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KarishmaB
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Which of the following sets includes ALL of the solutions of 02 Jul 2013, 21:14
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jjack0310
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Ok, so how did you plug in? I an ignoring options (A) and (B) and focusing on just (C)

Option (C) {-4, 0, 1, 4, 5, 6}

|x-2|-|x-3|=|x-5|
You put x = -4. It didn't satisfy
You put x = 0. It didn't satisfy
You put x = 1. It didn't satisfy
You put x = 4. It satisfied
You put x = 5. It didn't satisfy
You put x = 6. It satisfied

How can you say there is no other value of x that satisfies this equation?
How can you say that all solutions of this equation are included in (C)?


Calm down. Dont get excited.

I meant I plugged in all the values in all the options (a,b,c,d and e)
Option C had two values that satisfy the equations and the other options didnt have two values. So I just went with C and got the correct answer. I did say I got lucky - maybe/maybe not.

"How can you say that all solutions of this equation are included in (C)"

lol Pretty ironic - all solutions

Precisely my point that the wordings of this problem can be easily misinterpreted. Just like you did mine.
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The questions at the end 'How can you say...' were for you to mull over, not me pointing fingers. One could read them as 'How can one say ...'
Usually people stay detached and logical on this forum - they don't get excited/irritated so they stay productive.
I guess it's quite easy to misinterpret statements made online. For all the benefits online communication provides, this is certainly a folly it has.

And yes, one does need to read the question very closely to understand the meaning. The question is framed differently, not in the usual 'Which of the following represents all solutions of so and so...'
The word 'includes' helps. Which set "includes" all solutions. So all elements of the set may not be the solution but all solutions need to be in the set!
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Which of the following sets includes ALL of the solutions of 19 May 2014, 15:17
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Easy. Key points are 2,3,5. Let's try with 6. It works. Only C,D are left. Let's look for a number that can let us choose between these two. How about 8? Doesn't work. Then D is out.

C is the correct answer

Answer: C

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Which of the following sets includes ALL of the solutions of 20 May 2014, 07:33
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ratinarace
the equation has 3 intervals 2, 3 , and 5. i.e 4 segments x<2, 2<x<3, 3<x<5, 5<x.

1) When X>5 none of the brackets will change the sign and hence
(x-2)-(x-3) = (x-5)-------> x=6 ..We accept this solution as it satisfies the equation.

Now since we know x=6 is one of the solution we know the answer could be C or D

2) When 3<x<5 here (x-5) will change the sign
(x-2)-(x-3) = -(x-5) ------>x=4 ..We accept the solution as 3<4<5

We can stop solving here as the only set consisting both 6 and 4 is set C hence correct answer is C
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Hi,

I have a doubt on this

How can we say X>5 none of the brackets will change the sign and When 3<x<5 here (x-5) will change the sign?

Wat is it to define this?
Please help me in this

Thanks in advance.
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Which of the following sets includes ALL of the solutions of 20 May 2014, 08:34
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rrsnathan
ratinarace
the equation has 3 intervals 2, 3 , and 5. i.e 4 segments x<2, 2<x<3, 3<x<5, 5<x.

1) When X>5 none of the brackets will change the sign and hence
(x-2)-(x-3) = (x-5)-------> x=6 ..We accept this solution as it satisfies the equation.

Now since we know x=6 is one of the solution we know the answer could be C or D

2) When 3<x<5 here (x-5) will change the sign
(x-2)-(x-3) = -(x-5) ------>x=4 ..We accept the solution as 3<4<5

We can stop solving here as the only set consisting both 6 and 4 is set C hence correct answer is C

Hi,

I have a doubt on this

How can we say X>5 none of the brackets will change the sign and When 3<x<5 here (x-5) will change the sign?

Wat is it to define this?
Please help me in this

Thanks in advance.
Show more

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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Which of the following sets includes ALL of the solutions of 17 Jul 2014, 14:18
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WholeLottaLove
Why can we not plug in the answer choices into the question stem to solve that way?
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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.
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Which of the following sets includes ALL of the solutions of 17 Jul 2014, 21:22
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bhidu
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Why can we not plug in the answer choices into the question stem to solve that way?


