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}

52 - x = pounds of 60% aluminum alloy

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

and

≈ 21.4 pounds of 60% aluminum alloy

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