College Algebra (11th Edition)

$\left( -\infty,-\dfrac{2}{3} \right)\cup \left( 4,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $\left|5-3x \right|\gt7 ,$ remove the absolute value sign using the properties of absolute value inequality. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ For any $a\gt0,$ $|x|\ge a$ implies $x\ge a$ OR $x\le -a$ (Note that the symbols $\ge$ may be replaced with $\gt$ and the symbol $\le$ with $\lt.$) Hence, the inequality above is equivalent to \begin{array}{l}\require{cancel} 5-3x\gt7 \\\\\text{OR}\\\\ 5-3x\lt-7 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 5-3x\gt7 \\\\ -3x\gt7-5 \\\\ -3x\gt2 \\\\\text{OR}\\\\ 5-3x\lt-7 \\\\ -3x\lt-7-5 \\\\ -3x\lt-12 .\end{array} Dividing by a negative number and consequently reversing the sign results to \begin{array}{l}\require{cancel} x\lt\dfrac{2}{-3} \\\\ x\lt-\dfrac{2}{3} \\\\\text{OR}\\\\ x\gt\dfrac{-12}{-3} \\\\ x\gt4 .\end{array} Hence, the solution is the interval $\left( -\infty,-\dfrac{2}{3} \right)\cup \left( 4,\infty \right) .$