For what value of x, is |x – 3| + |x + 1| + |x| = 10?

Going the algebra way, you have missed two cases. There are 3 transition points: -1, 0, 3

When x > 3,
\(|x - 3| + |x + 1| + |x| = 10\)
\(x - 3 + x + 1 + x = 10\)
x = 4
Satisfies

When 0 < x < 3
\(|x - 3| + |x + 1| + |x| = 10\)
\(-(x - 3) + (x + 1) + x = 10\)
x = 6 (Not possible since 0 < x < 3)

When -1 < x < 0
\(|x - 3| + |x + 1| + |x| = 10\)
\(-(x - 3) + -(x + 1) + x = 10\)
x = -8 (Not possible since -1 < x < 0)

When x < -1
\(|x - 3| + |x + 1| + |x| = 10\)
-(x - 3) - (x+1) - x = 10
x = -8/3
Satisfies

Better is the number line method.

The question has been picked from here. Mind you, I have discussed two different questions:
1. For what value of x, is |x – 3| + |x + 1| + |x| = 10?
2. For how many values of x, is |x – 3| + |x + 1| + |x| = 10?

The method to be used is different in the two cases. As Ron mentioned above, just plug in the options in the first question. In the second question, you can use the number line to quickly solve it. Check out the post to avoid confusion.
_________________

I think you meant for the answer to be D....


|4 - 3| + |4 + 1| + |4|

|1| + |5| + |4| = 10

‘For how many values of x, is |x – 3| + 3|x + 1| + |x| = 4?’ The answer is 0. Why?

Answer cannot be C. The above equation only works with option D i.e. x=4.

This should define x as and integer. I would be careful with your sources...

Often on these types of questions you're better off just plugging in answer choices and seeing which one is correct. Once you plug in 3 and get (3-3) + (3+1) + 3 = 7, you can figure out that if you added one more to X, you'd get a new answer that's 3 higher than your old answer. However, even if you don't overthink anything and just plug the choices in in succession, you still get the right answer in well under 2 minutes. If you're struggling with these types of questions, keep it simple and you'll be fine.

Hope this helps!
-Ron

Here you can:

a) test each answer choice OR
b) take the algebra way

Both are fast but it depends how you are comfortable but if you start thinking about the positive side of absolute value ---> \(x - 3 + x + 1 + x = 10\) ---> \(3x = 12 x = 4\) OR\(- x + 3 -x - 1 - x = 10\) ----> \(- 3x = 8\) NOT POSSIBLE

Only D takes in account. in no more than 50 seconds - 60 at maximum

regards
_________________

There is no reason to define x as an integer.
There is no value of x which satisfies this equation - integer or real number.
_________________

There are various different questions in this post which is creating confusion. To get the correct answer of each one of them, kindly check my post and the responses below it:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/01 ... s-part-ii/
_________________

Thanks Mod to point out a detailed explanatio.

I know that for each absolute value you have to considere a sign positive and negative. However, as soon as possible you realize that on two value there is 4 (among the answers) and -8/3 is almost impossible to have another solution for two reasons:

1) we are in gmat land and each question is constructed in a certain way. (but of course)

2) is a intuitive method: doing a lot of practice then you know how the things work because is always the same thing, actually.

I mean: it's like in CR: I always read the stimulus and after the question stem but often when I finish the stimulus I know from the tone of the question, from how the same unfold and so on what will be the question: weaken and so on......a do not read the stem at all, simply because for the logic of the question, the same must go in one direction instead of another.

Anyway, Thanks . It's always useful to repeat over and over again the concepts (and I'm a huge fan of you).
_________________

Hi karisma
Whats the difference between for what value of x &for how many values of x.........

I think this question is not hard for us to explain: when x >3, when 0 < x < 3, .... If we write exam, we don't have much time to do like this.
I think we should use the answers into equation, we have the answer faster than divide into some parts (x > 3, 0 < x < 3)
Answer is 4

'For what value of x....' implies you are looking for one particular value of x. There could be 2 or more values satisfying this equation but you are looking for only one - the one which is included in your options.

x can take two values here: 4 and -8/3

'For what value of x....' will be answered by 4. 4 is one of the values that satisfy this equation. The other one, -8/3, is not in the options which is logical since only one option can be correct. This question is generally simpler since you can plug in the options to see which value satisfies the equation.

'For how many values of x ...' means 'how many values can x take?' The answer here will be 2. x can take 2 values: 4 and -8/3. This question is usually more complex. You need to find the number of values x can take. The options will be something like: 0, 1, 2, 3, 'More than 3'. Here, you cannot plug in the options since the options do not give you the values x can take. They only give you the number of values. So you actually need to solve the equation to get how many values x can take.
_________________

For what value of x, is |x – 3| + |x + 1| + |x| = 10?

It's easiest just to plug in answer choices:

(D): 4
|x – 3| + |x + 1| + |x| = 10?
|4 - 3| + |4 + 1| + |4| = 10?
|1| + |5| + |4| = 10

(D)

Karishma - could you elaborate on the transition points please? What are they and why it is helpful to find them. Thanks in advance.

When you are dealing with |x - a|, |x - b| etc, the transition points are a, b etc. Why are they called transition points? Because |x - a| behaves differently when x < a and when x >= a.

To know how and why exactly they do that, check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/06 ... -the-gmat/
_________________

Sol: The question asks about the sum of magnitudes. I have tried to accumulate all the scenarios below. Hope this helps

I just put all the Absolute value as Positive since the addition of these three Absolute value belongs to positive.Hence , my answer is D

substitute the value of x from the options. That is the shortest way to solve this question.