# Homework Help: Absolute value Quest,

1. Sep 18, 2011

### Nelo

Absolute value Quest, urgent

1. The problem statement, all variables and given/known data
a) |x-2| + |6-3x| = 12x

2. Relevant equations

3. The attempt at a solution

I have a question about y this negetive becomes a positive.. ::

Steps ::
|x-2| + -3|x-2|
|x-2| +3 |x-2| = 12x > this step if i factored out.. |x-2|+[-3]|x-2| , shouldnt it be -3 +1 =2?? why does it become 4
4|x-2| = 12/4
|x-2| = 3x
If i factor out a -3, then add the "1" on the left side it should be -2 no? in my notes its written this way and is correct in the book, why did that -3 become positive?

2. Sep 18, 2011

### Nelo

Re: Absolute value Quest, urgent

anyone....?

3. Sep 18, 2011

### ehild

Re: Absolute value Quest, urgent

You can not replace |6-3x| by -3|x-2|. The absolute value is never negative, and you made it never positive. Factor out 3, so your equation becomes |x-2| + 3|2-x| = 12x

Replace |x-2| by 2-x if x-2<0. At the same time, 2-x>0 so |2-x|=2-x.
What have you do when x-2 >0?

ehild

4. Sep 18, 2011

### Nelo

Re: Absolute value Quest, urgent

Um... okay.. so you cant take out a negetive..

Then..

:: |x-2| + |6-3x| = 12x
:: |x-2| + 3|2-x| = 12x
:: 4|x-2| |2-x| = 12x/4
|x-2| |2-x| = 3x.

Now your telling me to replace 2-x with x-2? because they both need a value of 2 yes?

So get rid of 2-x and it becomes |x-2|=3x , is that right??

5. Sep 18, 2011

### Nelo

Re: Absolute value Quest, urgent

??? anyone

6. Sep 18, 2011

### ehild

Re: Absolute value Quest, urgent

I do not understand your questions.

|x-2|=|2-x|, so the sum is 4|x-2|=12x , so |x-2|=3x.

You have two possibilities: x<2 and x≥2. What is |x-2| in both cases?

ehild

7. Sep 18, 2011

### Nelo

Re: Absolute value Quest, urgent

The two possibilities is not My question. My question is what is happening algebraically to this question inorder for there to only be one |x+1| BRACKET. What happends do other bracket? do they join to become one?

:: |x-2| + |6-3x| = 12x
:: |x-2| + 3|2-x| = 12x
:: 4|x-2| =12x, |2-x| = 12x/4
|x-2| |2-x| = 3x???. < Why does one of the absolute brackets go away? how do you know which one?

Now your telling me to replace 2-x with x-2? because they both need a value of 2 yes?

So get rid of 2-x and it becomes |x-2|=3x , is that right??

Let me pose another question.

b) 7|x+2| = 2|x+2| +15

does this simplify into..

5|x+2| = 15

The two brackets were the same so they added and joined, but the two brackets in the first question arent the same, so why would they join and become one? why did it vanish

I know all about the cases, thats not what im asking, just tryin to figure out the algebra stuff

Last edited: Sep 18, 2011
8. Sep 18, 2011

### HallsofIvy

Re: Absolute value Quest, urgent

|6- 3x|= |-3(x- 2)|= 3|x- 2|
(That's what meant when he said "you can't take out a negatve"- but you can take out the
"3". The "negative became a positive" because you are taking the absolute value: |-3|= 3.

With that, your equation becomes "4|x- 2|= 12x" so that |x- 2|= 3x.
Of course |x-2|=|2- x|. There is no difference between them because absolute value "strips away the sign". More accurately, if $x-2\ge 0$, so that |x- 2|= x-2, then $2- x\le 0$ so that |2- x|= -(2- x)= x- 2 also. And, of course, if x- 2< 0 so that |x- 2|= -(x- 2)= 2- x, then 2- x> 0 so that |2- x= 2- x. Either way, |x- 2|= |2- x|.
More generally, |x|= |-x|.

Now, to complete the problem, consider cases:
If $x- 2\ge 0$, |x- 2|= x- 2= 4x. Solve that. Is x> 2?
If x- 2< 0, |x- 2|= -(x- 2)= 2- x= 4x. Solve that. Is x< 2