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 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.

Answer 1

$\huge\mathfrak\blue{Answer:-}$.

Let the legionary's cost be x

Let the archer's cost be y

Then ATQ to $\huge\mathtt\purple{1st\:Condition}$.

3x+3y=10

And ATQ to $\huge\mathtt\red{2nd\:Condition}$

x+y= 3

Answer 2

Answer:

No; the system has no solution.

Step-by-step 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.

No; the system has no solution.