SOLUTION: solve the following equation {{{ |2x-6|=|x+1| }}} I remember that the absol splits into positive and negative but then I get lost. Thank you in advanced

Algebra ->  Absolute-value -> SOLUTION: solve the following equation {{{ |2x-6|=|x+1| }}} I remember that the absol splits into positive and negative but then I get lost. Thank you in advanced       Log On


   

Question 868819: solve the following equation
+%7C2x-6%7C=%7Cx%2B1%7C+
I remember that the absol splits into positive and negative but then I get lost.
Thank you in advanced

Found 2 solutions by josgarithmetic, KMST:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Rendering tags do not work on this way: |2x-6|=|x+1|
Use "abs()" instead of the vertical bars.
abs%282x-6%29=abs%28x%2B1%29 which is abs(2x-6)=abs(x+1).

This might be less efficient logic, but you may have four conditions to examine.
2x-6>=0 and x+1>=0;
2x-6<0 and x+1>=0;
2x-6>=0 and x+1<0;
2x-6<0 and x+1<0.


2x-6%3E=0 and x%2B1%3E=0;
The abs values simply give 2x-6=x+1.
x-6=1
x=6+1
x=7

2x-6%3C0 and x%2B1%3E=0;
abs%282x-6%29=abs%28x%2B1%29 gives then
-(2x-6)=x+1
-2x+6=x+1
-2x-x+6=1
-3x=1-6
-3x=-5
x=5%2F3

2x-6%3E=0 and x%2B1%3C0;
This combination will give
2x-6=-(x+1)
2x-6=-x-1
3x-6=-1
3x=-1+6
3x=5
x=5%2F3

2x-6%3C0 and x%2B1%3C0.
If both these expressions are negative, then the absolute value equation produces this ...
-(2x-6)=-(x+1)
-(2x-6)(-1)=-(x+1)(-1)
2x-6=x+1
which goes exactly the way the first combination, with both expressions positive or zero, went.
x=7

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ANSWER: highlight%28x=7%29 or highlight%28x=5%2F3%29 ****
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Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
TIP: Use "abs(2x-6)" to get abs%282x-6%29 .

abs%282x-6%29=abs%28x%2B1%29
2x-6%3E=0<--->x%3E=3 and x%2B1%3E=0<--->x%3E=-1

For the cases x%3E=3 and x%3C-1 both expressions have the same sign,
and we look for solutions to 2x-6=x%2B1
2x-6=x%2B1
2x-x=1%2B6
x=7
As x=7 is within one of the cases above.
highlight%28x=7%29 is a solution.

For the case -1%3C=x%3C3 ,
x%2B1%3E=0 so abs%28x%2B1%29=x%2B1,
but 2x-6%3C0 so {{abs(2x-6)=-(2x-6=-2x+6}}} ,
and we look for solutions to
x%2B1=-2x%2B6
x%2B2x=6-1
3x=5
x=5%2F3
Since x=5%2F3=1%261%2F3 is within the case -1%3C=x%3C3 ,
highlight%28x=5%2F3%29 is a solution.

Verification:
For x=7:
abs%282x-6%29=abs%282%2A7-6%29=abs%2814-6%29=abs%288%29=8 and
abs%28x%2B1%29=abs%287%2B1%29=abs%288%29=8

For x=5%2F3:
and
abs%28x%2B1%29=abs%285%2F3%2B1%29=abs%285%2F3%2B3%2F3%29=abs%288%2F3%29=8%2F3