## College Algebra (11th Edition)

$\left( -\infty, \dfrac{3}{2} \right] \cup \left[ \dfrac{7}{2}, \infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|10-4x|+1\ge5 ,$ use the properties of equality to isolate the absolute value expression. Then use the properties of absolute value inequality. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the statement above is equivalent to \begin{array}{l}\require{cancel} |10-4x|\ge5-1 \\\\ |10-4x|\ge4 .\end{array} Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} 10-4x\ge4 \\\\\text{OR}\\\\ 10-4x\le-4 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 10-4x\ge4 \\\\ -4x\ge4-10 \\\\ -4x\ge-6 \\\\ x\le\dfrac{-6}{-4} \text{ (reverse the sign)}\\\\ x\le\dfrac{3}{2} \\\\\text{OR}\\\\ 10-4x\le-4 \\\\ -4x\le-4-10 \\\\ -4x\le-14 \\\\ x\ge\dfrac{-14}{-4} \text{ (reverse the sign)}\\\\ x\ge\dfrac{7}{2} .\end{array} Hence, the solution set is the interval $\left( -\infty, \dfrac{3}{2} \right] \cup \left[ \dfrac{7}{2}, \infty \right) .$