1. **State the problem:** We are given demand and supply equations for maize and rice and need to find the equilibrium prices ($P_M$, $P_R$) and quantities ($Q_M$, $Q_R$). 2. **Equilibrium condition:** At equilibrium, demand equals supply for each good: $$Q_M^d = Q_M^s$$ $$Q_R^d = Q_R^s$$ 3. **Write the equations:** - Maize demand: $$Q_M^d = 60 - 2P_M + P_R$$ - Maize supply: $$Q_M^s = 20 + P_M$$ - Rice demand: $$Q_R^d = 50 - P_R + P_M$$ - Rice supply: $$Q_R^s = 10 + 2P_R$$ 4. **Set demand equal to supply for maize:** $$60 - 2P_M + P_R = 20 + P_M$$ Simplify: $$60 - 20 + P_R = 2P_M + P_M$$ $$40 + P_R = 3P_M$$ Rewrite: $$3P_M - P_R = 40 \quad (1)$$ 5. **Set demand equal to supply for rice:** $$50 - P_R + P_M = 10 + 2P_R$$ Simplify: $$50 - 10 + P_M = 2P_R + P_R$$ $$40 + P_M = 3P_R$$ Rewrite: $$P_M - 3P_R = -40 \quad (2)$$ 6. **Solve the system of equations:** From (1): $$3P_M - P_R = 40$$ From (2): $$P_M - 3P_R = -40$$ Multiply (2) by 3: $$3P_M - 9P_R = -120$$ Subtract (1) from this: $$(3P_M - 9P_R) - (3P_M - P_R) = -120 - 40$$ $$3P_M - 9P_R - 3P_M + P_R = -160$$ $$-8P_R = -160$$ $$P_R = 20$$ Substitute $P_R=20$ into (1): $$3P_M - 20 = 40$$ $$3P_M = 60$$ $$P_M = 20$$ 7. **Find equilibrium quantities:** - Maize quantity: $$Q_M = Q_M^s = 20 + P_M = 20 + 20 = 40$$ - Rice quantity: $$Q_R = Q_R^s = 10 + 2P_R = 10 + 2 \times 20 = 50$$ **Final answer:** $$P_M = 20, \quad P_R = 20, \quad Q_M = 40, \quad Q_R = 50$$