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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?
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 Ask the House
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