The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier? Choose 1 answer: (A) Yes; a legionary costs 1 denarius per day to support, and an archer costs 2 denarii per day to support. (B) Yes; a legionary costs 2 denarii per day to support, and an archer costs (4)/(3) denarii per day to support. (C) No; the system has many solutions. (D) No; the system has no solution.

Question

The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier? Choose 1 answer: (A) Yes; a legionary costs 1 denarius per day to support, and an archer costs 2 denarii per day to support. (B) Yes; a legionary costs 2 denarii per day to support, and an archer costs  (4)/(3) denarii per day to support. (C) No; the system has many solutions. (D) No; the system has no solution.

Bytelearn · Accepted Answer

Define Variables: Let's define the variables for the cost to support each type of soldier. Let L be the cost to support a legionary per day, and A be the cost to support an archer per day.Scenario 1: We are given two scenarios. In the first scenario, it costs 10 denarii to support 4 legionaries and 4 archers. This can be represented by the equation: 4L + 4A = 10Scenario 2: In the second scenario, it costs 5 denarii to support 2 legionaries and 2 archers. This can be represented by the equation: 2L + 2A = 5Simplify Equation: We can simplify the second equation by dividing each term by 2, which gives us: L + A = 2.5 This is a simpler equation that we can use to solve the system.System of Equations: Now we have a system of two equations: 4L + 4A = 10 L + A = 2.5 We can use the second equation to solve for one of the variables in terms of the other.Solve for L: Let's solve the second equation for L: L = 2.5 - A Now we have an expression for L that we can substitute into the first equation.Substitute into First Equation: Substitute L = 2.5 - A into the first equation: 4(2.5 - A) + 4A = 10 Now, distribute the 4: 10 - 4A + 4A = 10Canceling Terms: We notice that the terms -4A and +4A cancel each other out, leaving us with: 10 = 10 This means that the two equations are not independent; they are actually the same equation when simplified. Therefore, the system does not have a unique solution.No Unique Solution: Since the system does not have a unique solution, we cannot determine the individual costs for a legionary and an archer based on the given information. The correct answer is: (C) No; the system has many solutions.

Full solution

More problems from Solve a system of equations using substitution: word problems

Related topics