1. Problem 1: Given the demand function $Q_{dx} = 30 - 3P$. a. Find the quantity demanded when $P=5$ and $P=9$. Step 1: Substitute $P=5$ into the demand function: $$Q_{dx} = 30 - 3 \times 5 = 30 - 15 = 15$$ Step 2: Substitute $P=9$ into the demand function: $$Q_{dx} = 30 - 3 \times 9 = 30 - 27 = 3$$ b. Find the price when the quantity demanded is 25 units and 100 units. Step 1: For $Q_{dx} = 25$: $$25 = 30 - 3P \implies 3P = 30 - 25 = 5 \implies P = \frac{5}{3} \approx 1.67$$ Step 2: For $Q_{dx} = 100$: $$100 = 30 - 3P \implies -3P = 100 - 30 = 70 \implies P = -\frac{70}{3} \approx -23.33$$ Since price cannot be negative, this value is not economically feasible. c. Graph of the demand function is a straight line $Q_{dx} = 30 - 3P$. 2. Problem 2: Find the demand function given two points $(P_1=160, Q_1=12)$ and $(P_2=100, Q_2=25)$. Step 1: Assume the demand function is linear: $Q = a - bP$. Step 2: Create system of equations: $$12 = a - b \times 160$$ $$25 = a - b \times 100$$ Step 3: Subtract second from first: $$12 - 25 = (a - 160b) - (a - 100b) \implies -13 = -60b \implies b = \frac{13}{60} \approx 0.217$$ Step 4: Substitute $b$ into one equation to find $a$: $$25 = a - 0.217 \times 100 \implies 25 = a - 21.7 \implies a = 46.7$$ Step 5: The demand function is: $$Q = 46.7 - 0.217P$$ 3. Problem 3: Given supply function $Q_{sx} = 2P - 1$. a. Find quantity supplied when $P=2$ and $P=10$. $$Q_{sx}(2) = 2 \times 2 - 1 = 4 - 1 = 3$$ $$Q_{sx}(10) = 2 \times 10 - 1 = 20 - 1 = 19$$ b. Find price when quantity supplied is 25 and 100. For $Q_{sx} = 25$: $$25 = 2P - 1 \implies 2P = 26 \implies P = 13$$ For $Q_{sx} = 100$: $$100 = 2P - 1 \implies 2P = 101 \implies P = 50.5$$ c. Supply function graph is $Q_{sx} = 2P -1$. 4. Problem 4: Given $Q_d = 10 - 3P$ and $Q_s = 2P - 1$, find market equilibrium. a. Set $Q_d = Q_s$: $$10 - 3P = 2P - 1 \implies 10 + 1 = 2P + 3P \implies 11 = 5P \implies P = \frac{11}{5} = 2.2$$ Substitute $P=2.2$ into $Q_d$: $$Q = 10 - 3 \times 2.2 = 10 - 6.6 = 3.4$$ So equilibrium price is $2.2$ and quantity is $3.4$ units. b. Graph equilibrium curves $Q_d = 10 - 3P$ and $Q_s = 2P - 1$ and mark intersection at $(P=2.2, Q=3.4)$. Final answers: - Problem 1a: $Q=15$ when $P=5$, $Q=3$ when $P=9$. - Problem 1b: $P=1.67$ for $Q=25$, no feasible price for $Q=100$. - Problem 2: Demand function $Q=46.7 - 0.217P$. - Problem 3a: $Q=3$ at $P=2$, $Q=19$ at $P=10$. - Problem 3b: $P=13$ for $Q=25$, $P=50.5$ for $Q=100$. - Problem 4a: Equilibrium at $P=2.2$, $Q=3.4$.