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.
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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.
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Which of the following sets includes ALL of the solutions of 04 Jun 2018, 04:07
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ratinarace
the equation has 3 intervals 2, 3 , and 5. i.e 4 segments x<2, 2<x<3, 3<x<5, 5<x.

1) When X>5 none of the brackets will change the sign and hence
(x-2)-(x-3) = (x-5)-------> x=6 ..We accept this solution as it satisfies the equation.

Now since we know x=6 is one of the solution we know the answer could be C or D

2) When 3<x<5 here (x-5) will change the sign
(x-2)-(x-3) = -(x-5) ------>x=4 ..We accept the solution as 3<4<5

We can stop solving here as the only set consisting both 6 and 4 is set C hence correct answer is C
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Can anyone please explain why we have changed the sign? I am seriously struggling with this issue in most of the similar problems.
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Which of the following sets includes ALL of the solutions of 04 Jun 2018, 04:21
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ratinarace


2) When 3<x<5 here (x-5) will change the sign
(x-2)-(x-3) = -(x-5) ------>x=4 ..We accept the solution as 3<4<5

We can stop solving here as the only set consisting both 6 and 4 is set C hence correct answer is C
Show more

101mba101

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

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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Which of the following sets includes ALL of the solutions of 26 Sep 2020, 15:04
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mbaiseasy
1) Get the checkpoints using the absolute value expressions.
|x-2| tells us that x=2 as one checkpoint
|x-3| tells us that x=3 as another checkpoint
|x-5| tells us that x=5 as last checkpoint

2) 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

3) Let us test for x where 2 < x < 3

(x-2) = -(x-5) -(x-3)
x-2 = -x+5 -x +3
3x = 10
x = 10/3 Invalid since x=3.3 is not within x = (2,3)

4) Let us test for x where 3 < x < 5

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

5) Let us test for x where x > 5

x-2 = x-3 + x - 5
-x = -6
x=6 Valid

6) Now let's look for 4 and 6 in the sets of answer choices.

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

Only C has both x=4 and x=6.

Answer: C
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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.
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Which of the following sets includes ALL of the solutions of 27 Sep 2020, 03:33
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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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Which of the following sets includes ALL of the solutions of 07 Oct 2020, 09:38
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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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Which of the following sets includes ALL of the solutions of 08 Feb 2023, 03:36
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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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Which of the following sets includes ALL of the solutions of 08 Feb 2023, 11:02
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puneetfitness
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

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

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}

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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Which of the following sets includes ALL of the solutions of 17 May 2024, 05:36
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KarishmaB
Usually, plugging in numbers will be the best option but things are a little complicated here. Let me point out one thing:

"Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?" means you need to find ALL the solutions and some members of the set may not be a solution. So say, you check for 8 and see that it doesn't satisfy the equation, it doesn't mean that options (A) and (D) are out of the running. It may be one of the numbers which do not satisfy the equation but the set (A) or (D) may still contain all the solutions. With a lot of different numbers, it could get confusing and complicated.

Anyway, it all depends on your comfort with the various methods.
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­Bunuel and KarishmaB

So I was working with this problem, 

Wanted to check if the below is the correct method to find the solution where I want to see at what values of x we have LHS = RHS. 

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

1st. (x - 2) - (x - 3) = (x - 5)

Solving it will give me x =  6. 

2nd. (x - 2) - (x - 3) = - (x - 5)

Solving it will give me x = 4. 

