In 20 oz. of one alloy there are 6 oz. ofcopper, 4 oz. of zine and 10 oz. of lea

Question

In 20 oz. of one alloy there are 6 oz. ofcopper, 4 oz. of zine and 10 oz. of lead. In 20 oz. of a
second alloy there are 12 oz. of copper, 5 oz. of zinc and 3 oz oflead, while in 20 oz. of of a third
alloy there are 8 oz. of copper, 6 oz. of zinc and 6 oz. of lead.How many ounces of each alloy
should be combined to make a new alloy containing 34 oz. of copper,17 oz. of zinc and 19 oz.
of lead? In 20 oz. of one alloy there are 6 oz. ofcopper, 4 oz. of zine and 10 oz. of lead. In 20 oz. of a
second alloy there are 12 oz. of copper, 5 oz. of zinc and 3 oz oflead, while in 20 oz. of of a third
alloy there are 8 oz. of copper, 6 oz. of zinc and 6 oz. of lead.How many ounces of each alloy
should be combined to make a new alloy containing 34 oz. of copper,17 oz. of zinc and 19 oz.
of lead?

Explanation / Answer

This is easily solved using the method of undeterminedcoefficients. Let:    cu = copper    zn = zinc    pb = lead    a1 = ounces of alloy #1    a2 = ounces of alloy #2    a3 = ounces of alloy #3 Now we want to satisfy this equation:    a1((6cu+4zn+10pb)/20)+a2 ((12cu+5zn+3pb)/20)+a3 ((8cu+6zn+6pb)/20) =34cu+17zn+19pb Using the left hand side of this equation find thecoefficients of cu, zn, and pb respectivley. Expanding the leftside these are:    coefficient of cu = (3 a1)/10+(3 a2)/5+(2 a3)/5    coefficient of zn = a1/5+a2/4+(3 a3)/10    coefficient of pb = a1/2+(3 a2)/20+(3 a3)/10 Now set these coefficients equal to the right hand side ofyour equation. This gives three equations in three unknown asfollows:    (3a1)/10+(3 a2)/5+(2 a3)/5 = 34, a1/5+a2/4+(3 a3)/10 = 17, a1/2+(3 a2)/20+(3 a3)/10 =19 Solve these three linear equations and you get the answer a1 =20, a2 = 40, a3 = 10 Using the left hand side of this equation find thecoefficients of cu, zn, and pb respectivley. Expanding the leftside these are:    coefficient of cu = (3 a1)/10+(3 a2)/5+(2 a3)/5    coefficient of zn = a1/5+a2/4+(3 a3)/10    coefficient of pb = a1/2+(3 a2)/20+(3 a3)/10 Now set these coefficients equal to the right hand side ofyour equation. This gives three equations in three unknown asfollows:    (3a1)/10+(3 a2)/5+(2 a3)/5 = 34, a1/5+a2/4+(3 a3)/10 = 17, a1/2+(3 a2)/20+(3 a3)/10 =19 Solve these three linear equations and you get the answer a1 =20, a2 = 40, a3 = 10

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