How do you solve #abs(2x+7)- abs(6-3x)= 8#?

1 Answer
Apr 1, 2015

#x=7/5,5#

Lets split the absolute value operators:

#1) abs(2x+7)#
#2)abs(6-3x)#

There are 4 possibilities:
- #1)# can be positive when #2)# is positive.
- #1)# can be positive when #2)# is negative.
- #1)# can be negative when #2# is positive.
- #1)# can be negative when #2)# is negative.

We have to check all these possibilities to find the solution set.

Lets start checking:

#2x+7>=0#
#6-3x>=0#

Both should be satisfied for our first condition.
When we solve these inequalities:

#A: 2>=x>=(-7/5)#

We will need to remember that.

Since we assume both absolute values are positive:

#2x+7-(6-3x)=8#
#5x=7#
#x=7/5#

We assumed that both absolute values are positive. For this to happen, #x# must have a value in #[-7/5,2]#. (Look at #A:#)
#7/5# is in the specified range. So it is a member of our solution set.

Lets continue to our work. Our second possibility is: #1)# is positive when #2)# is negative.

So lets find the range of #x#

#2x+7>=0#
#6-3x<0#
#B: x>2#

#2x+7-(-1)*(6-3x)=8#
#2x+7+6-3x=8#
#-x=-5#
#x=5#

#x# is in the range #B: (2,+oo)#. So it is in our solution set.

Be patient, there are 2 possibilities left.

When #1)# is negative, #2)# is positive (we assume).

So:

#2x+7<0#
#6-3x>=0#
#2>=x#
#x<(-7/5)#

As you can see #x# cannot be greater than #2# while it is less than #(-1,4)#. This means #1)# and #2)# cannot be negative and positive respectively. There is no value of #x# to satisfy this condition.

Our final condition: #1)# and #2)# are both negative.

#2x+7<0#
#6-3x<0#

#D: 2 < x < (-7/5)#

#(-1) * (2x + 7) - (-1) * (6 - 3x) = 8#
#-2x - 7 + 6 - 3x = 8#
#-5x=9#
#x=(-9/5)=-1.8#

#x# is not in the range #D: (-1.4,2)#. So #1)# and #2)# cannot be both negative. There is no value of #x# to satisfy this condition.

So the solution set is:

#SS = {7/5, 5 }#