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M08#36

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shrive555
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M08#36 [#permalink] New post   31 Dec 2011, 19:37
What is \(X\) ?

\(X^2 - 1 = X + 1\)
\(X + 3\) is a prime number

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficien

why 1 is not sufficient :roll:

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Re: M08#36 [#permalink] New post   05 Jan 2012, 01:46
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Using (1), X^2 = X
=> X can be 0 or 1 [not sufficient]

Using (2), X+3 is a primer number
=> X can be -1,0,2,........ [not sufficient]

Combining (1) and (2), X is 0 [sufficient]

(C) it is.
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Re: M08#36 [#permalink] New post   06 Jan 2012, 13:59
Thanks:
Thats how i did.
(X+1)(X-1) = (X+1)
i cancelled out X+1...
X-1 = 1
X= 2
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Re: M08#36 [#permalink] New post   17 Sep 2012, 15:00
GyanOne wrote:
Using (1), X^2 = X
=> X can be 0 or 1 [not sufficient]

Using (2), X+3 is a primer number
=> X can be -1,0,2,........ [not sufficient]

Combining (1) and (2), X is 0 [sufficient]

(C) it is.


I understand that x can be 0 or 1 in the first statement. However, I'm just wondering algebraically why you couldn't cancel out (x+1) in the equation (x+1)(x-1)=x+1 ? There must be some rule broken if you do, because it will give you a definite solution, which is wrong. Any thoughts?
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Re: M08#36 [#permalink] New post   17 Sep 2012, 15:59
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dandarth1 wrote:
GyanOne wrote:
Using (1), X^2 = X
=> X can be 0 or 1 [not sufficient]

Using (2), X+3 is a primer number
=> X can be -1,0,2,........ [not sufficient]

Combining (1) and (2), X is 0 [sufficient]

(C) it is.


I understand that x can be 0 or 1 in the first statement. However, I'm just wondering algebraically why you couldn't cancel out (x+1) in the equation (x+1)(x-1)=x+1 ? There must be some rule broken if you do, because it will give you a definite solution, which is wrong. Any thoughts?


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

So, if you divide (reduce) \(x^2=x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution.

Hope it's clear.
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Re: M08#36 [#permalink] New post   18 Sep 2012, 19:47
Bunuel wrote:
dandarth1 wrote:
GyanOne wrote:
Using (1), X^2 = X
=> X can be 0 or 1 [not sufficient]

Using (2), X+3 is a primer number
=> X can be -1,0,2,........ [not sufficient]

Combining (1) and (2), X is 0 [sufficient]

(C) it is.


I understand that x can be 0 or 1 in the first statement. However, I'm just wondering algebraically why you couldn't cancel out (x+1) in the equation (x+1)(x-1)=x+1 ? There must be some rule broken if you do, because it will give you a definite solution, which is wrong. Any thoughts?


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

So, if you divide (reduce) \(x^2=x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution.

Hope it's clear.


Hi Bunnel,

How will x=0 satisfy in the 1st equation the possible values for 1st statement are -1 and 2....so not sufficient

statement 2

From statement 2 the possible values are -1, 2 0 .....so not sufficient

Even after combining we have 2 values satisfying the above statements i.e -1 and 2 so Not sufficient
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Re: M08#36 [#permalink] New post   18 Sep 2012, 23:07
GyanOne wrote:
Using (1), X^2 = X
=> X can be 0 or 1 [not sufficient]

Using (2), X+3 is a primer number
=> X can be -1,0,2,........ [not sufficient]

Combining (1) and (2), X is 0 [sufficient]

(C) it is.


Your equation \(x^2 = x\) is incorrect. It seems that you missed the signs, as in the lefthand side you have \(-1\) and in the righthand side you have \(+1\).
The two don't cancel out.

The correct equation is \(x^2-x-2=0\) or \((x+1)(x-2)=0\), which has two solutions, \(x=-1\) and \(x=2.\)
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Re: M08#36 [#permalink] New post   19 Sep 2012, 14:01
mydreammba wrote:
Bunuel wrote:
GyanOne wrote:
Using (1), X^2 = X
=> X can be 0 or 1 [not sufficient]

Using (2), X+3 is a primer number
=> X can be -1,0,2,........ [not sufficient]

Combining (1) and (2), X is 0 [sufficient]

(C) it is.


I understand that x can be 0 or 1 in the first statement. However, I'm just wondering algebraically why you couldn't cancel out (x+1) in the equation (x+1)(x-1)=x+1 ? There must be some rule broken if you do, because it will give you a definite solution, which is wrong. Any thoughts?


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

So, if you divide (reduce) \(x^2=x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution.

Hope it's clear.

Hi Bunnel,

How will x=0 satisfy in the 1st equation the possible values for 1st statement are -1 and 2....so not sufficient

statement 2

From statement 2 the possible values are -1, 2 0 .....so not sufficient

Even after combining we have 2 values satisfying the above statements i.e -1 and 2 so Not sufficient


x=0 won't satisfy the first equation. if ur referring to the rule to not divide by zero that bunuel just mentioned, it applies to the whole expression (x+1) in this problem. you cant divide by (x+1) because it could equal zero, and it does when x=-1...x=0 does not satisfy the equation, so u are correct

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Re: M08#36 [#permalink]
19 Sep 2012, 14:01
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Bunuel
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