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The inequality you provided is:

\[
\frac{4}{2x - 1} \geq \frac{1}{x + 1}
\]

To solve for \( x \), follow these steps:

### Step 1: Cross-multiply
First, we eliminate the denominators by cross-multiplying, but we must take care that the denominators are non-zero. We'll discuss the restrictions later. The cross-multiplication gives:

\[
4(x + 1) \geq 1(2x - 1)
\]

### Step 2: Expand both sides
Expand both sides of the inequality:

\[
4x + 4 \geq 2x - 1
\]

### Step 3: Isolate \( x \)
Now, move all the terms involving \( x \) to one side, and constants to the other:

\[
4x - 2x \geq -1 - 4
\]

This simplifies to:

\[
2x \geq -5
\]

### Step 4: Solve for \( x \)
Now, divide both sides by 2:

\[
x \geq -\frac{5}{2}
\]

### Step 5: Check the restrictions
To ensure the solution is valid, we need to check that the original denominators are not zero. This gives two conditions:

1. \( 2x - 1 \neq 0 \) implies \( x \neq \frac{1}{2} \)
2. \( x + 1 \neq 0 \) implies \( x \neq -1 \)

### Final Answer
The solution is \( x \geq -\frac{5}{2} \), but we must also exclude \( x = \frac{1}{2} \) and \( x = -1 \). Therefore, the solution set is:

\[
x \in \left[ -\frac{5}{2}, \infty \right) \setminus \left\{ -1, \frac{1}{2} \right\}
\]

Would you like more details or have any questions?

### Related Questions
1. How do you determine the domain of rational inequalities?
2. What happens if the inequality was strict (e.g., \( > \) instead of \( \geq \))?
3. Can we use graphing techniques to solve rational inequalities?
4. What is the importance of cross-multiplication in inequalities?
5. How do vertical asymptotes relate to restrictions on \( x \)?

### Tip
Always check for restrictions from denominators in rational equations to avoid invalid solutions.

Learn how to solve the rational inequality 4/(2x - 1) >= 1/(x + 1) with a detailed step-by-step solution. The problem covers core algebraic concepts such as cross-multiplication, solving inequalities, and analyzing domain restrictions. This guide is suitable for high school students in Grades 9-12.

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