# How do you solve 5x – 4 = |2x + 1| and find any extraneous solutions?

Oct 25, 2017

$x = \frac{5}{3}$

#### Explanation:

The solutions for 5x – 4 = |2x + 1|

are contained in the set of solutions for ${\left(5 x - 4\right)}^{2} - {\left(2 x + 1\right)}^{2} = 0$

but

${\left(5 x - 4\right)}^{2} - {\left(2 x + 1\right)}^{2} = \left(3 x - 5\right) \left(7 x - 3\right) = 0$ so

$x = \left\{\frac{3}{7} , \frac{5}{3}\right\}$ and the feasible solution is $x = \frac{5}{3}$ because it verifies the original relationship.