How do you solve #abs(8 - 3x) = 11#?

1 Answer
Jose Luis F.
Jul 3, 2015

The answer is #x=-1# or #x=19/3#.

Explanation:

First of all we must remember the definition of the absolute value, which is done by cases:
If #x>0 => abs(x) = x#
 If #x<0 => abs(x) = -x#
Applying this to our question, we obtain the following:
If #(8-3x)>0 => abs(8-3x) = 8-3x#
Then, #abs(8-3x)=11 => 8-3x=11 => -3x=3 => x=-1#
 If #(8-3x)<0 => abs(8-3x) = -(8-3x) = 3x-8#
Then, #abs(8-3x)=11 => 3x-8=11 => 3x=19 => x=19/3#

Related questions
See all questions in Absolute Value
Impact of this question
1839 views around the world
You can reuse this answer
Creative Commons License