# How do you solve the inequality abs(x-3)-abs(2x+1)<0 and write your answer in interval notation?

Sep 13, 2017

Assuming x is real, use the alternate form:

$\sqrt{{\left(x - 3\right)}^{2}} - \sqrt{{\left(2 x + 1\right)}^{2}} < 0$

This implies that:

${\left(x - 3\right)}^{2} < {\left(2 x + 1\right)}^{2}$

Expand the squares and solve the resulting quadratic inequality.

#### Explanation:

Starting with the above:

${\left(x - 3\right)}^{2} < {\left(2 x + 1\right)}^{2}$

Expand the squares

${x}^{2} - 2 x + 9 < 4 {x}^{2} + 4 x + 1$

Combine like terms:

$0 < 3 {x}^{2} + 6 x - 8$

The quadratic will be greater than 0 to the left and to the right of the roots, therefore, we should find the roots:

$0 = 3 {x}^{2} + 6 x - 8$

Factor:

$0 = \left(3 x - 2\right) \left(x + 4\right)$

$x = \frac{2}{3}$ and $x = - 4$

The intervals to the left and to the right of these roots are:

$\left(- \infty , - 4\right)$ and $\left(\frac{2}{3} , \infty\right)$

Here is a graph of $y = | x - 3 | - | 2 x + 1 |$

Please observe that it drops below the $y = 0$ line at $x = - 4$ and $x = \frac{2}{3}$