# How do you solve x+3=-abs(3x-1)?

Apr 7, 2015

Consider two possibilities which are significant for the absolute value term

Possibility 1: $\left(x \le \frac{1}{3}\right)$
In this case $\left(3 x - 1\right) \le 0$
and the equation can be re-written
$x + 3 = - \left(- \left(3 x - 1\right)\right)$
$x + 3 = 3 x - 1$
$- 2 x = - 4$
$x = 2$
Note that this is an extraneous solution since it does not exist within the range of values for Possibility 1: $\left(x \le \frac{1}{3}\right)$

**Possibility 2: $\left(x > \frac{1}{3}\right)$
In this case $\left(3 x - 1\right) > 0$
and the equation can be re-written
$x + 3 = - \left(3 x - 1\right)$
$4 x = - 2$
$x = - \frac{1}{2}$
Note that once again we have an extraneous solution sin it does not exist within the range of values for Possibility 2: ($x > \frac{1}{3}$)

To see why this happens consider a re-arrangement of the original equation into the form:
$x + 3 + \left\mid 3 x - 1 \right\mid = 0$
The graph of the left side of this equation is shown below. Notice that it does not intersect the X-axis (that is it is never equal to $0$).
graph{x+3+abs(3x-1) [-16.01, 16.02, -8.01, 8]}