1. **Problem Statement:** We have three types of cars: Alto, Suzuki, and City. Their rental rates per day are unknown and denoted as $A$, $S$, and $C$ respectively. 2. **Given Information:** - Three Alto, two Suzuki, and four City cars cost 106 per day: $$3A + 2S + 4C = 106$$ - Two Alto, four Suzuki, and three City cars cost 107 per day: $$2A + 4S + 3C = 107$$ - Four Alto, three Suzuki, and two City cars cost 102 per day: $$4A + 3S + 2C = 102$$ 3. **Goal:** Find the values of $A$, $S$, and $C$. 4. **Method:** Solve the system of linear equations using substitution or elimination. 5. **Step 1: Write the system:** $$\begin{cases} 3A + 2S + 4C = 106 \\ 2A + 4S + 3C = 107 \\ 4A + 3S + 2C = 102 \end{cases}$$ 6. **Step 2: Eliminate one variable.** Multiply equations to align coefficients and subtract. Multiply equation 1 by 2 and equation 3 by 4 to eliminate $C$: $$6A + 4S + 8C = 212$$ $$16A + 12S + 8C = 408$$ Subtract first from second: $$16A + 12S + 8C - (6A + 4S + 8C) = 408 - 212$$ $$10A + 8S = 196$$ Simplify by dividing by 2: $$5A + 4S = 98 \, \,(1)$$ 7. **Step 3: Eliminate $C$ using equations 2 and 3.** Multiply equation 2 by 2 and equation 3 by 3: $$4A + 8S + 6C = 214$$ $$12A + 9S + 6C = 306$$ Subtract first from second: $$12A + 9S + 6C - (4A + 8S + 6C) = 306 - 214$$ $$8A + S = 92 \, \,(2)$$ 8. **Step 4: Solve the system of two equations (1) and (2):** $$\begin{cases} 5A + 4S = 98 \\ 8A + S = 92 \end{cases}$$ From (2), express $S$: $$S = 92 - 8A$$ Substitute into (1): $$5A + 4(92 - 8A) = 98$$ $$5A + 368 - 32A = 98$$ $$-27A = 98 - 368$$ $$-27A = -270$$ $$A = \frac{-270}{-27} = 10$$ 9. **Step 5: Find $S$ using $A=10$:** $$S = 92 - 8(10) = 92 - 80 = 12$$ 10. **Step 6: Find $C$ using $A=10$, $S=12$ in one original equation, e.g., equation 1:** $$3(10) + 2(12) + 4C = 106$$ $$30 + 24 + 4C = 106$$ $$54 + 4C = 106$$ $$4C = 106 - 54 = 52$$ $$C = \frac{52}{4} = 13$$ 11. **Final Answer:** - Rental rate for Alto ($A$) = 10 per day - Rental rate for Suzuki ($S$) = 12 per day - Rental rate for City ($C$) = 13 per day