 
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Which of the following sets includes ALL of the solutions of 11 Nov 2024, 01:08
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To solve the equation |x - 2| - |x - 3| = |x - 5|, we consider cases based on the values where each absolute value term changes, at x = 2, x = 3, and x = 5.
Case 1: x < 2
In this range, |x - 2| = 2 - x, |x - 3| = 3 - x, and |x - 5| = 5 - x.
Substitute into the equation:
(2 - x) - (3 - x) = 5 - x
Simplify: -1 = 5 - x
Solving this gives x = 6, which does not satisfy x < 2, so there is no solution in this range.
Case 2: 2 <= x < 3
In this range, |x - 2| = x - 2, |x - 3| = 3 - x, and |x - 5| = 5 - x.
Substitute into the equation:
(x - 2) - (3 - x) = 5 - x
Simplify: 2x - 5 = 5 - x
Solving this gives x = 10/3, which does not satisfy 2 <= x < 3, so there is no solution in this range.
Case 3: 3 <= x < 5
In this range, |x - 2| = x - 2, |x - 3| = x - 3, and |x - 5| = 5 - x.
Substitute into the equation:
(x - 2) - (x - 3) = 5 - x
Simplify: 1 = 5 - x
Solving this gives x = 4, which satisfies 3 <= x < 5. So x = 4 is a solution.
Case 4: x >= 5
In this range, |x - 2| = x - 2, |x - 3| = x - 3, and |x - 5| = x - 5.
Substitute into the equation:
(x - 2) - (x - 3) = x - 5
Simplify: 1 = x - 5
Solving this gives x = 6, which satisfies x >= 5. So x = 6 is also a solution.
Final Solution Set: The solutions are x = 4 and x = 6.
The set that includes both solutions x = 4 and x = 6 is: {-4, 0, 1, 4, 5, 6}
Answer: {-4, 0, 1, 4, 5, 6}

imhimanshu
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

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

(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}
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Which of the following sets includes ALL of the solutions of 11 Nov 2024, 01:09
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To solve the equation |x - 2| - |x - 3| = |x - 5|, we consider cases based on the values where each absolute value term changes, at x = 2, x = 3, and x = 5.
Case 1: x < 2
In this range, |x - 2| = 2 - x, |x - 3| = 3 - x, and |x - 5| = 5 - x.
Substitute into the equation:
(2 - x) - (3 - x) = 5 - x
Simplify: -1 = 5 - x
Solving this gives x = 6, which does not satisfy x < 2, so there is no solution in this range.
Case 2: 2 <= x < 3
In this range, |x - 2| = x - 2, |x - 3| = 3 - x, and |x - 5| = 5 - x.
Substitute into the equation:
(x - 2) - (3 - x) = 5 - x
Simplify: 2x - 5 = 5 - x
Solving this gives x = 10/3, which does not satisfy 2 <= x < 3, so there is no solution in this range.
Case 3: 3 <= x < 5
In this range, |x - 2| = x - 2, |x - 3| = x - 3, and |x - 5| = 5 - x.
Substitute into the equation:
(x - 2) - (x - 3) = 5 - x
Simplify: 1 = 5 - x
Solving this gives x = 4, which satisfies 3 <= x < 5. So x = 4 is a solution.
Case 4: x >= 5
In this range, |x - 2| = x - 2, |x - 3| = x - 3, and |x - 5| = x - 5.
Substitute into the equation:
(x - 2) - (x - 3) = x - 5
Simplify: 1 = x - 5
Solving this gives x = 6, which satisfies x >= 5. So x = 6 is also a solution.
Final Solution Set: The solutions are x = 4 and x = 6.
The set that includes both solutions x = 4 and x = 6 is: {-4, 0, 1, 4, 5, 6}
Answer: {-4, 0, 1, 4, 5, 6}
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Which of the following sets includes ALL of the solutions of 12 Nov 2025, 23:30
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imhimanshu

Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

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

Case 1: x < 2
(2-x) - (3-x) = 5-x
x = 6; Not feasible since x<2

Case 2: 2<x<3
(x-2) - (3-x) = 5-x
2x - 5 = 5- x
3x = 10
x = 10/3 = 3.33 > 3 : Not feasible since 2<x<3

Case 3: 3<x<5
(x-2) - (x-3) = 5-x
x = 4; Feasible

Case 4: x>5
(x-2) - (x-3) = x-5
x = 6; Feasible

x = {4, 6}

IMO C
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