College Algebra 7th Edition

$(-\infty, \frac{1}{3}]$ Refer to the image below for the graph.
Multiply the LCD of $12$ to both sides of the inequality to eliminate the fractions: $\begin{array}{ccc} &6\left(\dfrac{2}{3} - \dfrac{1}{2}x\right) &\ge &6\left(\dfrac{1}{6} +x\right) \\&6\left(\dfrac{2}{3}\right) - 6\left(\dfrac{1}{2}x\right) &\ge &6\left(\dfrac{1}{6}\right)+6(x) \\&4 - 3x &\ge &1+6x \end{array}$ Add $3x$ and subtract $1$to both sides of the inequality to obtain: $\begin{array}{ccc} &4-3x+3x -1&\ge & 1+6x+3x - 1 \\&3 &\ge &9x \end{array}$ Divide $9$ to both sides of the inequality to obtain: $\begin{array}{ccc} \\&\dfrac{3}{9} &\ge & \dfrac{9x}{9} \\&\dfrac{1}{3} &\ge &x \\&x &\le &\dfrac{1}{3} \end{array}$ Thus, the solution set is $(-\infty, \frac{1}{3}]$. To graph this solution set, plot a solid dot at $\dfrac{1}{3}$ then shade the region to its left. (refer to the attached image in the answer part above for the graph)