<u>the correct question is</u>
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?
Let
x-------> the cost to support a legionary per day
y-------> the cost to support an archer per day
we know that
4x+4y=10 ---------> equation 1
2x+2y=5 ---------> equation 2
If you multiply equation 1 by 2
2*(2x+2y)=2*5-----------> 4x+4y=10
so
equation 1 and equation 2 are the same
The system has infinite solutions-------> Is a consistent dependent system
therefore
<u>the answer is</u>
We cannot solve for a unique cost for each soldier, because there are infinite solutions.
The answer 5 (1/4) because if you add 1 and a quarter to it then that would make it 6 and a half
Oh boy, here we go again
First we must convert 1 2/3 to an improper fraction. By doing this, we get 5/3 (3/3 + 2/3)
So now we have 273 / 5/3
To divide this easier we can do something that when I learned it was called (keep, change, flip) which basically means keep the first fraction, change the sign from division to multiplication, and flip the second fraction
This now turns into: 273/1 * 3/5
Combine 273 and 3/5
273⋅3/5
Multiply 273
by 3
819/5
is your answer
Multiplication, addition, subtraction
The common denominator is already there 12 is the common denominator!! :)
