When we have equation of a type: |x|=3x-2, we should consider two cases:

1. x<0 --> |x|=-x --> -x=3x-2 --> x=\frac{1}{2}, not a vaild solution for this range as we are considering x<0. Which means that when x<0 equation |x|=3x-2 has no real roots.
2. x\geq{0} --> |x|=x --> x=3x-2 --> x=1, good as x>0.
Final answer: equation |x|=3x-2 has one root x=1.

BUT the problem above can be solved in another way:
You should notice following: 3x-2 is equal to "something" and that "something", as it is an absolute value, can not be negative. So, we conclude that 3x-2 can not be negative. 3x-2\geq{0} --> x\geq{\frac{2}{3}}. After that when we definitely know that x\geq{\frac{2}{3}}>0, we can check only ONE range for x, which is x\geq{\frac{2}{3}}>0 and write |x|=x. So, equation will become x=3x-2 --> x=1.

But if it were: x=|3x-2|, we can definitely say that x\geq{0}, but we can not write x=3x-2. We should still check two ranges:
1. 0\leq{x}\leq{\frac{2}{3}} --> x=-3x+2 --> x=\frac{1}{2};
2. \frac{2}{3}<x -->x=3x-2 --> x=1.
So, this equation has two roots 1/2 and 1.

Third case: |x|=|3x-2|. In this case we should consider 3 ranges:
1. x<0 --> -x=-3x+2 --> x=1, not good as x<0;
2. 0\leq{x}\leq{\frac{2}{3}} --> x=-3x+2 --> x=\frac{1}{2};
3. \frac{2}{3}<x --> x=3x-2 --> x=1.
Again two roots 1/2 and 1.


Back to our original question:
Is |x|+|x-1|=1?
(1) x ≥ 0
(2) x ≤ 1

Absolute value, |x|, can not be negative, BUT x in it can be. |x|\geq{0} but x can be -3<0 --> |-3|=3>0.

Basically here we have the sum of two absolute values, or the sum of two non negative values totaling 1. But again x in it can take negative values and still these two can give us 1 as their sum.

Q: is |x|+|x-1|=1?
This equation can be true in some ranges of x and false in another, or can be true/false in all of the ranges. Thus we should check ALL ranges for x to answer the question. How to do this? There are two crucial points when absolute values |x| and |x -1| flip signs, two check points 0 (x=0) and 1 (x-1=0, x=1). Thus three ranges must be checked:

1. x<0 --> -x-x+1=1 --> x=0. Not good as x<0;

2. 0\leq{x}\leq{1} --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0\leq{x}\leq{1}, equation |x|+|x-1|=1 holds true.

3. x>1 --> x+x-1=1 --> x=1. Not good as x>1.

Note that we aren't considering the statements (1) and (2) yet. We are just checking in which range the equation |x|+|x-1|=1 holds true. And from 1. 2. and 3. we get that it's true ONLY in the range 0\leq{x}\leq{1} and not true in all other ranges. But at this stage we still don't know in which range x is.

Moving to the statements:
(1) x ≥ 0. Not sufficient, as x must be also <=1
(2) x ≤ 1. Not sufficient, as x must be also >=0.

(1)+(2) 0\leq{x}\leq{1}, exactly the range we needed. Sufficient.

Answer: C.

Hope it's clear.
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I got C by substituting some numbers:

Is |x| + |x -1| = 1? lets call this the equation.

(1) x ≥ 0
(2) x ≤ 1

S1) X>=0 when x= 1, equation holds true, when x =0 equation holds true,but when x = 3 equation doesn't hold true - insuff

S2) x<=1 when x=0 equation holds true, x=-1 equation doesn't hold true - insuff

Combining S1 and S2, we have 0<=x<=1

and using 0, 1 for x we know equation holds true, try 0.5, 0.3 and 0.7 for equation , still holds true - hence C is correct.

I like Bunuel's explanation though - as almost always..
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Hi Bunuel,
can you please explain:
0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true.

we cannot derive anything in this interval, does it mean that all values in this interval satisfy the equation ? This is something new for me...do you have any links for this? I thought, since we cannot derive anything, this interval is also out of scope.

Though, I got the answer by some quick number substitutions.

bunuel /economist,

Can you please clear my concept here:-

Is |x| + |x -1| = 1? i thought that an absolute can not be negative , so x can not be less than 0

like in case of |x| = 3x – 2 , absolute value can not be negative

why did you consider x <0 ? can you please explain or point me to some document , may be this is a basic question , but damn i am confused
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Always tag your question

x=|3x-2|

Left Hand Side is x;
RHS is |3x-2|;

RHS is absolute value, which means that it's never negative. If RHS\geq{0} (more than or equal to zero), hence LHS, x, must be also \geq{0}. x\geq{0}.

Actually we should check x=|3x-2| for two ranges.
1. 3x-2\leq{0} --> x\leq{\frac{2}{3}} (Note I'm saying that 3x-2\leq{0}, not |3x-2|\leq{0}, the something that is in || can be negative, but the absolute value of 3x-2 which is |3x-2| can not.)
2. 3x-2>0 --> x>\frac{2}{3} (I put = in first, because |3x-2| can be zero thus zero must be included in either of ranges we check)

So TWO ranges x\leq{\frac{2}{3}} and x>\frac{2}{3}.

BUT since we already determined that x\geq{0}, we could narrow the first range x\leq{\frac{2}{3}} to 0\leq{x}\leq{\frac{2}{3}}. In fact if you write just x\leq{\frac{2}{3}} it won't affect anything in this case, I just wanted to demonstrate what you can notice at first glance when looking at the equations of these type. Meaning, when we have x=|3x-2|, it should be clear that as RHS is absolute value, thus LHS, x must be more or equal to zero, x\geq{0}.

The way you've written is almost right:
3x-2>0 ---> x>\frac{2}{3};
3x-2<0 ---> x<\frac{2}{3};

You should just add = sign in one of them, as to include possibility that 3x-2 can be equal to 0.
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could it get any better. Thank a lot for this explaination

We know that for |x|:
When x\leq{0}, then |x|=-x;
When x\geq{0}, then |x|=x.

We have |x| + |x -1| = 1.

Now for the range: 0\leq{x}\leq{1} --> |x|=x (as x in given range is positive) and |x-1|=-(x-1)=-x+1 (as expression x-1 in the given range is negative, to check this try some x from this range, let x=-0.5 then x-1=0.5-1=-0.5=negative). So |x| + |x -1| = 1 in this range becomes: x-x+1=1 --> 1=1, which is true. That means that for ANY value from the range 0\leq{x}\leq{1}, equation |x| + |x -1| = 1 holds true.

Hope it's clear.
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Are you referring to the original question? If yes then given equation |x| + |x -1| = 1 holds true ONLY in the range 0\leq{x}\leq{1}, so it's not necessary for x to be an integer, it can be for example: 1/2 or 3/4, basically any fraction in the range (0,1).
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I got C. I plugged in numbers for each statement.
If x=0, then true. If x=1, then true. If x=2, then not true. S1 not sufficient
If x=1, then true. If x=-2, then not true. S2 not sufficient
if x is between 0 and 1 inclusive that means we plug in fractions (plus we already know that it's true for 0 and 1). No matter what fraction x represents 1-x will always give the value needed to add to x to make it = 1. Thus C is sufficient.
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Is |x|+|x-1|=1?
(1) x>=0
(2) x<=1

Question is from GWD. This should a middle I would like to know if there is any other approach.