How do you solve |3x + 2| = 5 and find any extraneous solutions?

Jun 11, 2017

See a solution process below:

Explanation:

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1)

$3 x + 2 = - 5$

$3 x + 2 - \textcolor{red}{2} = - 5 - \textcolor{red}{2}$

$3 x + 0 = - 7$

$3 x = - 7$

$\frac{3 x}{\textcolor{red}{3}} = - \frac{7}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x}{\cancel{\textcolor{red}{3}}} = - \frac{7}{3}$

$x = - \frac{7}{3}$

Solution 2)

$3 x + 2 = 5$

$3 x + 2 - \textcolor{red}{2} = 5 - \textcolor{red}{2}$

$3 x + 0 = 3$

$3 x = 3$

$\frac{3 x}{\textcolor{red}{3}} = \frac{3}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x}{\cancel{\textcolor{red}{3}}} = 1$

$x = 1$

The solutions are: $x = - \frac{7}{3}$ and $x = 1$

Both solution solve the problem so neither of them are extraneous