Solve the rational inequality. Express your answer in interval notation.
$\frac{3}{x + 3} \geq \frac{2 x}{x + 3}$

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Thus, the solution is $x \in (- \infty, - 3) \cup (\frac{3}{2}, \infty)$ .

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Step 1 ：First, we subtract $\frac{2 x}{x + 3}$ from both sides of the inequality to get $\frac{3}{x + 3} - \frac{2 x}{x + 3} \geq 0$ .

Step 2 ：Then, we combine the fractions on the left side to get $\frac{3 - 2 x}{x + 3} \geq 0$ .

Step 3 ：We can build a sign chart for $\frac{3 - 2 x}{x + 3}$ :

Step 4 ： $\begin{array}{cccc} x < - 3 & - 3 < x < \frac{3}{2} & \frac{3}{2} < x \\ x + 3 & - & + & + \\ 3 - 2 x & + & - & - \\ \frac{3 - 2 x}{x + 3} & - & - & + \end{array}$

Step 5 ：Also, $\frac{3 - 2 x}{x + 3} = 0$ for $x = \frac{3}{2}$ .

Step 6 ：Thus, the solution is $x \in (- \infty, - 3) \cup (\frac{3}{2}, \infty)$ .

Solve the rational inequality. Express your answer in interval notation.3x+3≥2xx+3

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Solve the rational inequality. Express your answer in interval notation.
$\frac{3}{x + 3} \geq \frac{2 x}{x + 3}$