1. **State the problem:** We are given demand and supply functions for two commodities: $$Q_{d1} = 100 - 2P_1 + P_2, \quad Q_{s1} = P_1 - 10$$ $$Q_{d2} = 5 + 2P_1 - 3P_2, \quad Q_{s2} = 6P_2 - 5$$ We need to find the equilibrium prices $P_1$ and $P_2$ where demand equals supply for both commodities: $$Q_{d1} = Q_{s1}, \quad Q_{d2} = Q_{s2}$$ 2. **Set up the system of equations:** From $Q_{d1} = Q_{s1}$: $$100 - 2P_1 + P_2 = P_1 - 10$$ Rearranging: $$100 + 10 = P_1 + 2P_1 - P_2$$ $$110 = 3P_1 - P_2$$ Rewrite as: $$3P_1 - P_2 = 110$$ From $Q_{d2} = Q_{s2}$: $$5 + 2P_1 - 3P_2 = 6P_2 - 5$$ Rearranging: $$5 + 5 = 6P_2 + 3P_2 - 2P_1$$ $$10 = 9P_2 - 2P_1$$ Rewrite as: $$-2P_1 + 9P_2 = 10$$ 3. **Fill in the missing entries in the system:** The system is: $$3P_1 - P_2 = 110$$ $$-2P_1 + 9P_2 = 10$$ The problem states the system as: $$-3P_1 + P_2 = -110$$ $$2P_1 - 9P_2 = -10$$ which is the same system multiplied by $-1$ for both equations. 4. **Solve the system:** Using matrix form: $$\begin{bmatrix}3 & -1 \\ -2 & 9\end{bmatrix} \begin{bmatrix}P_1 \\ P_2\end{bmatrix} = \begin{bmatrix}110 \\ 10\end{bmatrix}$$ The inverse matrix is: $$\frac{1}{(3)(9) - (-1)(-2)} \begin{bmatrix}9 & 1 \\ 2 & 3\end{bmatrix} = \frac{1}{27 - 2} \begin{bmatrix}9 & 1 \\ 2 & 3\end{bmatrix} = \frac{1}{25} \begin{bmatrix}9 & 1 \\ 2 & 3\end{bmatrix}$$ Multiply inverse by constants: $$\begin{bmatrix}P_1 \\ P_2\end{bmatrix} = \frac{1}{25} \begin{bmatrix}9 & 1 \\ 2 & 3\end{bmatrix} \begin{bmatrix}110 \\ 10\end{bmatrix} = \frac{1}{25} \begin{bmatrix}9 \times 110 + 1 \times 10 \\ 2 \times 110 + 3 \times 10\end{bmatrix} = \frac{1}{25} \begin{bmatrix}990 + 10 \\ 220 + 30\end{bmatrix} = \frac{1}{25} \begin{bmatrix}1000 \\ 250\end{bmatrix}$$ Simplify: $$P_1 = \frac{1000}{25} = 40$$ $$P_2 = \frac{250}{25} = 10$$ 5. **Calculate equilibrium quantities:** Using $Q_{s1} = P_1 - 10$: $$Q_1 = 40 - 10 = 30$$ Using $Q_{s2} = 6P_2 - 5$: $$Q_2 = 6 \times 10 - 5 = 60 - 5 = 55$$ **Final answers:** $$P_1 = 40, \quad P_2 = 10$$ $$Q_1 = 30, \quad Q_2 = 55$$