The denarius was a unit of currency in ancient rome. suppose it costs the roman government 1 0 1010 denarius per day to support 3 33 legionaries and 3 .33 archers. it only costs 3.33 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?

assuming you copied and pasted this from khan academy and the original numbers are 10, 3, 3 and 3, make two equations:

x+y=3 and 3x+3y=10

x+y=3 -> solve for y -> y=-x+3

plug into 3x+3y=10

3x+3(-x+3)=10

3x-3x+9=10 -> the 3x's cancel out

9=10

however 9 does not equal 10 and therefore, there is no solution

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)