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To solve the rational linear inequality:

\[
\frac{2x}{5+x} \leq \frac{5}{3}
\]

### Step 1: Eliminate the denominator
Multiply both sides by \( 5 + x \), keeping in mind that we need to check the sign of \( 5 + x \) since it could change the direction of the inequality:
- If \( 5 + x > 0 \), then the inequality remains the same.
- If \( 5 + x < 0 \), the inequality flips direction.

For now, let's proceed as if \( 5 + x > 0 \), and we'll check later for other cases.

\[
2x \leq \frac{5}{3} \cdot (5 + x)
\]

### Step 2: Distribute and simplify
Distribute \( \frac{5}{3} \) on the right-hand side:

\[
2x \leq \frac{25}{3} + \frac{5}{3}x
\]

### Step 3: Eliminate fractions
Multiply through by 3 to get rid of the fractions:

\[
6x \leq 25 + 5x
\]

### Step 4: Isolate \( x \)
Subtract \( 5x \) from both sides:

\[
x \leq 25
\]

### Step 5: Check for restrictions
The original inequality has a denominator of \( 5 + x \), which means that \( x \neq -5 \) (because it would make the denominator zero). Additionally, we considered \( 5 + x > 0 \), which implies \( x > -5 \).

### Final solution:
So, the solution is:

\[
-5 < x \leq 25
\]

This is the inequality in interval notation:
\[
x \in (-5, 25]
\]

Would you like more details on any of the steps? 

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Here are 5 related questions to expand on this:

1. How does the solution change if the inequality sign is reversed (i.e., \( \geq \))?
2. What if the inequality involved absolute values—how would the process change?
3. How do we handle cases where multiplying by an unknown might reverse the inequality?
4. Can you show how the inequality would be solved graphically?
5. How do you test solutions in an inequality with restrictions on the denominator?

**Tip**: Always remember to consider domain restrictions when dealing with rational inequalities!

Solve the rational linear inequality: \frac{2x}{5+x} \leq \frac{5}{3}

Learn how to solve the rational inequality 2x/(5+x) ≤ 5/3 with a detailed step-by-step guide. This problem covers key algebraic concepts, including inequality properties, rational expressions, and domain restrictions. Suitable for students in Grades 10-12.

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