To solve the inequality $\frac{x + 1}{x + 3} \leq 2$, follow these steps:

Step 1: Move all terms to one side

Subtract 2 from both sides of the inequality: $\frac{x + 1}{x + 3} - 2 \leq 0$

Step 2: Combine into a single fraction

Express $2$ as $\frac{2(x + 3)}{x + 3}$ to have a common denominator: $\frac{x + 1 - 2(x + 3)}{x + 3} \leq 0$ Simplify the numerator: $\frac{x + 1 - 2x - 6}{x + 3} \leq 0$ $\frac{-x - 5}{x + 3} \leq 0$

Step 3: Analyze the inequality

To solve $\frac{-x - 5}{x + 3} \leq 0$, we need to determine where the expression is zero and where it is negative or positive.

Find where the fraction is zero:

Set the numerator equal to zero: $-x - 5 = 0 \implies x = -5$

Identify the critical points:

The critical points occur where the numerator or denominator is zero:

• Numerator: $x = -5$
• Denominator: $x = -3$ (Note: $x = -3$ is a vertical asymptote, where the expression is undefined.)

Step 4: Test intervals

Test the sign of the expression in each interval determined by the critical points $x = -5$ and $x = -3$:

1. $x < -5$ (e.g., $x = -6$): $\frac{-(-6) - 5}{-6 + 3} = \frac{1}{-3}$, which is negative.
2. $-5 < x < -3$ (e.g., $x = -4$): $\frac{-(-4) - 5}{-4 + 3} = \frac{-1}{-1} = 1$, which is positive.
3. $x > -3$ (e.g., $x = 0$): $\frac{-(0) - 5}{0 + 3} = \frac{-5}{3}$, which is negative.

Step 5: Conclusion

The inequality $\frac{-x - 5}{x + 3} \leq 0$ holds when the expression is less than or equal to zero. This occurs in the intervals $(-\infty, -5] \cup (-3, \infty)$.

Therefore, the solution to the inequality is: $x \in (-\infty, -5] \cup (-3, \infty)$

Do you want more details or have any questions? Here are 5 related questions you might consider:

1. How does the sign of the fraction change with different intervals of $x$?
2. What happens if the inequality is strict ($<$) instead of $\leq$)?
3. How would you solve the inequality if it were $\geq$ instead of $\leq$?
4. How can this method be applied to more complex rational inequalities?
5. How do you graph the solution set on a number line?

Tip: Always check for points where the denominator could be zero, as these values make the expression undefined and are critical in determining the solution set.