SOLUTION: A person wishes to mix coffee worth $12 per lb with coffee worth $6 per lb to get 90 lb of a mixture worth $8 per lb. How many pounds of the $12 and the $6 coffees will be needed?
Algebra ->
Equations
-> SOLUTION: A person wishes to mix coffee worth $12 per lb with coffee worth $6 per lb to get 90 lb of a mixture worth $8 per lb. How many pounds of the $12 and the $6 coffees will be needed?
Log On
Question 1004583: A person wishes to mix coffee worth $12 per lb with coffee worth $6 per lb to get 90 lb of a mixture worth $8 per lb. How many pounds of the $12 and the $6 coffees will be needed?
You can put this solution on YOUR website! x+y= 90
Subtract x on both sides:
y= 90-x In the next equation we will substitute y with this value:
----------------------
We want to get 90lbs of coffee at $8 per pound, we want 90*8= 720:
12x+6y= 8(90) In this equation, substitute the value of y with 90-y:
--------------
12x+6(90-x)= 720
12x+540-6x= 720
6x= 180
x= 30
You need 30lb of $12 coffee with 90-30= 60 lbs of the $6 coffee
