1. **State the problem:** Solve the equation $$\frac{2}{3}(x - 7)^2 + 5\left(x - \frac{1}{3}\right) + \frac{4}{6} = \frac{2}{3}(x - 1)(x + 1) + \frac{129}{6}$$ for $x$. 2. **Recall formulas and rules:** - Expand squares and products. - Combine like terms. - Use common denominators to simplify fractions. - Isolate $x$ terms to one side. - When dividing both sides by a factor, show cancellation. 3. **Expand and simplify each side:** - Left side: $$\frac{2}{3}(x - 7)^2 = \frac{2}{3}(x^2 - 14x + 49) = \frac{2}{3}x^2 - \frac{28}{3}x + \frac{98}{3}$$ - The term $5\left(x - \frac{1}{3}\right) = 5x - \frac{5}{3}$ - The term $\frac{4}{6} = \frac{2}{3}$ So left side sum: $$\frac{2}{3}x^2 - \frac{28}{3}x + \frac{98}{3} + 5x - \frac{5}{3} + \frac{2}{3}$$ Combine constants: $$\frac{98}{3} - \frac{5}{3} + \frac{2}{3} = \frac{98 - 5 + 2}{3} = \frac{95}{3}$$ Left side becomes: $$\frac{2}{3}x^2 - \frac{28}{3}x + 5x + \frac{95}{3}$$ Convert $5x$ to thirds: $$5x = \frac{15}{3}x$$ So left side: $$\frac{2}{3}x^2 - \frac{28}{3}x + \frac{15}{3}x + \frac{95}{3} = \frac{2}{3}x^2 - \frac{13}{3}x + \frac{95}{3}$$ 4. **Right side:** $$\frac{2}{3}(x - 1)(x + 1) + \frac{129}{6} = \frac{2}{3}(x^2 - 1) + \frac{129}{6} = \frac{2}{3}x^2 - \frac{2}{3} + \frac{129}{6}$$ Convert constants to common denominator 6: $$-\frac{2}{3} = -\frac{4}{6}$$ Sum constants: $$-\frac{4}{6} + \frac{129}{6} = \frac{125}{6}$$ Right side becomes: $$\frac{2}{3}x^2 + \frac{125}{6}$$ 5. **Set equation:** $$\frac{2}{3}x^2 - \frac{13}{3}x + \frac{95}{3} = \frac{2}{3}x^2 + \frac{125}{6}$$ 6. **Subtract $\frac{2}{3}x^2$ from both sides:** $$\frac{2}{3}x^2 - \frac{13}{3}x + \frac{95}{3} - \frac{2}{3}x^2 = \frac{2}{3}x^2 + \frac{125}{6} - \frac{2}{3}x^2$$ Intermediate step with cancellation: $$\cancel{\frac{2}{3}x^2} - \frac{13}{3}x + \frac{95}{3} - \cancel{\frac{2}{3}x^2} = \cancel{\frac{2}{3}x^2} + \frac{125}{6} - \cancel{\frac{2}{3}x^2}$$ Simplifies to: $$- \frac{13}{3}x + \frac{95}{3} = \frac{125}{6}$$ 7. **Subtract $\frac{95}{3}$ from both sides:** $$- \frac{13}{3}x + \frac{95}{3} - \frac{95}{3} = \frac{125}{6} - \frac{95}{3}$$ Intermediate step with cancellation: $$- \frac{13}{3}x + \cancel{\frac{95}{3}} - \cancel{\frac{95}{3}} = \frac{125}{6} - \frac{95}{3}$$ Simplify right side by converting $\frac{95}{3}$ to sixths: $$\frac{95}{3} = \frac{190}{6}$$ So: $$\frac{125}{6} - \frac{190}{6} = -\frac{65}{6}$$ Equation becomes: $$- \frac{13}{3}x = -\frac{65}{6}$$ 8. **Solve for $x$ by dividing both sides by $-\frac{13}{3}$:** $$x = \frac{-\frac{65}{6}}{-\frac{13}{3}} = \frac{-65}{6} \times \frac{3}{-13}$$ Intermediate step with cancellation: $$x = \frac{\cancel{-65}}{6} \times \frac{3}{\cancel{-13}} = \frac{5}{6} \times 3 = \frac{15}{6}$$ Simplify fraction: $$\frac{15}{6} = \frac{5}{2}$$ **Final answer:** $$x = \frac{5}{2}$$