Question

The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 101010 denarius per day to support 333 legionaries and 333 archers. It only costs 333 denarius per day to support one legionary and one archer. Use a system of linear equations in two variables.

Answer 1

No; the system has no solution.

Explanation:

We'll use L to represent the cost per day of a legionary, and we'll use A to represent the cost per day of an archer. So we have the following system of equations:

3L+3A=10

L+A=3

If we multiply both sides of the second equation by 3, we get the equation 3L+3A=9

But this indicates that the cost to support 3 legionaries and 3 archers is 9 denarius. The first equation indicates that it costs 10 denarius to support 3 legionaries and 3 archers.

The cost cannot be 9 denarius and 10 denarius at the same time.

Henceforth the answer is:

No; the system has no solution.

Answer 2

Complete question :

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 3archers. 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?

Answer:

No unique solution

Explanation:

Given that:

Cost of supporting 3 legion aries and 3 archers = 10 denarius daily

Cost of supporting one legionary and one archer = 3 denarius

Let:

Legionaries = l ; archers = a

Equation for the first sentence :

3l + 3a = 10 - - - (1)

Second sentence :

l + a = 3 - - - - - (2)

From (2)

l = 3 - a

Substituting l = 3 - a into (1)

3(3 - a) + 3a = 10

9 - 3a + 3a = 10

9 - 0 = 10

The variable cancels out, Hence, ( there is no unique solution to find the cost of each soldier)