# How do you solve |x-1| - |x-3| >= 5?

Oct 13, 2017

See below.

#### Explanation:

The solution for

$\sqrt{{\left(x - 1\right)}^{2}} - \sqrt{{\left(x - 3\right)}^{2}} = 5 + {e}^{2}$ is contained in the solutions for

${\left(x - 1\right)}^{2} + {\left(x - 3\right)}^{2} - 2 \sqrt{{\left(x - 1\right)}^{2} {\left(x - 3\right)}^{2}} = {\left(5 + {e}^{2}\right)}^{2}$

and also in the solutions for

${\left({\left(x - 1\right)}^{2} + {\left(x - 3\right)}^{2} - {\left(5 + {e}^{2}\right)}^{2}\right)}^{2} = {\left(x - 1\right)}^{2} {\left(x - 3\right)}^{2}$ or

$\left(3 + {e}^{2}\right) \left(7 + {e}^{2}\right) \left(9 + {e}^{2} - 2 x\right) \left(1 + {e}^{2} + 2 x\right) = 0$ and then we obtain the conditions

$\left\{\begin{matrix}9 + {e}^{2} - 2 x = 0 \\ 1 + {e}^{2} + 2 x = 0\end{matrix}\right.$

which is equivalent to

$2 x \ge 9$ and $2 x \le - 1$

Thus we conclude that the inequality has no solution.