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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
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If (y+3)(y-1) (y-2)(y-1) = r(y-1), what is the value of y? [#permalink] New post 18 Jun 2008, 14:25
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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


I want to divide everything by (y-1), but it doesn't seem to work out. Any suggestions?
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Re: DS algebra [#permalink] New post 19 Jun 2008, 03:22
I think it should be E.

Quote:
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


Yes, you are right in that it would be wrong to divide by (y-1) – at least at this stage. So, let’s first simplify the expression:

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

We can rewrite the last one in the form: (y-1)*(5-r)=0. From here, we have two possibilities: y=1 or r=5. Now, it’s time to look at our statements.

From the first, we can’t say whether r=5 (it could be 5 or -5). Thus, we can’t say whether y=1 => statement is insufficient.

From the second, if r=5 then y could be any number. So, again, it’s insufficient.

Thus, E.
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Re: DS algebra [#permalink] New post 19 Jun 2008, 04:23
But why can't you divide everything by (y-1) in the beginning?

from

(y+3)(y-1) – (y-2)(y-1) = r(y-1)

to

((y+3)(y-1)/(y-1)) – ((y-2)(y-1)/(y-1)) = (r(y-1)/(y-1))

to

(y+3)-(y-2)=r

to

5=r

Why is this wrong?
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Re: DS algebra [#permalink] New post 19 Jun 2008, 06:00
When solving the equation, the general rule is not to divide it by anything that could possibly be the solution, because there is a risk that we may lose some roots.

In this problem, for example, when you divided the equation by (y-1) you lost the possibility that (y-1) can be 0. And this was crucial, because here y=1 was indeed one of the possible solutions.

Had we smth that is never be zero - e.g. (y^2+2) on both sides, it would be all right to divide the equation by this expression.
Re: DS algebra   [#permalink] 19 Jun 2008, 06:00
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