## College Algebra 7th Edition

$(\frac{16}{3}, +\infty)$ Refer to the image below for the graph.
Solving the inequality will be easier of there are no fractions involved. To get rid of the denominators, multiply the LCD of $6$ on both sides of the inequality to obtain: $6(\frac{1}{2}x-\frac{2}{3}) \gt 2(6) \\\frac{6}{2}x-\frac{12}{3|} \gt 12 \\3x - 4 \gt 12$ Add $4$ on both side of the inequality to obtain: $3x \gt 16$ Divide $3$ on both sides to obtain: $x \gt \dfrac{16}{3}$ The solution set includes all real numbers greater than $\dfrac{16}{3}$. In interval notation, the solution set is: $(\frac{16}{3}, +\infty)$ To graph the solution set, plot a hole (or hollow dot) at $x=\frac{16}{3}$ then shade the region to its right. (refer to the attached image in the answer portion above)