1. **State the problem:** Determine the type of system and the number of solutions for the system of equations: $$2x + 3y = 9$$ $$6y = 4x - 6$$ 2. **Rewrite the second equation in standard form:** $$6y = 4x - 6 \implies 4x - 6y = 6$$ 3. **Compare the two equations:** Equation 1: $$2x + 3y = 9$$ Equation 2: $$4x - 6y = 6$$ 4. **Check if the system is consistent and determine the number of solutions by comparing ratios of coefficients:** Calculate the ratios: $$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$$ $$\frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2}$$ $$\frac{c_1}{c_2} = \frac{9}{6} = \frac{3}{2}$$ 5. **Interpretation:** Since $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$, the lines are not parallel. Therefore, the system has **one unique solution**. 6. **Find the solution:** From equation 1: $$2x + 3y = 9 \implies 2x = 9 - 3y \implies x = \frac{9 - 3y}{2}$$ Substitute into equation 2: $$6y = 4x - 6$$ $$6y = 4 \times \frac{9 - 3y}{2} - 6$$ Simplify: $$6y = 2(9 - 3y) - 6 = 18 - 6y - 6 = 12 - 6y$$ Add $$6y$$ to both sides: $$6y + 6y = 12$$ $$12y = 12$$ Divide both sides by 12: $$\cancel{12}y = \frac{12}{\cancel{12}} \implies y = 1$$ Substitute $$y=1$$ back into $$x = \frac{9 - 3y}{2}$$: $$x = \frac{9 - 3(1)}{2} = \frac{9 - 3}{2} = \frac{6}{2} = 3$$ **Final answer:** The system has one unique solution at $$\boxed{(3,1)}$$.