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

There are multiple ways of solving this.
You can directly solve for x:
\(\frac{1}{x} - \frac{1}{(x+1)} = \frac{1}{(x+4)}\)

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

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

Cross multiply to get,
x + 4 = x(x+1)
x^2 - 4 = 0
(x-2)(x+2) = 0
x = 2 or -2

Though a much faster approach would be to use the options.
You can see that x cannot be 0/-1/-4 because then one of the terms will have 0 in the denominator. A fraction cannot have 0 in the denominator hence x must be either -2 or -3.
Put x = -2 and check:
\(\frac{1}{(-2)} - \frac{1}{(-2+1)} = \frac{1}{(-2+4)}\)
1 - (1/2) = 1/2
1/2 = 1/2

So x = -2 satisfies and hence, is the answer.

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. 0
Plug 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. -2
Plug 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: Algebra
One 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

Hi,

I think the first step is to get the variables up to the numerator. To do this we should find the common factor on the left side:

1/x - 1/(x+1) => becomes (x+1)/(x)(x+1) - (x)/(x)(x+1)
=> combines to (x+1) - (x) / (x)(x+1)
=> notice that the top two Xs cancel out 1 / (x)(x+1)

We now have the full equation: 1 / (x)(x+1) = 1 / (x+4)

We can simply swap the denominators around to have: (x+4) = x(x+1)

After we expand the equation we get (x+4) = x^2+x
Notice that the two Xs cancel out leaving us with: 4 = x^2
Thus we have the solution x = 2 or -2
ANSWER is C

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

=> 1/(x(x+1)) = 1/(x+4)

=> x^2 +x= x+ 4 => x^2 =4 => x = 2 or -2

we dont have 2 in the answer , so x = -2 .

Answer C.

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

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

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

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"

okay thanks a lot guys I now understand this method to solving it. The method that I'm still stuck on in the book is how they get it from factoring

Denominators are eliminated leading to this:
(x+1)(x+4)-x(x+4)=x(x+1)

I don't understand how it gets from that to the following:
(x+4)(x+1-x)=x(x+1)
(x+4)(1)=x(x+1)

########
a*b-a*z=k
a(b-z)=k;;;;;;;;;; Taking "a" as common from a*b and a*z
########

(x+1)(x+4)-x(x+4)=x(x+1)
(x+1)*(x+4)-x*(x+4)=x*(x+1)
Taking (x+4) common
(x+4){x+1-x} = x*(x+1)
(x+4)*1=x*(x+1)

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.

Thanks, thats actually very simple, or a lot more simple than I made it out to be.

Here too..!!
the easier and faster(depends on how much one is compatible with algebra) is to use the LCM process..

Again Personal Opinion

regards
_________________

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.
_________________

Using the arithmetic approach we can see that x can be "+" or "-" 2.

The only answer choice matches our results is C.

My 50 Cents here.

I think its much faster here to Substitute the answer choices

A. Will make \(\frac{1}{x}\) as Infinity. Not Possible.

B. Will make \(\frac{1}{x+1}\) as infinity. Not Possible.

C. \(\frac{-1}{2}\) - (-\(\frac{1}{1}\)) = \(\frac{1}{2}\). Turns out \(\frac{1}{2}\) = \(\frac{1}{2}\). Possible.

D. \(\frac{-1}{3}\)- (-\(\frac{1}{2}\)) = 1. Not Possible

E. -\(\frac{1}{4}\) - (-\(\frac{1}{3}\)) = \(\frac{1}{8}\). Not possible.

Final Answer = C.
_________________

We can try each of the options. Only (-2) will satisfy the requirment, So 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

It can be done with Algebra approach, but that will be time consuming.

As we know denominator can never be "0", then any answer choice that makes any of the three denominators "0". We can eliminate A, B, & E. Now we can test C & D.

C gives the right answer.
_________________

Hi, I am really bad in math (specially with fractions ), so I tried to solve this question but I don't know what was my mistake.

I did Like this

1/x - 1/X+1 = 1/x+4

1/x = 1/x+4 + 1/X+1

I didn't know what to do after this... so I tried to calculate it as it was.

MMC (I don't know if I can do this)
X+4,X+1 | X+4
1,X+1 | X+1
1,1 | (X+4)*(X+1) = X²+5x +4

1/x = (X²+5x +4) + (X²+5x +4) / X²+5x +4 >>> 2X²+10x+8/X²+5x +4 = 2 (I tought this equation would be just like 4/2, so = 2)

Now I have 1/x = 2 >>> 2x = 1 >>> X = 0,5...

\(x \neq 0, -1, -4\)
If x=-2
-1/2 -(-1)=1/2=1/(4-2)=1/2

IMO C

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