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

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If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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Updated on: 17 Jul 2014, 06:56
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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

Originally posted by ArmorPierce on 09 Mar 2011, 19:23.
Last edited by Bunuel on 17 Jul 2014, 06:56, edited 2 times in total.
Edited the question.
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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09 Mar 2011, 19:48
12
8
ArmorPierce wrote:
Hey guys, I'm having trouble with the below problem. Can anyone explain it to me? please explain it like you would a child step by step. Thanks.

If (1/x) - (1/(x+1)) = (1/(x+4)), then x could be
a 0
b -1
c -2
d -3
e -4

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.
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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09 Mar 2011, 20:00
1
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
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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09 Mar 2011, 20:24
(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 .

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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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09 Mar 2011, 20:34
1
$$\frac{1}{x} - \frac{1}{x+1} = \frac{1}{x+4}$$

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

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

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"
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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09 Mar 2011, 21:33
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)
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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09 Mar 2011, 22:14
1
ArmorPierce wrote:
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)
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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09 Mar 2011, 23:32
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 !
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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10 Mar 2011, 04:46
3
gmat1220 wrote:
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.
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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10 Mar 2011, 17:49
fluke wrote:
########
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)

Thanks, thats actually very simple, or a lot more simple than I made it out to be.
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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08 Mar 2016, 08:01
VeritasPrepKarishma wrote:
ArmorPierce wrote:
Hey guys, I'm having trouble with the below problem. Can anyone explain it to me? please explain it like you would a child step by step. Thanks.

If (1/x) - (1/(x+1)) = (1/(x+4)), then x could be
a 0
b -1
c -2
d -3
e -4

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.

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
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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15 Jul 2016, 04:40
1
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

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

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!)

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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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18 May 2017, 09:52
Using the arithmetic approach we can see that x can be "+" or "-" 2.

The only answer choice matches our results is C.
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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19 Aug 2017, 07:22
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.

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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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23 Aug 2017, 18:07
Top Contributor
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

Plugging in the answer choices is definitely faster, but here's an algebraic solution...

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: 1/x - 1/(x+1) = 1/(x+4)
Multiply both sides by (x)(x+1)(x+4) to get: (x)(x+1)(x+4)[1/x - 1/(x+1)] = (x)(x+1)(x+4)[1/(x+4)]
Simplify: (x+1)(x+4) - (x)(x+4) = (x)(x+1)
Expand: [x² + 5x + 4] - [x² + 4x] = x² + x
Simplify: x + 4 = x² + x
Rearrange: x² = 4
Solve: x = 2 OR x = -2

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If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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22 Jan 2018, 09:53
We can try each of the options. Only (-2) will satisfy the requirment, So answer is C
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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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16 Mar 2018, 10:49
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

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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be  [#permalink]

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16 Mar 2018, 23:07
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

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.

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Re: If 1/x - 1/(x+1) = 1/(x+4), then x could be &nbs [#permalink] 16 Mar 2018, 23:07
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