## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\left( 7,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $\dfrac{1}{2}(2x+2b)\gt\dfrac{1}{3}(21+3b) ,$ remove first the fraction by multiplying both sides by the $LCD.$ Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $\{ 2,3 \},$ is $6$ since this is the least number that can be evenly divided (no remainder) by all the denominators. Multiplying both sides by the $LCD,$ the given inequality is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{2}(2x+2b)\gt\dfrac{1}{3}(21+3b) \\\\ 6\left[ \dfrac{1}{2}(2x+2b) \right]\gt6\left[ \dfrac{1}{3}(21+3b) \right] \\\\ 3(2x+2b)\gt2(21+3b) .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the inequality above is equivalent to \begin{array}{l}\require{cancel} 3(2x+2b)\gt2(21+3b) \\\\ 3(2x)+3(2b)\gt2(21)+2(3b) \\\\ 6x+6b\gt42+6b .\end{array} Using the properties of inequality, the given is equivalent to \begin{array}{l}\require{cancel} 6x+6b\gt42+6b \\\\ 6x\gt42+6b-6b \\\\ 6x\gt42 \\\\ x\gt\dfrac{42}{6} \\\\ x\gt7 .\end{array} Hence, the solution set is $\left( 7,\infty \right) .$