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Tough and Tricky questions: Algebra.
If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?
A. -2
B. -1
C. 0
D. 1
E. 2
Kudos for a correct solution.
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82vkgmat
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Bunuel
Answer = C.
IMO simple way to look at this is if x = 2, -2 or 1, then one of the fractions listed becomes invalid. So none of these can be answers. That leaves us with -1 or 0.
Plug in each of these. I started with 0 first as it is easiest
LHS = -1/2
RHS = 1/2-1/1 = -1/2.
LHS = RHS, hence x=0.
Answer: C.
PareshGmat
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Answer = C = 0
\(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\)
\(\frac{1}{x-1} = \frac{1}{x-2} - \frac{1}{x+2}\)
\(\frac{1}{x-1} = \frac{x+2-(x-2)}{x^2 - 4}\)
\(\frac{1}{x-1} = \frac{4}{x^2 - 4}\)
\(x^2 - 4 = 4x - 4\)
x = 0 OR x = 2
If we take x = 2; LHS becomes void
hence x = 0
\(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\)
\(\frac{1}{x-1} = \frac{1}{x-2} - \frac{1}{x+2}\)
\(\frac{1}{x-1} = \frac{x+2-(x-2)}{x^2 - 4}\)
\(\frac{1}{x-1} = \frac{4}{x^2 - 4}\)
\(x^2 - 4 = 4x - 4\)
x = 0 OR x = 2
If we take x = 2; LHS becomes void
hence x = 0
General Discussion
Posts: 32
Just put the value of x from the given options and check
we can verify equation is valid only for x=0
Hence C
we can verify equation is valid only for x=0
Hence C
Official Solution:
If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?
A. -2
B. -1
C. 0
D. 1
E. 2
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:
\(\frac{1}{-3} = \frac{1}{1} + \frac{1}{-2}\)?
\(-\frac{1}{3} = 1 - \frac{1}{2}\)?
This is not true.
However, if \(x = 0\), then we have the following:
\(\frac{1}{-2} = \frac{1}{2} + \frac{1}{-1}\)?
\(-\frac{1}{2} = \frac{1}{2} - 1\)?
\(-\frac{1}{2} = -\frac{1}{2}\)?
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.
\((x - 1)(x + 2) = (x - 2)(x - 1) + (x - 2)(x + 2)\)
\(x^2 + x - 2 = x^2 - 3x + 2 + x^2 - 4\)
\(0 = x^2 - 4x\)
\(0 = x(x - 4)\)
\(x = 0\) or \(x = 4\)
Answer: C.
_________________
If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?
A. -2
B. -1
C. 0
D. 1
E. 2
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:
\(\frac{1}{-3} = \frac{1}{1} + \frac{1}{-2}\)?
\(-\frac{1}{3} = 1 - \frac{1}{2}\)?
This is not true.
However, if \(x = 0\), then we have the following:
\(\frac{1}{-2} = \frac{1}{2} + \frac{1}{-1}\)?
\(-\frac{1}{2} = \frac{1}{2} - 1\)?
\(-\frac{1}{2} = -\frac{1}{2}\)?
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.
\((x - 1)(x + 2) = (x - 2)(x - 1) + (x - 2)(x + 2)\)
\(x^2 + x - 2 = x^2 - 3x + 2 + x^2 - 4\)
\(0 = x^2 - 4x\)
\(0 = x(x - 4)\)
\(x = 0\) or \(x = 4\)
Answer: C.
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There is always 2 method to solve these question in most cases either putting values or solving equation...
I need expert opinion which is the fastest way to do it
I usually find putting values fastest not sure about precision what you people say?
I need expert opinion which is the fastest way to do it
I usually find putting values fastest not sure about precision what you people say?
Posts: 177
An easy approach to solve the problem is process elimination.
If we use 0 for x
1/0 - 2 = 1/0+2 + 1/0-1
1/-2 = 1/2 + 1/-1
-1/2 = -1/2
Try the other answers and you will see the answers do not coincide.
The answer is C
If we use 0 for x
1/0 - 2 = 1/0+2 + 1/0-1
1/-2 = 1/2 + 1/-1
-1/2 = -1/2
Try the other answers and you will see the answers do not coincide.
The answer is C
Bunuel
By substituting the value we 0 will satisfy the given equation
Hence C
danbrown9445
Joined: 01 Feb 2021
Last visit: 11 Nov 2023
Posts: 46
GMAT 1: 600 Q36 V38
GMAT 2: 710 Q49 V36
Products:
GMAT 2: 710 Q49 V36
Posts: 46
PareshGmat
I dont see how x = 2.
\(x^2 - 4 = 4x - 4\)
\( x^2 = 4x \)
\(x(x-4) = 0\)
x = 0 or x = 4 NOT x = 2 or x = 0
GMAT 1: 650 Q49 V31
Posts: 16
An easy way will be to go with the options.
Put the values of x from the options and verify for the correct answer.
Put the values of x from the options and verify for the correct answer.
vichitravir
Also, to add - you only have to evaluate 2 options - B and C. All the other options would make one of the terms undefined (0 in the denominator).
I'm no expert but here's my train of thoughts:
Right from the get go, I noticed that x can not equal 2; -2; 1 because that would result in operation where we divide by zero which is undefined.
This effectively leaves us with only 2 viable options --> we plug in values.
If we plug 0 for x in the equation, we get:
-1/2 = 1/2 - 1/1
-1/2 = -1/2 which means we found a valid solution. Plugging in -1 for x would result in invalid solution such as -1/3 = 1/2, hence, 0 is the correct answer found within a minute.
Right from the get go, I noticed that x can not equal 2; -2; 1 because that would result in operation where we divide by zero which is undefined.
This effectively leaves us with only 2 viable options --> we plug in values.
If we plug 0 for x in the equation, we get:
-1/2 = 1/2 - 1/1
-1/2 = -1/2 which means we found a valid solution. Plugging in -1 for x would result in invalid solution such as -1/3 = 1/2, hence, 0 is the correct answer found within a minute.
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