Why Does the Negative Sign Become Positive in Absolute Value Equations?

[Nelo]

Absolute value Quest, urgent

Homework Statement

a) |x-2| + |6-3x| = 12x

Homework Equations

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| , shouldn't 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?

 

[Nelo]

anyone...?

 

[ehild]

Homework Statement

a) |x-2| + |6-3x| = 12x


Steps ::
|x-2| + -3|x-2|

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

 

[Nelo]

Um... okay.. so you can't 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??

 

[Nelo]

? anyone

 

[ehild]

Um... okay.. so you can't take out a negetive..

Then..

:: |x-2| + |6-3x| = 12x
:: |x-2| + 3|2-x| = 12x
:: 4|x-2| =12x, |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??

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

 

[Nelo]

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, that's not what I am asking, just tryin to figure out the algebra stuff

 

[HallsofIvy]

|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