To solve the rational inequality:

$\frac{2x}{x + 3} \leq 1$

Step-by-Step Solution:

1. Move all terms to one side: Subtract 1 from both sides to set the inequality to 0:

   $\frac{2x}{x + 3} - 1 \leq 0$

2. Combine the terms: To combine the terms, rewrite $1$ as $\frac{x+3}{x+3}$ so that both terms have the same denominator:

   $\frac{2x}{x+3} - \frac{x+3}{x+3} \leq 0$

3. Simplify the numerator: Combine the numerators:

   $\frac{2x - (x+3)}{x+3} \leq 0$

   Simplify the numerator:

   $\frac{2x - x - 3}{x+3} = \frac{x - 3}{x+3} \leq 0$

4. Find the critical points: Set the numerator and the denominator equal to zero separately:

   • Numerator: $x - 3 = 0 \Rightarrow x = 3$
   • Denominator: $x + 3 = 0 \Rightarrow x = -3$

   These are the critical points, $x = 3$ and $x = -3$. The critical points divide the real line into intervals.

5. Test the intervals: The critical points divide the number line into three intervals: $(- \infty, -3)$, $(-3, 3)$, and $(3, \infty)$.

   Let's test a point in each interval to determine the sign of $\frac{x-3}{x+3}$.

   • For $x = -4$ (in the interval $(- \infty, -3)$): $\frac{-4 - 3}{-4 + 3} = \frac{-7}{-1} = 7 > 0$ (positive)

   • For $x = 0$ (in the interval $(-3, 3)$): $\frac{0 - 3}{0 + 3} = \frac{-3}{3} = -1 < 0$ (negative)

   • For $x = 4$ (in the interval $(3, \infty)$): $\frac{4 - 3}{4 + 3} = \frac{1}{7} > 0$ (positive)

6. Determine the solution: We are looking for where the expression is less than or equal to zero. This happens when the expression is negative or zero. The expression is negative in the interval $(-3, 3)$. Also, the expression equals zero at $x = 3$.

   • At $x = 3$, $\frac{x - 3}{x + 3} = 0$, so include $x = 3$.
   • At $x = -3$, the expression is undefined because it would involve division by zero, so exclude $x = -3$.

   Therefore, the solution is:

   $(-3, 3]$

Final Answer:

The solution to the inequality is $x \in (-3, 3]$.

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Would you like more details on any of the steps or concepts? Here are five related questions:

1. What is the difference between solving rational inequalities and solving rational equations?
2. How can we graph the solution to a rational inequality?
3. What happens if the inequality sign is reversed?
4. How can we check whether our solution is correct by plugging in test values?
5. What are some tips for handling undefined points in rational expressions?

Tip: Always check for points where the denominator becomes zero, as these make the function undefined and affect the solution set.