Solve the rational inequality. Express your answer in interval notation.
\[
\frac{3}{x+3} \geq \frac{2 x}{x+3}
\]
Answer
Thus, the solution is \(x \in \boxed{(-\infty,-3) \cup (\frac{3}{2},\infty)}\).
Step 1 ：First, we subtract \(\frac{2x}{x+3}\) from both sides of the inequality to get \(\frac{3}{x+3} - \frac{2x}{x+3} \geq 0\).
Step 2 ：Then, we combine the fractions on the left side to get \(\frac{3-2x}{x+3} \geq 0\).
Step 3 ：We can build a sign chart for \(\frac{3-2x}{x+3}\):
Step 4 ：\[\begin{array}{c|ccc}& x < -3 & -3 < x < \frac{3}{2} & \frac{3}{2} < x \\ \hline x + 3 & - & + & + \\ 3-2x & + & - & - \\ \frac{3-2x}{x + 3} & - & - & +\end{array}\]
Step 5 ：Also, \(\frac{3-2x}{x + 3} = 0\) for \(x = \frac{3}{2}\).
Step 6 ：Thus, the solution is \(x \in \boxed{(-\infty,-3) \cup (\frac{3}{2},\infty)}\).
