1. Stating the problem: Solve the equation $$\frac{2}{3} + \frac{1}{a - b} = \frac{2}{3}a$$ for $a$. 2. Write down the equation: $$\frac{2}{3} + \frac{1}{a - b} = \frac{2}{3}a$$ 3. To solve for $a$, first eliminate the fractions by multiplying both sides by the common denominator $3(a - b)$: $$3(a - b) \left( \frac{2}{3} + \frac{1}{a - b} \right) = 3(a - b) \cdot \frac{2}{3}a$$ 4. Simplify each term: $$3(a - b) \cdot \frac{2}{3} = 2(a - b)$$ $$3(a - b) \cdot \frac{1}{a - b} = 3$$ $$3(a - b) \cdot \frac{2}{3}a = 2a(a - b)$$ 5. Substitute back: $$2(a - b) + 3 = 2a(a - b)$$ 6. Expand the left side: $$2a - 2b + 3 = 2a^2 - 2ab$$ 7. Bring all terms to one side: $$0 = 2a^2 - 2ab - 2a + 2b - 3$$ 8. Rearrange: $$2a^2 - 2ab - 2a + 2b - 3 = 0$$ 9. This is a quadratic equation in $a$: $$2a^2 - 2a(b + 1) + (2b - 3) = 0$$ 10. Use the quadratic formula: $$a = \frac{2(b + 1) \pm \sqrt{(-2(b + 1))^2 - 4 \cdot 2 \cdot (2b - 3)}}{2 \cdot 2}$$ 11. Simplify inside the square root: $$\Delta = 4(b + 1)^2 - 8(2b - 3) = 4(b^2 + 2b + 1) - 16b + 24 = 4b^2 + 8b + 4 - 16b + 24 = 4b^2 - 8b + 28$$ 12. Final solution: $$a = \frac{2(b + 1) \pm \sqrt{4b^2 - 8b + 28}}{4} = \frac{b + 1 \pm \frac{1}{2}\sqrt{4b^2 - 8b + 28}}{2}$$ 13. This gives two possible values for $a$ depending on the sign chosen. Final answer: $$a = \frac{b + 1 \pm \frac{1}{2} \sqrt{4b^2 - 8b + 28}}{2}$$