## Precalculus (6th Edition) Blitzer

$(-\displaystyle \infty,-\frac{1}{3}]\cup[3,\infty)$
3 times a number is subtracted from 4 is: $4-3x$ the absolute value of the difference $|4-3x|$ is at least 5$\qquad...\qquad |4-3x|\geq 5$ $|4-3x|=|3x-4|\qquad$, because $|a-b|=|b-a|$ So, we solve $|3x-4|\geq 5$ $|u| \gt c$ is equivalent to ($u\lt -c$) or ($u\gt c$) $\begin{array}{lllll} 3x-4\leq-5 & /+4 & ...or... & 3x-4\geq 5 & /+4\\ 3x\leq-1 & /\div 3 & & 3x \geq 9 & /\div 3\\ x\leq-1/3 & & & x\geq 3 & \\ x\in(-\infty,-1/3] & & or & x\in[3,\infty) & \\ & & & & \end{array}$ Solution set: $(-\displaystyle \infty,-\frac{1}{3}]\cup[3,\infty)$