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16 Sep 2014, 01:51
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C
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Difficulty:
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Question Stats:
86% (01:07) correct
14%
(01:13)
wrong
based on 35
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If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?
A. -2
B. -1
C. 0
D. 1
E. 2
A. -2
B. -1
C. 0
D. 1
E. 2
16 Sep 2014, 01:51
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Official Solution:
If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?
A. -2
B. -1
C. 0
D. 1
E. 2
The fastest way to solve this problem is first to recognize that an algebraic approach will take a little time. Essentially, we will have to multiply through by the product \((x - 2)(x + 2)(x - 1)\), then simplify.
If, instead, we glance at the answer choices, we see that 3 of them make one of the denominators zero, a result that is not allowed (we cannot divide by zero). Specifically, \(x\) cannot be -2 because one denominator is \(x + 2\); likewise, \(x\) cannot be 1 or 2, since we have \(x - 1\) and \(x - 2\) as denominators as well.
Thus, the only two possible answers are -1 and 0. We try each in turn.
If \(x = -1\), then we have the following:
\(\frac{1}{-3} = \frac{1}{1} + \frac{1}{-2}\)?
\(-\frac{1}{3} = 1 - \frac{1}{2}\)?
This is not true.
However, if \(x = 0\), then we have the following:
\(\frac{1}{-2} = \frac{1}{2} + \frac{1}{-1}\)?
\(-\frac{1}{2} = \frac{1}{2} - 1\)?
\(-\frac{1}{2} = -\frac{1}{2}\)?
This is true, so \(x\) can be equal to 0.
Alternatively, we could take the algebraic approach.
First, we multiply through by the product \((x - 2)(x + 2)(x - 1)\) to eliminate denominators.
\((x - 1)(x + 2) = (x - 2)(x - 1) + (x - 2)(x + 2)\)
\(x^2 + x - 2 = x^2 - 3x + 2 + x^2 - 4\)
\(0 = x^2 - 4x\)
\(0 = x(x - 4)\)
\(x = 0\) or \(x = 4\)
Answer: C
If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?
A. -2
B. -1
C. 0
D. 1
E. 2
The fastest way to solve this problem is first to recognize that an algebraic approach will take a little time. Essentially, we will have to multiply through by the product \((x - 2)(x + 2)(x - 1)\), then simplify.
If, instead, we glance at the answer choices, we see that 3 of them make one of the denominators zero, a result that is not allowed (we cannot divide by zero). Specifically, \(x\) cannot be -2 because one denominator is \(x + 2\); likewise, \(x\) cannot be 1 or 2, since we have \(x - 1\) and \(x - 2\) as denominators as well.
Thus, the only two possible answers are -1 and 0. We try each in turn.
If \(x = -1\), then we have the following:
\(\frac{1}{-3} = \frac{1}{1} + \frac{1}{-2}\)?
\(-\frac{1}{3} = 1 - \frac{1}{2}\)?
This is not true.
However, if \(x = 0\), then we have the following:
\(\frac{1}{-2} = \frac{1}{2} + \frac{1}{-1}\)?
\(-\frac{1}{2} = \frac{1}{2} - 1\)?
\(-\frac{1}{2} = -\frac{1}{2}\)?
This is true, so \(x\) can be equal to 0.
Alternatively, we could take the algebraic approach.
First, we multiply through by the product \((x - 2)(x + 2)(x - 1)\) to eliminate denominators.
\((x - 1)(x + 2) = (x - 2)(x - 1) + (x - 2)(x + 2)\)
\(x^2 + x - 2 = x^2 - 3x + 2 + x^2 - 4\)
\(0 = x^2 - 4x\)
\(0 = x(x - 4)\)
\(x = 0\) or \(x = 4\)
Answer: C
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