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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 
(4)/(3) denarius per day to support.
(C) No; the system has many solutions.
D 
No; the system has no solution.

The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 1010 denarius per day to support 33 legionaries and 33 archers. It only costs 33 denarius per day to support one legionary and one archer. Use a system of linear equations in two variables.\newlineCan we solve for a unique cost for each soldier?\newlineChoose 11 answer:\newline(A) Yes; a legionary costs 11 denarius per day to support, and an archer costs 22 denarius per day to support.\newline(B) Yes; a legionary costs 22 denarius per day to support, and an archer costs 43 \frac{4}{3} denarius per day to support.\newline(C) No; the system has many solutions.\newline(D) No; the system has no solution.

Full solution

Q. The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 1010 denarius per day to support 33 legionaries and 33 archers. It only costs 33 denarius per day to support one legionary and one archer. Use a system of linear equations in two variables.\newlineCan we solve for a unique cost for each soldier?\newlineChoose 11 answer:\newline(A) Yes; a legionary costs 11 denarius per day to support, and an archer costs 22 denarius per day to support.\newline(B) Yes; a legionary costs 22 denarius per day to support, and an archer costs 43 \frac{4}{3} denarius per day to support.\newline(C) No; the system has many solutions.\newline(D) No; the system has no solution.
  1. Define Variables: Let's define two variables: let LL be the cost to support one legionary per day, and AA be the cost to support one archer per day. We can then create two equations based on the information given.
  2. First Equation: The first equation comes from the statement that it costs the Roman government 1010 denarius per day to support 33 legionaries and 33 archers. This can be written as:\newline3L+3A=103L + 3A = 10
  3. Second Equation: The second equation comes from the statement that it costs 33 denarius per day to support one legionary and one archer. This can be written as: L+A=3L + A = 3
  4. Solve System: Now we have a system of two linear equations:\newline11) 3L+3A=103L + 3A = 10\newline22) L+A=3L + A = 3\newlineWe can solve this system using substitution or elimination. Let's use the elimination method.
  5. Simplify First Equation: First, we can simplify the first equation by dividing every term by 33, which gives us:\newlineL+A=103L + A = \frac{10}{3}\newlineBut 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.
  6. 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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