Answer
$\left\{ x|x \gt \dfrac{39}{11} \right\}
\text{ or }
\left( \dfrac{39}{11},\infty \right)
$
Work Step by Step
Using the properties of inequality, the given inequality, $
\dfrac{2}{3}(6-x)\lt\dfrac{1}{4}(x+3)
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2}{3}(6)+\dfrac{2}{3}(-x)\lt\dfrac{1}{4}(x)+\dfrac{1}{4}(3)
\\\\
4-\dfrac{2}{3}x\lt\dfrac{1}{4}x+\dfrac{3}{4}
\\\\
12\left( 4-\dfrac{2}{3}x \right) \lt \left( \dfrac{1}{4}x+\dfrac{3}{4} \right)12
\\\\
48-4(2x) \lt x(3)+3(3)
\\\\
48-8x \lt 3x+9
\\\\
-8x-3x \lt 9-48
\\\\
-11x \lt -39
\\\\
x \gt \dfrac{-39}{-11}
\\\\
x \gt \dfrac{39}{11}
.\end{array}
Hence, the solution is $
\left\{ x|x \gt \dfrac{39}{11} \right\}
\text{ or }
\left( \dfrac{39}{11},\infty \right)
.$
