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The problem is to solve the equation:

\[
\frac{2x}{5} - \frac{x - 2}{3} = 4 \left(\frac{-x}{3}\right)
\]

### Step 1: Distribute and simplify
First, let's simplify the right-hand side:

\[
4 \left(\frac{-x}{3}\right) = \frac{-4x}{3}
\]

So the equation becomes:

\[
\frac{2x}{5} - \frac{x - 2}{3} = \frac{-4x}{3}
\]

### Step 2: Eliminate fractions by finding a common denominator
To eliminate the fractions, we multiply each term by the least common denominator (LCD) of 5 and 3, which is 15:

\[
15 \cdot \frac{2x}{5} - 15 \cdot \frac{x - 2}{3} = 15 \cdot \frac{-4x}{3}
\]

This simplifies to:

\[
3 \cdot 2x - 5 \cdot (x - 2) = 5 \cdot (-4x)
\]

\[
6x - 5(x - 2) = -20x
\]

### Step 3: Distribute and combine like terms
Now distribute the -5 on the left-hand side:

\[
6x - 5x + 10 = -20x
\]

This simplifies to:

\[
x + 10 = -20x
\]

### Step 4: Solve for \( x \)
Add \( 20x \) to both sides to get all \( x \) terms on one side:

\[
x + 20x + 10 = 0
\]

\[
21x + 10 = 0
\]

Subtract 10 from both sides:

\[
21x = -10
\]

Finally, divide by 21:

\[
x = \frac{-10}{21}
\]

### Final Answer:
\[
x = \frac{-10}{21}
\]

Would you like more details or have any questions?

Here are some related questions to explore:
1. How do you find the least common denominator of two fractions?
2. What are the steps to simplify expressions with multiple fractions?
3. Can you solve the equation if the right-hand side had a different constant multiplier?
4. How does multiplying both sides of an equation by the LCD help in solving the equation?
5. What would the solution be if the equation was slightly modified, e.g., different coefficients?

**Tip:** Always check your final answer by substituting it back into the original equation to ensure it satisfies the equation.

Solve the following equation: \( \frac{2x}{5} - \frac{x - 2}{3} = 4 \left(\frac{-x}{3}\right) \). Leave your answer as a single fraction or term. No mixed numbers.

Step-by-step guide to solving the linear equation \( \frac{2x}{5} - \frac{x - 2}{3} = 4 \left(\frac{-x}{3}\right) \) by simplifying and eliminating fractions. This problem involves key algebraic concepts like the distributive property and properties of equality, suitable for students in Grades 8-10.

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