ArmorPierce wrote:
If \(\frac{1}{x} - \frac{1}{x+1} = \frac{1}{x+4}\), then x could be
A. 0
B. -1
C. -2
D. -3
E. -4
STRATEGY: As with all GMAT Problem Solving questions, we should immediately ask ourselves, Can I use the answer choices to my advantage?
In this case, we can easily test each answer choice by plugging it into the given equation.
Now, we should give ourselves about 20 seconds to identify a faster approach.
In this case, solving the equation will be a total pain, so let's start testing the answer choices.
A. 0Plug in \(x = 0\) to get: \(\frac{1}{0} - \frac{1}{0+1} = \frac{1}{0+4}\)
At this point, we should recognize we have a problem.
Since \(\frac{1}{0}\) is undefined, there's no way \(x = 0\) can be a solution. So, we can eliminate answer choice A.
Similarly, \(x = -1\) and \(x = -4 \) also yield fractions with \(0\) in the denominator. So we can eliminate choices B and E as well. So, we're already down to answer choices C or D. In fact, we're only going to test one of those values. Here's why: if we test C and it satisfies the given equation, then the correct answer is C. If we test C and it doesn't satisfy the given equation, then the correct answer must be D.
C. -2Plug in \(x = 2\) to get: \(\frac{1}{2} - \frac{1}{2+1} = \frac{1}{2+4}\)
Simplify to get: \(\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)
Rewrite with common denominators: \(\frac{3}{6} - \frac{2}{6} = \frac{1}{6}\) . . . It works!
Answer: C
Alternate approach: AlgebraOne way to eliminate fractions is to multiply both sides by the least common multiple (LCM) of all the denominators.
The LCM of \(x\), \(x+1\) and \(x+4\) is \((x)(x+1)(x+4)\)
Given: \(\frac{1}{x} - \frac{1}{x+1} = \frac{1}{x+4}\)
Multiply both sides by \((x)(x+1)(x+4)\) to get: \((x)(x+1)(x+4)[\frac{1}{x} - \frac{1}{x+1}] = (x)(x+1)(x+4)[\frac{1}{x+4}]\)
Simplify: \((x+1)(x+4) - (x)(x+4) = (x)(x+1)\)
Expand: \([x^2 + 5x + 4] - [x^2 + 4x] = x^2 + x\)
Simplify: \(x + 4 = x^2 + x\)
Subtract \(x\) from both sides: \(4=x^2 \)
Solve: \(x = 2\) OR \(x = -2\)
Answer: C