# How do you solve abs(4-3x)= 4x + 6 and state the extraneous solutions?

Apr 22, 2015

$\left\mid 4 - 3 x \right\mid = 4 x + 6$

Consider the two possibilities for $\left(4 - 3 x\right)$

Possibility 1:
$\left(4 - 3 x\right) < 0$
$\rightarrow x > \frac{4}{3}$

$\left\mid 4 - 3 x \right\mid = 4 x + 6$
becomes
$3 x - 4 = 4 x + 6$
$x = - 10$
but $x > \frac{4}{3}$ for this condition to apply
so this solution is extraneous.

Possibility 2:
$\left(4 - 3 x\right) \ge 0$
$\rightarrow x \le \frac{4}{3}$

$\left\mid 4 - 3 x \right\mid = 4 x + 6$
becomes
$4 - 3 x = 4 x + 6$
$x = \frac{2}{7}$

Since $x = \frac{2}{7}$ satisfies the condition $x \le \frac{4}{3}$
this is a valid solution