The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 10 denarius per day to support 3 legionaries and 3 archers. It only costs 3 denarius per day to support one legionary and one archer. 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 denarius per day to support.(B) Yes; a legionary costs 2 denarius per day to support, and an archer costs 34 denarius per day to support.(C) No; the system has many solutions.(D) No; the system has no solution.
Q. The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 10 denarius per day to support 3 legionaries and 3 archers. It only costs 3 denarius per day to support one legionary and one archer. 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 denarius per day to support.(B) Yes; a legionary costs 2 denarius per day to support, and an archer costs 34 denarius per day to support.(C) No; the system has many solutions.(D) No; the system has no solution.
Define Variables: Let's define two variables: let L be the cost to support one legionary per day, and A be the cost to support one archer per day. We can then create two equations based on the information given.
First Equation: The first equation comes from the statement that it costs the Roman government 10 denarius per day to support 3 legionaries and 3 archers. This can be written as:3L+3A=10
Second Equation: The second equation comes from the statement that it costs 3 denarius per day to support one legionary and one archer. This can be written as: L+A=3
Solve System: Now we have a system of two linear equations:1) 3L+3A=102) L+A=3We can solve this system using substitution or elimination. Let's use the elimination method.
Simplify First Equation: First, we can simplify the first equation by dividing every term by 3, which gives us:L+A=310But we notice that this equation is essentially the same as the second equation multiplied by a constant. This means that the two equations are not independent; they represent the same line.
Infinitely Many Solutions: Since both equations represent the same line, there are infinitely many solutions to this system. This means that we cannot determine a unique cost for each soldier based on the information given.
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