The inequality is:

$\frac{2x}{x+1} \geq 1$

Step 1: Move all terms to one side

First, subtract $1$ from both sides:

$\frac{2x}{x+1} - 1 \geq 0$

Step 2: Combine the terms

We need a common denominator to combine the terms. Rewrite $1$ as $\frac{x+1}{x+1}$:

$\frac{2x}{x+1} - \frac{x+1}{x+1} \geq 0$

Now combine the fractions:

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

Simplify the numerator:

$\frac{2x - x - 1}{x+1} = \frac{x - 1}{x+1} \geq 0$

Step 3: Solve the inequality

Now we need to solve:

$\frac{x - 1}{x+1} \geq 0$

This inequality will be zero when $x = 1$. It will change signs when $x+1 = 0$, i.e., $x = -1$. So, we need to check the sign of the expression in the intervals determined by these two critical points: $x = -1$ and $x = 1$.

Step 4: Analyze the intervals

• For $x < -1$: Both $x - 1$ and $x + 1$ are negative, so $\frac{x-1}{x+1} > 0$.
• For $-1 < x < 1$: $x - 1$ is negative, and $x + 1$ is positive, so $\frac{x-1}{x+1} < 0$.
• For $x > 1$: Both $x - 1$ and $x + 1$ are positive, so $\frac{x-1}{x+1} > 0$.

Step 5: Final solution

The inequality is satisfied when $\frac{x-1}{x+1} \geq 0$. Therefore, the solution is:

$x \in (-\infty, -1) \cup [1, \infty)$

Note:

• $x = -1$ is not included because the expression is undefined at $x = -1$.
• $x = 1$ is included because the inequality allows equality.

Do you want more details or have any questions?

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Relative questions:

1. How do you solve rational inequalities?
2. What happens if both sides of the inequality contain fractions?
3. How do we handle inequalities involving absolute values?
4. Can inequalities be solved using a graphing approach?
5. How does interval notation differ from set notation?

Tip: Always check for points where the expression is undefined (like division by zero) when solving inequalities!