A metal worker has several 1-kilogram bars of a metal alloy that contains 23%copper and several 1-kilogram bars that contain 79% cooper. How many bars of each type of alloy should be melted and combined to create 48 kilograms of a 44% copper alloy?

"a" - one kilo bar, which contains 23% copper. Mass of copper in "a" bars: 0.23a kg

"b" - on kilo bar, which contains 79% copper. Mass of copper in "b" bars: 0.79b kg

Sum of bars: 48 (because 48kg) so a+b=48

Sum of mass of copper: 44% from 48 = 0.44 * 48 = 21.12 so 0.23a + 0.79b = 21.12

From first equation you've got a=48-b. Substitute it to the second equation, you've got

0.23 (48-b) + 0.79b=21.12

11.04-0.23b+0.79b=21.12

0.56b=10.08 |:0.56

b=18

So a = 48-b=48-18=30

So he has to mix 30 bars that contain 23%copper and 18 bars that contain 79% copper

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