An alloy of copper is 10% copper and weighs 25 pounds. a second alloy is 18% copper. how much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 13% mixture.

  let  the weight  of the  second  alloy  be  y
convert the  percentage  to  fraction, that  is  10/100=0.1,  18/100=0.18,  13/100=0.13
 Total weight  if  the  mixture is  0.1  x 25+0.18y=0.13(25+y)
2.5+0.18y=3.25+0.13y
  like  terms  together
0.18y +0.13y =  3.25-2.5
 0.05y=0.75
divide both  side  by  0.05
 Y  is  therefore  15 pound  of 18%  copper is needed

Ibelieve the answer would be 7, 13, 23