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.

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?

No unique solution

Step-by-step 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)

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3