SOLUTION: The owner of a candy store wants to mix some peanuts worth $2 per pound, some cashews worth $10 per pound, and some Brazil nuts worth $10 per pound to get 50 pounds of a mixture th

Question 827575: The owner of a candy store wants to mix some peanuts worth $2 per pound, some cashews worth $10 per pound, and some Brazil nuts worth $10 per pound to get 50 pounds of a mixture that will sell for $6.80 per pound. She uses 5 fewer pounds of cashews than peanuts. How many pounds of each did she use?
Found 2 solutions by lwsshak3, josgarithmetic:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The owner of a candy store wants to mix some peanuts worth $2 per pound, some cashews worth $10 per pound, and some Brazil nuts worth $10 per pound to get 50 pounds of a mixture that will sell for $6.80 per pound. She uses 5 fewer pounds of cashews than peanuts. How many pounds of each did she use?
***
let x=amt of peanuts used
x-5=amt of cashews used
50-(x-5)-x=50-2x+5=55-2x=amt of Brazil nuts used
2x+10(x-5)+10(55-2x)=50(6.8)
2x+10x-50+550-20x=340
8x=160
x=20
..
amt of peanuts used=20pounds
amt of cashews used=15pounds
amt of Brazil nuts used=15pounds
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
p, c, and b, the amounts of peanuts, cashews, and brazil nuts.
for the "five fewer" part.
for the 50 pound wanted part.
'
accounts for mixture price.

This appears to be a system of three equations in three unknowns, but proceed this way:

Multiply the mixture price account equation by 50 and then divide by 2 to simplify it. Substitute for c in both the quantity sum equation and the simplified price account equation,and simplify these two; so you now have a system of two equations in just the unknowns, p and b.
Solve this simpler system for p and b; and then find the value for c.
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