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605-655 (Medium)|   Algebra|      
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audiogal101
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What is the value of x? Updated on: 29 Nov 2013, 04:06
Originally posted by audiogal101 on 29 Nov 2013, 04:03.
Last edited by Bunuel on 29 Nov 2013, 04:06, edited 2 times in total.
Edited the question.
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E
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Difficulty:

45% (medium)

Question Stats:

53% (01:17) correct 47% (01:14) wrong based on 95 sessions

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What is the value of x?

(1) x^2 - 1 = x + 1
(2) x + 3 is a prime number
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E
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Bunuel
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What is the value of x? 29 Nov 2013, 04:09
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What is the value of x?

(1) x^2 - 1 = x + 1 --> \(x^2-x-2=0\). Solving for x gives \(x=-1\) or \(x=2\). Not sufficient.

(2) x + 3 is a prime number. Clearly insufficient, consider x=2 and x=4.

(1)+(2) Both values from (1) satisfy the condition from (2): \(-1+3=2=prime\) and \(2+3=5=prime\), thus we still have two values of x. Not sufficient.

Answer: E.
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What is the value of x? 29 Nov 2013, 04:53
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Bunuel
What is the value of x?

(1) x^2 - 1 = x + 1 --> \(x^2-x-2=0\). Solving for x gives \(x=-1\) or \(x=2\). Not sufficient.

(2) x + 3 is a prime number. Clearly insufficient, consider x=2 and x=4.

(1)+(2) Both values from (1) satisfy the condition from (2): \(-1+3=2=prime\) and \(2+3=5=prime\), thus we still have two values of x. Not sufficient.

Answer: E.
Show more


I definitely understand the solution presented. However, when I did the problem,

using statement 1 ---> (x+1) (x-1) = x+1
Solving I get , x-1 = x+1/x+1 , therefore x=2 ; so I figured there was one definite solution. So should be A as answer.

Now if I were to cancel the (x+1) factor on each side of the equation, I would have gotten x = 1 as solution. But I am thinking that's not algebraically correct!

Please explain, why I should have made the equation a quadratic before I solved for x, which rightfully gives 2 values...and hence E as answer. Thanks
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What is the value of x? 29 Nov 2013, 05:00
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audiogal101
Bunuel
What is the value of x?

(1) x^2 - 1 = x + 1 --> \(x^2-x-2=0\). Solving for x gives \(x=-1\) or \(x=2\). Not sufficient.

(2) x + 3 is a prime number. Clearly insufficient, consider x=2 and x=4.

(1)+(2) Both values from (1) satisfy the condition from (2): \(-1+3=2=prime\) and \(2+3=5=prime\), thus we still have two values of x. Not sufficient.

Answer: E.


I definitely understand the solution presented. However, when I did the problem,

using statement 1 ---> (x+1) (x-1) = x+1
Solving I get , x-1 = x+1/x+1 , therefore x=2 ; so I figured there was one definite solution. So should be A as answer.

Now if I were to cancel the (x+1) factor on each side of the equation, I would have gotten x = 1 as solution. But I am thinking that's not algebraically correct!

Please explain, why I should have made the equation a quadratic before I solved for x, which rightfully gives 2 values...and hence E as answer. Thanks
Show more

If you divide (reduce) \((x+1) (x-1) = x+1\) by x+1 you assume, with no ground for it, that x+1 does not equal to zero thus exclude a possible solution (notice that both x=-1 AND x=2 satisfy the equation).

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.

What you could do is:
\((x+1) (x-1) = x+1\) --> \((x+1) (x-1)-(x+1)=0\) --> factor out x+1: \((x+1) (x-1-1)=0\) --> \((x+1) (x-2)=0\) --> \(x=-1\) or \(x=2\).

As you can see it's possible to solve the equation without making it a quadratic.

Hope it's clear.
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What is the value of x? 29 Nov 2013, 05:58
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Bunuel
audiogal101
Bunuel
What is the value of x?

(1) x^2 - 1 = x + 1 --> \(x^2-x-2=0\). Solving for x gives \(x=-1\) or \(x=2\). Not sufficient.

(2) x + 3 is a prime number. Clearly insufficient, consider x=2 and x=4.

(1)+(2) Both values from (1) satisfy the condition from (2): \(-1+3=2=prime\) and \(2+3=5=prime\), thus we still have two values of x. Not sufficient.

Answer: E.


I definitely understand the solution presented. However, when I did the problem,

using statement 1 ---> (x+1) (x-1) = x+1
Solving I get , x-1 = x+1/x+1 , therefore x=2 ; so I figured there was one definite solution. So should be A as answer.

Now if I were to cancel the (x+1) factor on each side of the equation, I would have gotten x = 1 as solution. But I am thinking that's not algebraically correct!

Please explain, why I should have made the equation a quadratic before I solved for x, which rightfully gives 2 values...and hence E as answer. Thanks

If you divide (reduce) \((x+1) (x-1) = x+1\) by x+1 you assume, with no ground for it, that x+1 does not equal to zero thus exclude a possible solution (notice that both x=-1 AND x=2 satisfy the equation).

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.

What you could do is:
\((x+1) (x-1) = x+1\) --> \((x+1) (x-1)-(x+1)=0\) --> factor out x+1: \((x+1) (x-1-1)=0\) --> \((x+1) (x-2)=0\) --> \(x=-1\) or \(x=2\).

As you can see it's possible to solve the equation without making it a quadratic.

Hope it's clear.
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Thanks a ton. Makes sense. Hopefully i will remember to not make that mistake on the test! ...
Btw I sent u a PM regarding another question...don't see an answer to that one yet!
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What is the value of x? 28 Oct 2020, 09:07
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