1. **State the problem:** Solve the equation $$\frac{4}{3b-3} = \frac{9}{b+3}$$ for $b$. 2. **Rewrite the equation:** We have two fractions set equal: $$\frac{4}{3b-3} = \frac{9}{b+3}$$. 3. **Cross-multiply to eliminate fractions:** Multiply both sides by the denominators to get rid of the fractions: $$4(b+3) = 9(3b - 3)$$ 4. **Expand both sides:** Left side: $$4(b+3) = 4b + 12$$ Right side: $$9(3b - 3) = 27b - 27$$ So the equation becomes: $$4b + 12 = 27b - 27$$ 5. **Isolate variable terms:** Subtract $4b$ from both sides: $$12 = 27b - 27 - 4b$$ $$12 = 23b - 27$$ 6. **Isolate $b$:** Add 27 to both sides: $$12 + 27 = 23b$$ $$39 = 23b$$ 7. **Solve for $b$:** Divide both sides by 23: $$b = \frac{39}{23}$$ **Final answer:** $$\boxed{b = \frac{39}{23}}$$ This is the solution for $b$ that satisfies the original equation.