To solve the inequality $\frac{x + 1}{x + 3} \leq 2$, we need to consider the following steps:

Step 1: Move all terms to one side

First, subtract $2$ from both sides to bring all terms to one side of the inequality:

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

Step 2: Combine the fractions

To combine the fractions, express $2$ as $\frac{2(x + 3)}{x + 3}$:

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

Simplify the numerator:

$\frac{x + 1 - 2x - 6}{x + 3} \leq 0$

$\frac{-x - 5}{x + 3} \leq 0$

Step 3: Analyze the inequality

Now, we need to determine when the fraction $\frac{-x - 5}{x + 3}$ is less than or equal to zero.

The fraction is equal to zero when the numerator $-x - 5 = 0$, so:

$-x - 5 = 0 \implies x = -5$

Next, we check the sign of the fraction in different intervals determined by the critical points $x = -5$ and $x = -3$ (where the denominator $x + 3 = 0$).

Step 4: Test intervals

The intervals to test are:

1. $x < -5$
2. $-5 < x < -3$
3. $x > -3$

• For $x < -5$, both $-x - 5$ and $x + 3$ are negative, so the fraction is positive.
• For $-5 < x < -3$, $-x - 5$ is negative and $x + 3$ is positive, so the fraction is negative.
• For $x > -3$, both $-x - 5$ and $x + 3$ are positive, so the fraction is negative.

Step 5: Solution

The inequality is satisfied when the fraction is less than or equal to zero:

$-5 \leq x < -3$

Thus, the solution to the inequality is:

$x \in [-5, -3)$

Do you have any questions, or would you like more details on any step?

Here are 5 related questions to consider:

1. How would the solution change if the inequality were strict $\frac{x + 1}{x + 3} < 2$?
2. How do you solve the inequality if the expression is $\frac{x - 1}{x + 3} \leq -1$?
3. What are the critical points for the function $\frac{x + 1}{x + 3}$?
4. How can the sign chart method help in solving inequalities like this?
5. What would happen if the denominator $x + 3$ was a squared term?

Tip: Always check the sign of both the numerator and denominator separately when solving inequalities involving fractions. This helps determine the intervals where the inequality holds.