Let [tex]x[/tex] be the amount of the 30% alloy and [tex]y[/tex] the amount of the 60% alloy the metalworker will use. However much is used, the final alloy will have a mass of
[tex]x+y=100[/tex]
kilograms. For each kg of the 30% alloy used, 0.3 kg is copper; similary, each kg of the 60% alloy contributes 0.6 kg, so that
[tex]0.3x+0.6y=0.54(x+y)=54[/tex]
Now,
[tex]x+y=100\implies y=100-x[/tex]
[tex]\implies0.3x+0.6(100-x)=54[/tex]
[tex]\implies60-0.3x=54[/tex]
[tex]\implies0.3x=6[/tex]
[tex]\implies x=20[/tex]
[tex]\implies y=100-20=80[/tex]
