Precalculus (6th Edition)

The solution is $\Big(-\infty,-\dfrac{2}{3}\Big)\cup\Big(4,\infty\Big)$
$|5-3x|\gt7$ Solving this absolute value inequality is equivalent to solving two separate inequalities, which are: $5-3x\gt7$ and $5-3x\lt-7$ $\textbf{Solve the first inequality:}$ $5-3x\gt7$ Take $5$ to the right side: $-3x\gt7-5$ $-3x\gt2$ Take $3$ to divide the right side and reverse the inequality sign: $x\lt-\dfrac{2}{3}$ $\textbf{Solve the first inequality:}$ $5-3x\lt-7$ Take $5$ to the right side: $-3x\lt-7-5$ $-3x\lt-12$ Take $3$ to divide the right side and reverse the inequality sign: $x\gt\dfrac{-12}{-3}$ $x\gt4$ Expressing the solution in interval notation: $\Big(-\infty,-\dfrac{2}{3}\Big)\cup\Big(4,\infty\Big)$