# How do you graph and solve |x+4| + |2x - 1| <4?

Oct 11, 2017

See below.

#### Explanation:

This formulation is equivalent to

$\sqrt{{\left(x + 4\right)}^{2}} + \sqrt{{\left(2 x - 1\right)}^{2}} + {e}^{2} = 4$ or

${\left(x + 4\right)}^{2} + {\left(2 x - 1\right)}^{2} + 2 \sqrt{{\left(x + 4\right)}^{2} {\left(2 x - 1\right)}^{2}} = {\left(4 - {e}^{2}\right)}^{2}$ or

$4 {\left(x + 4\right)}^{2} {\left(2 x - 1\right)}^{2} - {\left({\left(4 - {e}^{2}\right)}^{2} - {\left(x + 4\right)}^{2} - {\left(2 x - 1\right)}^{2}\right)}^{2} = 0$ or

$\left({e}^{2} - 3 x - 7\right) \left({e}^{2} - x + 1\right) \left({e}^{2} + x - 9\right) \left({e}^{2} + 3 x - 1\right) = 0$ or

$\left\{\begin{matrix}3 x + 7 \ge 0 \\ x - 1 \ge 0 \\ 9 - x \ge 0 \\ 1 - 3 x \ge 0\end{matrix}\right.$ or

$\left\{\begin{matrix}x \ge - \frac{7}{3} \\ x \ge 1 \\ x \le 9 \\ x \le \frac{1}{3}\end{matrix}\right.$

then we conclude that there is no solution for

$| x + 4 | + | 2 x - 1 | < 4$