#### Answer

$\left( 7,\infty \right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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)
.$