If 1/x - 1/(x+1) = 1/(x+4), then x could be

\(\frac{1}{x} - \frac{1}{x+1} = \frac{1}{x+4}\)

\(\frac{x+1-x}{x(x+1)} = \frac{1}{x+4}\)

\(\frac{1}{x(x+1)} = \frac{1}{x+4}\)

Don't need to factorize any further.

Just by looking at the denominator; we can say x can't be 0,-4 or -1; a,b,e are ruled out as the answer.

Let's substitute x=-2 in the equation;

\(\frac{1}{2} = \frac{1}{2}\)

No need to look for "d".

Ans: "c"

Start from option C.
LHS = a. RHS = b. a should be b
If the answer is too low, go down the options. If the answer is too high, go up the options. You will hit the answer pretty soon !

0, -1 and -4 cannot be the answers since denominator cannot be 0 and each of these values makes one of the denominator 0.
Just check for -2 and -3.

We must first eliminate the fractions in the equation 1/x - 1/(x+1) = 1/(x+4).

Thus, we will multiply the entire equation by the least common multiple of the denominators, which is:

x(x+1)(x+4)

We are now left with:

(x+1)(x+4) - x(x+4) = x(x+1)

x^2 + 5x + 4 - x^2 - 4x = x^2 + x

x + 4 = x^2 + x

4 = x^2

√4 = √x^2

x = 2 or x = -2

The answer is C.

Another option is to backsolve, substituting the answer choices into the given equation:

First, we can eliminate choices A, B and E because any one of them will make one of the denominators equal to 0 and we can’t have denominator = 0. Choice A will make the denominator of the first fraction on the left hand side of the equation equal to 0; choice B will make the denominator of the second fraction on the left hand side of the equation equal to 0; and choice E will make the denominator of the fraction on the right hand side of the equation equal to 0.

So we only need to test choices C and D.

C. -2

1/(-2) – 1/(-2+1) = 1/(-2+4) ?

-1/2 - 1/(-1) = 1/2 ?

-1/2 + 1 = 1/2 ?

1/2 = 1/2 (Yes!)

We see that C is the correct choice, but let’s show that D will not be the correct choice.

D. -3

1/(-3) – 1/(-3+1) = 1/(-3+4) ?

-1/3 - 1/(-2) = 1/1 ?

-1/3 + 1/2 = 1 ?

1/6 = 1 (No!)

The answer is C.

Hi All,

I'm a big fan of Brent's approach to this question (TESTing the ANSWERS); on certain Quant questions on the GMAT, strategically plugging in the answers to find the one that matches will get the correct answer faster than the traditional "math approach."

There is one additional Number Property in this question that meshes with what Brent already showed you: one involving Common Denominators….

Here, we're subtracting one fraction from another to get a third fraction.

Notice that the first two fractions are over "X" and over "X+1"…

If X = odd, then X+1 = even
If X = even, then X+1 = odd

With one even and one odd, we'll end up with a common denominator that is EVEN…

So, X+4 = EVEN

Thus, X MUST be EVEN

Combining this deduction with Brent's eliminations, the correct answer would have to be

GMAT assassins aren't born, they're made,
Rich

Sorry to bump a really old thread it seems my math foundations are lacking..

Is \frac{1}{-2} - \frac{1}{(-2+1)} becomes -\frac{1}{2} and -1? How do you get 1-(1/2)? I appreciate your insight.

\(\frac{1}{(-2)} - \frac{1}{(-2+1)} = \frac{1}{(-2+4)}\)

\(-\frac{1}{2} - \frac{1}{-1} = \frac{1}{2}\)

\(-\frac{1}{2} + 1 = \frac{1}{2}\)

\(\frac{1}{2}= \frac{1}{2}\)

Hope it helps.

Helps a lot, thank you.