Joined: 15 Nov 2007
Posts: 135
If (y+3)(y-1) (y-2)(y-1) = r(y-1), what is the value of y?  [#permalink]

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19 Jan 2008, 11:57
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If (y+3)(y-1) – (y-2)(y-1) = r(y-1), what is the value of y?

(1) r^2 = 25
(2) r = 5

[Reveal] Spoiler: My Take
i read a couple explanations on here that i didn't understand, like:

"(y+3)(y-1) – (y-2)(y-1) = r(y-1)
5(y-1)=r(y-1)

r=5 satisfied both conditions. " E

and

"5(y-1)=5(y-1) ==> y can be any number.

we found that: r(y-1)=5(y-1).

(1) r=+/- 5

if r=-5 -> y=1
if r=5 -> y= any number ==> we need one value only. Insuff.

(2) r=5 -> y= any number ==> we need one value only. Insuff. " E

but they don't make sense...
VP
Joined: 22 Nov 2007
Posts: 1080

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19 Jan 2008, 12:18
dominion wrote:
If (y+3)(y-1) – (y-2)(y-1) = r(y-1), what is the value of y?

(1) r^2 = 25
(2) r = 5

i read a couple explanations on here that i didn't understand, like:

"(y+3)(y-1) – (y-2)(y-1) = r(y-1)
5(y-1)=r(y-1)

r=5 satisfied both conditions. " E

and

"5(y-1)=5(y-1) ==> y can be any number.

we found that: r(y-1)=5(y-1).

(1) r=+/- 5

if r=-5 -> y=1
if r=5 -> y= any number ==> we need one value only. Insuff.

(2) r=5 -> y= any number ==> we need one value only. Insuff. " E

but they don't make sense...

I'll try:

the equality itself says that r=5. does this information help us find y? no.
let's see what 1 and 2 tell us.
1. r can be +-5, but we don't care about that, do you agree?
2. r=5. once again, we only have a repetition of what is stated in the text.

neither 1 nor 2 help find y. y can be any value. let's try:

y=1 and r=5:
we have 0=0 so it's ok

let's try with 2, same thing

let's try with 3: we find always r=5.

is that a good explanation?
Senior Manager
Joined: 02 Aug 2007
Posts: 346
Location: Greater New York City area
Schools: Tuck, Ross (R1), Duke, Tepper, ISB (R2), Kenan Flagler (R2)

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19 Jan 2008, 12:25
dominion wrote:
If (y+3)(y-1) – (y-2)(y-1) = r(y-1), what is the value of y?

(1) r^2 = 25
(2) r = 5

i read a couple explanations on here that i didn't understand, like:

"(y+3)(y-1) – (y-2)(y-1) = r(y-1)
5(y-1)=r(y-1)

r=5 satisfied both conditions. " E

and

"5(y-1)=5(y-1) ==> y can be any number.

we found that: r(y-1)=5(y-1).

(1) r=+/- 5

if r=-5 -> y=1
if r=5 -> y= any number ==> we need one value only. Insuff.

(2) r=5 -> y= any number ==> we need one value only. Insuff. " E

but they don't make sense...

Can I ask what is the source of this question ?
Director
Joined: 03 Sep 2006
Posts: 871

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19 Jan 2008, 23:51
$$5*(y-1) = r*(y-1)$$

From ( i )

$$r=+/-5$$

Insufficient

From ( ii )

r=5

y can be any value!

Therefore "E"
Re: geometry... kind of   [#permalink] 19 Jan 2008, 23:51
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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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