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Welcome back to this week’s Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 minutes and go!

Question

If 1/(x – 2) = 1/(x + 2) + 1/(x – 1), which of the following is a possible value of x?

(A) –2
(B) –1
(C) 0
(D) 1
(E) 2

Solution

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:
1/(–3) = 1/(1) + 1/(–2)?
–1/3 = 1 – 1/2?

This is not true.

However, if x = 0, then we have the following:
1/(–2) = 1/(2) + 1/(–1)?
–1/2 = 1/2 – 1?
–1/2 = –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 = – 3x + 2 + – 4
0 = – 4x
0 = x(x – 4)
x = 0 or x = 4

The correct answer is (C) 0.

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