SOLUTION: This is not exactly what my problem is but it's the closest thing I could understand. The problem says "Coffee worth $3.75 a pound was mixed with coffee worth $4.35 a pound to prod

Algebra ->  Graphs -> SOLUTION: This is not exactly what my problem is but it's the closest thing I could understand. The problem says "Coffee worth $3.75 a pound was mixed with coffee worth $4.35 a pound to prod      Log On


   



Question 116010: This is not exactly what my problem is but it's the closest thing I could understand. The problem says "Coffee worth $3.75 a pound was mixed with coffee worth $4.35 a pound to produce a blend worth $4.11 a pound. How much of each kind of coffee was used to produce 40 pounds of blended coffee?" My problem is finding the equations to solve by elimination because, this seems to need 3 variables in order to be made into the first equation. Hope I explained it well enough for you to help.
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
one variable is plenty

let x="$3.75 coffee", so 40-x="$4.35 coffee" ___ working in cents to avoid decimals

375(x)+435(40-x)=411(40) ___ 375x+17400-435x=16440 ___ -60x=-960