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The inequality you provided is:

\[
\frac{5}{3}(x - 15) \leq x - 20
\]

Let's solve this step by step:

### Step 1: Distribute \(\frac{5}{3}\) on the left side:
\[
\frac{5}{3}(x - 15) = \frac{5}{3}x - \frac{5}{3} \times 15
\]
\[
\frac{5}{3}x - 25
\]

So the inequality becomes:
\[
\frac{5}{3}x - 25 \leq x - 20
\]

### Step 2: Move all \(x\)-terms to one side and constants to the other:
\[
\frac{5}{3}x - x \leq 25 - 20
\]

### Step 3: Combine like terms:
\[
\frac{5}{3}x - \frac{3}{3}x = \frac{2}{3}x
\]
\[
\frac{2}{3}x \leq 5
\]

### Step 4: Multiply both sides by \(\frac{3}{2}\) to isolate \(x\):
\[
x \leq 5 \times \frac{3}{2} = \frac{15}{2}
\]

### Final Answer:
\[
x \leq \frac{15}{2} \quad \text{or} \quad x \leq 7.5
\]

Let me know if you want further details or have any questions!

Here are 5 related questions for practice:
1. Solve \(\frac{3}{4}(x - 8) > x - 6\).
2. Solve \(\frac{7}{2}(x + 4) \leq 3x + 10\).
3. Solve \(2(x - 5) + 3x \geq 4x - 10\).
4. Solve \(5 - 2x \leq x + 1\).
5. Solve \(x - \frac{1}{2} \geq 3x + \frac{5}{2}\).

**Tip**: When solving inequalities, remember that multiplying or dividing both sides by a negative number reverses the inequality sign!

Solve the inequality \( \frac{5}{3}(x - 15) \leq x - 20 \).

Learn how to solve the linear inequality \( \frac{5}{3}(x - 15) \leq x - 20 \) with a detailed step-by-step solution. This problem covers key algebraic concepts like the distributive property and solving linear inequalities. Suitable for Grades 9-10.

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