Intermediate Algebra, Chapter 2, 2.7, Section 2.7, Problem 30

Solve the inequality $|4x + 1| \geq 21$, and graph the solution set.

The absolute value inequality is rewritten as

$4x + 1 \geq 21$ or $4x + 1 \leq - 21$,

because $4x + 1$ must represent a number that is more than $21$ units from on either side of the number line. We can solve the compound inequality.


$\begin{equation} \begin{aligned} |4x + 1| \geq & 21 && \qquad \text{or} &&& |4x + 1| \leq & -21 && \\ 4x \geq & 20 && \qquad \text{or} &&& 4x \leq & -22 && \text{Subtract } 1 \\ x \geq & 5 && \qquad \text{or} &&& x \leq & \frac{-11}{2} && \text{Divide by } 4 \end{aligned} \end{equation}$


The solution set is $\displaystyle \left( - \infty, \frac{-11}{2} \right] \bigcup [5, \infty)$.