I believe there is something wrong with question because -4,5 & 8 does not satisfy the inequality.

I believe testing values is the best solution for this question.

Let me know if i am wrong.
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The given equation can be rewritten as |x-2|=|x-3|+|x-5|, which means that the distance between x and 2 is the sum of the distances between x and 3 plus the distance between x and 5.
Using the number line and visualizing the points, we can easily see that x cannot be smaller than 3.

If x>5, then x-2=x-3+x-5, from which x=6. We have to choose between answers C and D.
If 3<x<5, then x-2=x-3-x+5, which yields x=4.

Answer C
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The question reads "Which of the following sets includes ALL of the solutions of x that will satisfy the equation" it doesn't say that all the elements of the set are solutions of the given equation. When solving, you can find that there are exactly two values only which are solutions: 4 and 6. And they are both included only in the set of answer C.
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Responding to a pm:

|x-2|-|x-3|=|x-5|

First and foremost, 'the difference between the distances' is a concept which is a little harder to work with as compared with 'the sum of distances'. So try to get rid of the negative sign, if possible.

|x-2| = |x-3|+|x-5|

This equation tells you the following: x is a point whose distance from 2 is equal to the sum of its distance from 3 and its distance from 5. Plot the 3 points on the number line and run through the 4 regions.



Green: For every point to the left of 2, the distance of the point from 2 is less than the distance from 3. So there is no way any point here will satisfy the equation. Distance of x from 2 will always be less than the sum of distances from 3 and 5.

Blue: Between 2 and 3, 3 is farthest from 2 i.e. at a distance of 1 but 5 is at a distance of 2 from 3. So again, no point here can satisfy the equation. Distance of x from 2 will still be less than the sum of distances from 3 and 5.

Black: Now we are going a little farther from 2 and toward 5 so there might be a point where distance from 2 is equal to sum of distances from 3 and 5. Check what happens at 4. You see that 4 satisfies the equation. At 4, the distance of x from 2 is equal to the sum of distances from 3 and 5. After 4, distance of x from 2 will be more than the sum of distances from 3 and 5 till we reach 5.

Red: After 5, as x moves to the right, its distance from 3 as well as 5 increases. So it is possible that the distance of x from 2 is again equal to the sum of distances from 3 and 5. At 5, x is 3 units away from 2 and 2 units away from 3 and 5 together. Check what happens at 6. At 6, x is 4 away from 2 and 4 away from 3 and 5 together. Hence 6 satisfies too. As you move one step to the right of 6, distance from 2 increases by 1 unit and distance from 3 and 5 together increases by 2 units. So the distances will never be the same again.
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The equation is |x-2|=|x-3|+|x-5|.
Draw the number line and place 2, 3 and 5 on it.
If x is between 2 and 3, the distance between x and 2, which is |x-2|, is less than 1, while the distance between x and 5, which is |x-5|, is greater than 2. So, the left hand side is for sure less than the right hand side of the equality.
If x is less than 2, the distance between x and 2 is the smallest, is both less than distance between x and 3 and less than the distance between x and 5. Again, equality cannot hold.
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Guys.. help me out of this doubt..

|x-2| tells us that x=2 as one checkpoint

Let us test for x where x < 2

|x-2| = |x-5| + |x-3|
-(x-2)= -(x-5) - (x-3)
-x +2 = -x +5 -x +3

x=6 Invalid since x=6 is not x<2

This shows obviously that value of x will not lie in the region x<2.
But the Answer choice C contains (-1) as one of the solution in the set. (In fact all answer choices contains a negative value).

Is my understanding wrong ??

Second that.. By test 8,6,5,4 you could arrive at the solution in less than 30 secs.
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I didnt get this part at all
I am sorry, could you please elaborate a bit more, or give a link regarding this method ?
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