How do you solve #|2x - 4| + 2> 8#?

1 Answer
Mar 3, 2018

#x>5 or x<-1#

Explanation:

In order to solve:

#|2x−4|+2>8# ,

Find a way to get #x# on a side by itself. (Isolate the absolute value)

First, subtract #2# from both sides to cancel it from the left side.

#|2x−4|+cancel(2) -cancel(2)> 8 -2#

Giving you:

#|2x−4| > 6#

Now, set up an OR statement.
This looks like this:

(absolute value quantity) > (number on other side)
OR
(absolute value quantity) < -(number on other side)

Which would look like this when substituted:

#|2x−4| > 6#
OR
#|2x−4| < - 6#

---

Now, we solve for each by taking away the absolute value signs.

  1. #2x−4 > 6#
    [add 4 to both sides]
    #2xcancel(-4)cancel(+4) >6 +4#
    [divide by 2 on both sides]
    #(cancel(2)x)/cancel(2) > 10/2#
    [simplify]
    Solution : #x>5#
    .
    .

  2. #2x−4< - 6#
    [add 4 to both sides]
    #2xcancel(-4)cancel(+4) < -6 +4#
    [divide by 2 on both sides]
    #(cancel(2)x)/cancel(2) < (-2)/2#
    [simplify]
    Solution : #x<-1#

Now, we combine both solutions to get the final answer.

#x>5 or x<-1#