SOLUTION: I am having trouble setting up the problem/equations for this question... Deoraj Bharath wishes to mix coffee worth $6 per lb with coffee worth $3 per lb to get 90 lb of a mixture

Question 764425: I am having trouble setting up the problem/equations for this question... Deoraj Bharath wishes to mix coffee worth $6 per lb with coffee worth $3 per lb to get 90 lb of a mixture worth $4 per lb. How many pounds of the $6 and the $3 coffees will be needed? Can you please help me set up and solve?

Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Deoraj has two coffees, each of a different strength ---- this strength being PRICE. One coffee is L dollars per pound, and the other coffee is H dollars per pound. The person wants a M pounds of a blend which is T dollars per pound. The question is, how many pounds, u and v are needed? The u and v are the pounds of the L and the H coffees respectively.

ACCOUNT FOR PRICES:

ACCOUNT FOR POUNDAGE:

The two equations form a system. Solve for u and v.
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