To solve the inequality \( \frac{5 + x}{2x + 1} \leq 1 \), follow these steps:

### Step 1: Clear the fraction
Multiply both sides of the inequality by \( (2x + 1) \), but note that you need to consider when \( 2x + 1 \) is positive or negative, as this will affect the direction of the inequality.

1. **When \( 2x + 1 > 0 \)**, i.e., \( x > -\frac{1}{2} \), the inequality remains as is:
\[
5 + x \leq (2x + 1)
\]
Simplifying:
\[
5 + x \leq 2x + 1
\]
Subtract \( x \) from both sides:
\[
5 \leq x + 1
\]
Subtract 1 from both sides:
\[
4 \leq x
\]
Thus, \( x \geq 4 \) when \( x > -\frac{1}{2} \).

2. **When \( 2x + 1 < 0 \)**, i.e., \( x < -\frac{1}{2} \), the inequality flips:
\[
5 + x \geq 2x + 1
\]
Simplifying:
\[
5 + x \geq 2x + 1
\]
Subtract \( x \) from both sides:
\[
5 \geq x + 1
\]
Subtract 1 from both sides:
\[
4 \geq x
\]
Thus, \( x \leq 4 \) when \( x < -\frac{1}{2} \).

### Step 2: Check boundary conditions
We need to ensure the solution works within the domain of the inequality:
- The denominator \( 2x + 1 \) cannot be zero, which happens at \( x = -\frac{1}{2} \). Therefore, \( x = -\frac{1}{2} \) is excluded from the solution set.

### Step 3: Combine solutions
From the two cases:
- For \( x > -\frac{1}{2} \), \( x \geq 4 \).
- For \( x < -\frac{1}{2} \), \( x \leq 4 \).

Thus, the solution to the inequality is:
\[
x \in (-\infty, -\frac{1}{2}) \cup [4, \infty)
\]

### Final Solution:
\[
x \in (-\infty, -\frac{1}{2}) \cup [4, \infty)
\]

Would you like more details or have any questions about this solution?

Here are some related questions:
1. How would the solution change if the inequality were strict, i.e., \( < \) instead of \( \leq \)?
2. What happens if the coefficient of \( x \) in the denominator changes?
3. How can we solve the same inequality using a graphical approach?
4. What is the impact of adding a constant to both sides of the inequality?
5. How does this inequality compare to one where the denominator has an \( x^2 \) term?

**Tip:** Always check the sign of the denominator when multiplying or dividing inequalities to avoid errors.

Learn how to solve the rational inequality \( \frac{5 + x}{2x + 1} \leq 1 \) with a clear, step-by-step guide. This problem covers essential algebraic concepts like rational functions and inequalities, suitable for high school students in Grades 9-12.

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