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 }