Answer
The solution is $[2,3)$
Work Step by Step
$\frac{2k}{k-3} \leq \frac{4}{k-3}$
$\frac{2k}{k-3} - \frac{4}{k-3}\leq0$
First solve the corresponding equation
$\frac{2k}{k-3} - \frac{4}{k-3}=0$
$2k-4=0$
$k=2$
For $(-\infty,2)$, choose $0$: $\frac{2(0)}{0-3} - \frac{4}{0-3}=\frac{4}{3}\gt0$
For $(2,3)$, choose $2.5$: $\frac{2(2.5)}{2.5-3} - \frac{4}{2.5-3}=-2\lt0$
For $(3,\infty)$, choose $4:$, $\frac{2(4)}{4-3} - \frac{4}{4-3}=4\gt$
The solution is $[2,3)$
