mixtures

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mixtures

A metallurgist has one alloy containing 34% copper and another containing 48% copper. How many pounds of each alloy must he use to make 46 pounds of a third alloy containing 37% copper? (Round to two decimal places if necessary.)
AlgebraHouse AlgebraHouse
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Re: mixtures

x = pounds of 34% copper
46 - x = pounds of 48% copper

0.34x + 0.48(46 - x) = 0.37(46)   {pounds times percent value equals total}
0.34x + 22.08 - 0.48x = 17.02   {used distributive property}
-0.14x + 22.08 = 17.02   {combined like terms}
-0.14x = -5.06   {subtracted 22.08 from each side}
x ≈ 36.14   {divided by -0.14 and rounded to two places}
46 - x ≈ 9.86   {substituted 36.14, in for x,into 46 - x}

Approximately,
36.14 pounds of 34% copper
9.86 pounds of 48% copper


@AlgebraHouse