Intermediate Algebra, Chapter 2, 2.7, Section 2.7, Problem 28

Solve the inequality $|3r - 1| \geq 8$, and graph the solution set.

The absolute value inequality is rewritten as

$3r - 1 \geq 8$ or $3r - 1 \leq -8$,

because $3r - 1$ must represent a number that is more than $8$ units from on either side of the number line. We can solve the compound inequality.


$\begin{equation} \begin{aligned} 3r - 1 \geq & 8 \qquad \text{or} &&& 3r - 1 \leq & -8 && \\ 3r \geq & 9 \qquad \text{or} &&& 3r \leq & -7 && \text{Add } 1 \\ r \geq & 3 \qquad \text{or} &&& r \leq & \frac{-7}{3} && \text{Divide by } -3 \end{aligned} \end{equation}$


The solution set is $\displaystyle \left( - \infty, \frac{-7}{3} \right] \bigcup [3, \infty)$.