Alloy Mixture Problem

A metallurgist has one alloy containing 26% aluminum and another containing 60% aluminum. How many pounds of each alloy must he use to make 52 pounds of a third alloy containing 40% aluminum?

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x = pounds of 26% aluminum alloy
52 - x = pounds of 60% aluminum alloy   {there are 52 total pounds}

0.26(x) + 0.6(52 - x) = 0.4(52)
0.26x + 31.2 - 0.6x = 20.8   {used distributive property}
-0.34x + 31.2 = 20.8   {combined like terms}
-0.34x = -10.4   {subtracted 31.2 from each side}
x ≈ 30.6   {divided each side by -0.34}
52 - x ≈ 21.4   {substituted 30.6, in for x, into 52 - x}

≈ 30.6 pounds of 26% aluminum alloy
and
≈ 21.4 pounds of 60% aluminum alloy

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