|x-2|-|x-3|=|x-5| Extremely Difficult absolute question

[matt.lopz]

[question deleted because source couldn't be confirmed. If anyone can cite the source, please feel free to re-post the problem and confirm the source. Thanks!]

Answer: C

I have read the entire manhattan inequalities chapter. But I still dont have a clue to this question. Can somebody please help me with these kind of questions?

[AndreaDB]

Very difficult only if you watch at this from the wrong way.

I approached as simple minded way.. sometimes it is the quickiest way to find a solution and so, for paced Exams, also one of the right solution.

Start with brute force and always put an eye on the answers:

1step) we have 4 ranges to analyze: __a__|2|_b_|3|_c_|4|_d_|5|__e__
2Step) Start with the solution in the "e" interval: all the absolutes are as they're presented ==> X=6
3Step) interval d: invert the sign of the last module and leave the forst two as they're: ==> X=-4
4Step) So on...
as you can soon notice you see that you are excluding answers as if it would rain.
Ans C.
Andrea

[kanaks123]

Not sure if the Answer is C.

C falls apart when we take x=5 => |5-2| - |5-3| != |5-5|

The answer is A and I have a very convoluted way of solving the problem.

set-1: |x-2| => x > -2 and x > 2 => x belongs to (-infinity, -2) or (2, infinity)
set-2: |x-3| => x belongs to (-infinity, -3) or (3, infinity)

*Where as ( means not inclusive and ] means inclusive

now |x-2| - |x-3| belongs to (-infinity, -1] or [1, infinity)
To come at the above set, I picked the boundary numbers from set-1 and set-2 above :
-1 - (-3) = 2, -3 - (-4)=1, 4 - 3 =1,....

|x-5| = some number that belongs to (-infinity, -1] or [1, infinity)

so we have two cases
case 1: x-5 > 0, therefore x-5 belongs to [1, infinity)
case 2: x-5 < 0, therefore x-5 belongs to (-infinity, -1]

Taking both cases we can say, x belongs to (-infinity, 4] or [6, infinity). It is lot easier if you place these values on number line.

[frigorificos]

The best way to answer such question is to identify the domains.
We have 4 cases to consider out here
X < 2
2 < x < 3
3< x < 5
and x > 5

Case 1 X < 2

the equation reduces to -(x-2) - (-(x-3)) = -(x-5) -----> solve for x to get 6 but X cant be 6 coz x is less than 2 so this aint a soln

Case 2 2<x<3
The equation becomes x-2 - (-(x-3) = -(x-5) --- >. solve for x as -- 3.33 - this cant be a soln as 2<x<3

case 3 - 3 < x < 5

11the equation is x-2 -(x-3) = -(x-5) solve for x as x = - 10 this aint ok

case 4 x > 5

x-2 -(x-3) = x-5 solve for x as 6 - This is ok - so soln is x=6

[matt.lopz]

Andrea,

If I substitute value "0" from solution C among answer options in the complex absolute value equation, the result does not hold true.

Stacey/Ron/Instructors,

Can you guys give us the correct and quick approach to solve this problem?

Matt

[matt.lopz]

GMAT Instructors,

Can you please help with this one?

[AndreaDB]

Not sure if the Answer is C.

C falls apart when we take x=5 => |5-2| - |5-3| != |5-5|

The answer is A and I have a very convoluted way of solving the problem.

set-1: |x-2| => x > -2 and x > 2 => x belongs to (-infinity, -2) or (2, infinity)
set-2: |x-3| => x belongs to (-infinity, -3) or (3, infinity)

*Where as ( means not inclusive and ] means inclusive

now |x-2| - |x-3| belongs to (-infinity, -1] or [1, infinity)
To come at the above set, I picked the boundary numbers from set-1 and set-2 above :
-1 - (-3) = 2, -3 - (-4)=1, 4 - 3 =1,....

|x-5| = some number that belongs to (-infinity, -1] or [1, infinity)

so we have two cases
case 1: x-5 > 0, therefore x-5 belongs to [1, infinity)
case 2: x-5 < 0, therefore x-5 belongs to (-infinity, -1]

Taking both cases we can say, x belongs to (-infinity, 4] or [6, infinity). It is lot easier if you place these values on number line.

You're right.
The method perhaps works but the results not, ever paying attention on boundaries.

Also A presents problem in backsolving. Try 0 or -6 and so on.
I think the set in the answer stem aren't correct.

Invocating for instructor :-)
Andrea

[esledge]

I would solve this one visually on a number line, with the understanding that |x-a| is distance of x from a. If you think of the equation as "dist from 2 - dist from 3 = dist from 5" you can find the answer fairly quickly. (hint: there are exactly two solutions)

Can anyone confirm the source of this problem? Is it in one of our Strategy Guides? I didn't find it after a quick search, and for copyright reasons we must cite the author of all problems. If no author is cited in the next few days, we will have to delete the post.

[jsingh1970]

I think the answer is A. If you scan teh answer choices, you see the number 5 will yeild a total of 0 on the right side but not on the left. So we know for sure that number 5 does not work. The number five exists in all answer choices except A.

[kevinmarmstrong]

Remember that we are asked to identify a set that contains all of the solutions. Not every element of this set need be a solution.

[dlou0716]

there has to be a typo somewhere. it is impossible for any one of the choices to be correct, since all of them contain at least a negative #.

1. when x is a negative number, x-2 will always be smaller than x=3. In that case, it would be impossible for the left side of the equation to be positive, while the right side of the equation will never be negative.

There has to be an error somewhere.

Dion

[dlou0716]

Remember that we are asked to identify a set that contains all of the solutions. Not every element of this set need be a solution.

oh yeah!! you are right, now it makes sense! =)

then yes, c is the right answer.

first solve the equation as if there is no abs bracket. you get x=6. when you identify the sets that contain 6, try any # 3 or bigger. Anything less will produce a negative result on the left side of the equation.

[StaceyKoprince]

Question text was deleted. Emily looked for the problem in our guides but couldn't find it. She asked the original poster (or anyone) to confirm the specific source. Nobody did.

Remember, guys, that we do have to follow copyright laws!