1. **State the problem:** Solve the system of equations: $$\frac{2}{3} = 3x + \frac{1}{3}$$ $$17x = -10$$ 2. **Solve the first equation:** Start with: $$\frac{2}{3} = 3x + \frac{1}{3}$$ Subtract $\frac{1}{3}$ from both sides: $$\frac{2}{3} - \frac{1}{3} = 3x$$ Simplify the left side: $$\frac{1}{3} = 3x$$ Divide both sides by 3: $$x = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$ 3. **Solve the second equation:** $$17x = -10$$ Divide both sides by 17: $$x = \frac{-10}{17}$$ 4. **Interpretation:** The system gives two different values for $x$, so these are two separate equations, not a system to solve simultaneously. **Final answers:** $$x = \frac{1}{9}, \quad x = \frac{-10}{17}$$