1. **State the problem:** Solve the equation $$-\frac{4}{15} n + \frac{2}{3} = \frac{2}{5} n$$ for $n$. 2. **Write down the equation:** $$-\frac{4}{15} n + \frac{2}{3} = \frac{2}{5} n$$ 3. **Goal:** Isolate $n$ on one side. 4. **Move all $n$ terms to one side:** Add $\frac{4}{15} n$ to both sides: $$-\frac{4}{15} n + \frac{2}{3} + \frac{4}{15} n = \frac{2}{5} n + \frac{4}{15} n$$ Simplifies to: $$\frac{2}{3} = \frac{2}{5} n + \frac{4}{15} n$$ 5. **Combine like terms on the right:** Find common denominator for $\frac{2}{5}$ and $\frac{4}{15}$ which is 15: $$\frac{2}{5} n = \frac{6}{15} n$$ So: $$\frac{2}{3} = \frac{6}{15} n + \frac{4}{15} n = \frac{10}{15} n$$ 6. **Simplify fraction:** $$\frac{10}{15} = \frac{2}{3}$$, so: $$\frac{2}{3} = \frac{2}{3} n$$ 7. **Divide both sides by $\frac{2}{3}$ to solve for $n$:** $$n = \frac{\frac{2}{3}}{\frac{2}{3}}$$ Show cancellation: $$n = \frac{\cancel{\frac{2}{3}}}{\cancel{\frac{2}{3}}} = 1$$ 8. **Final answer:** $$n = 1$$ This means the value of $n$ that satisfies the equation is 1.