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

Apr 1, 2015

$x = \frac{7}{5} , 5$

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:

$2 x + 7 \ge 0$
$6 - 3 x \ge 0$

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

$A : 2 \ge x \ge \left(- \frac{7}{5}\right)$

We will need to remember that.

Since we assume both absolute values are positive:

$2 x + 7 - \left(6 - 3 x\right) = 8$
$5 x = 7$
$x = \frac{7}{5}$

We assumed that both absolute values are positive. For this to happen, $x$ must have a value in $\left[- \frac{7}{5} , 2\right]$. (Look at $A :$)
$\frac{7}{5}$ 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 $x$

$2 x + 7 \ge 0$
$6 - 3 x < 0$
$B : x > 2$

$2 x + 7 - \left(- 1\right) \cdot \left(6 - 3 x\right) = 8$
$2 x + 7 + 6 - 3 x = 8$
$- x = - 5$
$x = 5$

$x$ is in the range $B : \left(2 , + \infty\right)$. So it is in our solution set.

Be patient, there are 2 possibilities left.

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

So:

$2 x + 7 < 0$
$6 - 3 x \ge 0$
$2 \ge x$
$x < \left(- \frac{7}{5}\right)$

As you can see $x$ cannot be greater than $2$ while it is less than $\left(- 1 , 4\right)$. 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.

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

$D : 2 < x < \left(- \frac{7}{5}\right)$

$\left(- 1\right) \cdot \left(2 x + 7\right) - \left(- 1\right) \cdot \left(6 - 3 x\right) = 8$
$- 2 x - 7 + 6 - 3 x = 8$
$- 5 x = 9$
$x = \left(- \frac{9}{5}\right) = - 1.8$

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

So the solution set is:

$S S = \left\{\frac{7}{5} , 5\right\}$