# How do you solve the absolute value abs(3x-1) + 10 = 25?

May 13, 2015

Actually there are two conditions to look at:

(1) If $3 x - 1 \ge 0$ we have:

$\left(3 x - 1\right) + 10 = 25 \to 3 x = 25 + 1 - 10 = 16 \to$
$x = 16 / 3 = 5 \frac{1}{3}$

(2) If $3 x - 1 < 0$ the absolutes 'turn it around':

$- \left(3 x - 1\right) + 10 = 25 \to - 3 x = 25 - 1 - 10 = 14 \to$
$x = 14 / \left(- 3\right) = - 4 \frac{2}{3}$

Validation:
(allways do this, sometimes one of the answers is wrong)
(1) exists if $x \ge \frac{1}{3}$ which is consistent with the answer
(2) exists if $x < \frac{1}{3}$ which is also consistent with the answer

Final answer: $x = 5 \frac{1}{3} \mathmr{and} x = - 4 \frac{2}{3}$