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

In the equation above we see that (y-1) is the common factor in all terms. So let's take (y-1) common out of all the terms.

(y+3)(y-1) - (y-2)(y-1) - r(y-1) = 0
(y-1)[(y+3) -(y-2) - r] = 0
(y-1)(5 - r) = 0

Now, the product of these two is 0. This means that at least one of them has to be 0.
Either (y-1) = 0 or (5 - r) = 0 or both are 0.
So, either y = 1 or r = 5 or both.
Only if we know that r is not 5, then we can say that y must be 1. If r is 5, y may be 1 or may not be 1.

Stmnt 1: r^2 = 25
So r = +- 5
This statement tells us that r can be 5.
If r = 5, y may or may not be 1.
If r is not 5, y will be 1.
Since we do not know whether r is 5 or not, we cannot say what the value of y is. Not sufficient.

Stmnt 2: r = 5
If r = 5, y may or may not be 1.
Not sufficient.

Both together, r = 5. Again, not sufficient. Answer E

Correct answer Should be B

y-1 is the common factor in the question stem

Try to break the stem as much as you can

So, y-1[(y+3) - (y-2)] = r(y-1)

= y+3-y+2 = r

r=5. So the question becomes is r=5?

Statement 1 gives us two solution + and _ 5 So insufficient

Statement 2 is sufficient because its same as our new question stem which is r=5.

Hope this helps.

I am wrong. Sorry guys. Thanks Karishma.

should be E.
The questions becomes y-1[y+3 - y+2)] = r(y-1).

r=5. So, can not calculate the value of Y from both 1 and 2.

Rephrasing the stem provides us:

(y-1)[y+3-y+2] = r(y-1)
(y-1)5 = r(y-1)
(y-1)(5-r) = 0
y = 1 or r = 5

1. r = +/- 5, if r = -5, y = 1 and if r = 5, y = anything. Insuff.
2. r = 5, which means y = anything. Insuff.

Combined, only thing common is r = 5, which still provides y = anything. Insuff.

E.

Guys I'm bit lost here.


(y-1)[y+3-y+2] = r(y-1)
(y-1)5 = r(y-1)

(y-1) is common here

so 5= r

which is what stmt 2 tells us.

I think i should be B.
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Notice that we are asked to find the value of y not r. Also, I guess you reduced (y-1)5 = r(y-1) by y-1 which cannot be done here.

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) (y-1)5 = r(y-1) by y-1, you assume, with no ground for it, that y-1 does not equal to zero thus exclude a possible solution (notice that both y=1 AND r=5 satisfy the equation).

Hope it's clear.

Complete solution:

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

\((y+3)(y-1) - (y-2)(y-1) - r(y-1) = 0\);

\((y-1)[(y+3) -(y-2) - r] = 0\);

\((y-1)(5 - r) = 0\);

\(y=1\) or/and \(r=5\).


(1) \(r^2=25\) --> \(r=5\) or \(r=-5\). Not sufficient.

(2) \(r=5\) --> y may or may not be 1. Not sufficient.


(1)+(2) The same here: we have that r=5 but y still may or may not be 1. Not sufficient.


Answer: E.
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Hi Buneul, has my logic actually worked here?
correct me if i am wrong...


From the given statement:
(y-1)[Y+3-y+2]=r(y-1)
Divide both sides by (y-1)

y+3-y+2=r
r=5

Here I thought that in the given statement when Y does not exist it is impossible to calculate the value of Y.
Thus both statements are not sufficient.
Ans: E

Unfortunately that's not correct. Refer to my post above: if-y-3-y-1-y-2-y-1-r-y-1-what-is-the-value-of-y-119561.html#p1224280

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) (y-1)5 = r(y-1) by y-1, you assume, with no ground for it, that y-1 does not equal to zero thus exclude a possible solution (notice that both y=1 AND r=5 satisfy the equation).
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Oops by bad, Thanks Bunuel.

Given information=
(y+3)(y-1)-(y-2)(y-1)=r(y-1)

Question- value of y?

Inference from given information-

(y+3)(y-1)-(y-2)(y-1)=r(y-1)
(y+3)- (y-2)= r (y-1) is common at both the sides of the equation.
y+3-y+2=r
r=5
(y is cancelling out at both sides of equation, so most likely we need one of the statement to produce value of y )

(1) r^2=25
It does not give us a solution for y

(2) r=5
It does not give us a solution for y

E is the answer
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