document.write( "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? \n" ); document.write( "

Algebra.Com's Answer #498764 by josgarithmetic(39620) $\"\"$ $\"About$

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p, c, and b, the amounts of peanuts, cashews, and brazil nuts.
\n" ); document.write( " $\"c=p-5\"$ for the \"five fewer\" part.
\n" ); document.write( " $\"p%2Bc%2Bb=50\"$ for the 50 pound wanted part.
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\n" ); document.write( " $\"%282p%2B10c%2B10b%29%2F50=6.80\"$ accounts for mixture price.\r
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\n" ); document.write( "\n" ); document.write( "This appears to be a system of three equations in three unknowns, but proceed this way:\r
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\n" ); document.write( "\n" ); document.write( "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.
\n" ); document.write( "Solve this simpler system for p and b; and then find the value for c. \n" ); document.write( "

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