# How do you solve abs(2z -3) = 4z - 1?

Jun 24, 2016

$x = \frac{3}{2}$

#### Explanation:

It is equivalent to:

$\left(2 z - 3 \ge 0 \mathmr{and} 2 z - 3 = 4 z - 1\right) \mathmr{and} \left(2 z - 3 < 0 \mathmr{and} - 2 z + 3 = 4 z - 1\right)$

$\left(z \ge \frac{3}{2} \mathmr{and} 2 z = - 2\right) \mathmr{and} \left(z < \frac{3}{2} \mathmr{and} 6 z = 4\right)$

$\left(z \ge \frac{3}{2} \mathmr{and} z = - 1\right) \mathmr{and} \left(z < \frac{3}{2} \mathmr{and} z = \frac{2}{3}\right)$

the first condition is impossible, then since $\frac{2}{3} < \frac{3}{2}$:

the solution is $x = \frac{3}{2}$