Which of the following sets includes ALL of the solutions of

Hi rrsnathan,
The eqn :
|x-2|-|x-3|=|x-5|
is about modulus function.

The modulus always takes the magnitude i.e.
|2| = 2 and also |-2| = 2,
so you can see that for negative expression y ; |y | = - (y)
which is same as changing the sign , to make it positive

So,when x>5, all the modulus expression
|x-2|,|x-3|and |x-5| are positive,so we need not change sign

when 3<x<5,
|x-2| is positive , |x-3| is positive and |x-5| is negative,
so we need to change the sign for |x-5| i.e. |x-5| = - (x-5)

That's why the equation
|x-2|-|x-3|=|x-5|
becomes (x-2) -(x-3) = - (x-5) this is what Ratinarace meant, although explained confusingly

Hope above example clears your doubt

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Exactly!! you dont need to solve this one. Just plug in one option from each set and check if it works. Option A and D have 8 so plug it in and check that whether it works. It does not so discard A and D. Option B and E have 5 so plug it in. Again it does work. So we are left with option C. Dont even bother checking if all the values in option C work. Select C and move on.

I don't understand why you chose to ignore options (A) and (D) just because they had a value which is not a solution of x.

Read the question again: Which of the following sets includes ALL of the solutions of x that will satisfy the equation:... ?

This means the option which gives the correct answer "includes" all the solutions of x but it could also have other numbers which are not solutions of x.
The answer is (C) which includes both solutions of x: 4 and 6. But it also has other numbers.

The only way number plugging would work is if you put in every value of (A) - no solutions there. Then check every value of (B) - you get one solution: 4. Now look at the options which have 4. They are (C) and (E). Now check every value of (C). You find that 6 is also a solution. Now see whether (E) has 6. It doesn't. This means option (C) must have all solutions of x since one option has to be correct.

Can anyone please explain why we have changed the sign? I am seriously struggling with this issue in most of the similar problems.

The sign change is happening due to the assumed range of x.
As x is assumed to between 3 and 5, exclusive, (i.e. 3 < x < 5) the expression (x - 5) will be negative, as x is less than 5.
For that reason (x - 5) is written as - (x - 5).

Hope this answers your query.


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Good sir, what if the more than one option had the options 4 & 6, I know it had to have a zero. But other than that, getting 5 & 1. Trial and error at later stages are very tough and time consuming.

I do not understand the question and the solution at all. Why can't we check for the values by taking out x first?

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Algebra Solution/Opening the 3 Modulus

1st, reorganize the equation so it’s easier to negate the expressions inside the Modulus


[x-2] = [x - 5] + [x - 3]



the Critical Values at which the Expressions inside Each Modulus will Equal = 0 and the Barrier over which the Sign of the expression will Change ——-

Critical Values: X = 2 ; 3 ; 5

———-2——-3————5——


Scenario 1: Range 1: X >/= 5

For the Assumed X Values in this Range, all the “Input” expressions Inside Each Modulus will be NON-Negative



RULE for Opening the Modulus when the Input Expression Inside the Modulus is NON Negative:

When X > 0———- [x] = x

Thus;
(x - 2) = (x - 5) + (x - 3)

x = 6

This is a Valid Solution

Eliminate Answer Choices:

A ; B ; and E

Because they do NOT include x = 6


Scenario 2: Range 2:
3 =/< x < 5

For the Modulus [x - 5], the Input Expression is (-)Negative in this Range


RULE 2 for Opening the Modulus:

When x < 0 ———- [x] = -(x)

****why? Because the OUTPUT of an absolute value Modulus is always NON Negative, we must NEGATE the (-)Negative Input Expression to make it match the positive output.


(x - 2) = -1(x - 5) + (x - 3)

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

x = +4——VALID Solution.


Thus: 6 and 4 must be part of the Solution Set

Answer -C- is the only answer choice that contains both 6 and 4. You can eliminate every other answer.

Correct Answer must be -C-

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Hi Bunuel Bunuel

I tried to solve it this way.please advise if am wrong

(x-2)-(x-3)= -(x-5)
-1= -x+5
x=6

So no i eliminates set 1,2 and 5 as they don't have 6

(x-2)-(x-3)= (x-5)
-1=x-5
x=4

Hence i eliminated set 4 as it doesn't have 4

Now i am left with set 3 which has both 4&6

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You should have plugged back x = 6 and x = 4 to check whether they satisfy |x - 2| - |x - 3| = |x - 5| (they do). The point is that it's not clear why you considered only these two cases. By conventional approach you should consider four possible cases, \(x < 2\), \(2 \leq x \leq 3\), \(3 <x \leq 5\) and \(x > 5\), and expand the modulus according to those ranges.

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

When \(x < 2\), then we get \(-(x - 2) + (x - 3) = -(x - 5)\)\(^*\), which gives x = 6. Discard because x = 6 is out of the range we consider (\(x < 2\)).

*Because when \(x < 2\):
x - 2 is negative, thus |x - 2| = -(x - 2);
x - 3 is negative, thus |x - 3| = -(x - 3);
x - 5 is negative, thus |x - 5| = -(x - 5).

When \(2 \leq x \leq 3\), then we get \((x - 2) + (x - 3) = -(x - 5)\)\(^*\), which gives x = 10/3. Discard because x = 10/3 is out of the range we consider (\(2 \leq x \leq 3\)).

*Because when \(2 \leq x \leq 3\):
x - 2 is positive (not negative when x = 2), thus |x - 2| = x - 2;
x - 3 is negative (not positive when x = 3), thus |x - 3| = -(x - 3);
x - 5 is negative, thus |x - 5| = -(x - 5).

When \(3 <x \leq 5\), then we get \((x - 2) - (x - 3) = -(x - 5)\)\(^*\), which gives x = 4. This solutions is OK because x = 4 is in the range we consider (\(3 <x \leq 5\)).

*Because when \(3 <x \leq 5\):
x - 2 is positive, thus |x - 2| = x - 2;
x - 3 is positive, thus |x - 3| = x - 3;
x - 5 is negative (not positive when x = 5), thus |x - 5| = -(x - 5).

When \(x > 5\), then we get \((x - 2) - (x - 3) = (x - 5)\)\(^*\), which gives x = 6. This solutions is OK because x = 6 is in the range we consider (\(x > 5\)).

*Because when \(x > 5\):
x - 2 is positive, thus |x - 2| = x - 2;
x - 3 is positive, thus |x - 3| = x - 3;
x - 5 is positive, thus |x - 5| = x - 5.

So, two values of x satisfy \(|x - 2| - |x - 3| = |x - 5|\): x = 4 and x = 6.

The only set which includes both of these values is (C): {-4, 0, 1, 4, 5, 6}.

Answer: C.

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