document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=511177>Question 511177</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many kilograms of a 15% copper alloy must be melted with a 45% copper alloy to obtain 60 kilograms of a 32% copper alloy? <BR>\n" );
document.write( "I tried to do it this way: 0.45x-0.15x=.32(60), but it didn't work. I don't know how to do these kinds of problems. I don't know how to turn them into equations. Can somebody please help me? And can somebody show me how to do it the way that I just tried, with the decimals before them, please? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #342036 by </B><A HREF=/tutors/aboutme.mpl?userid=Maths68><B>Maths68(1474)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Maths68><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Maths68><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=342036\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Two mistakes in your approach to solve this problem\r<BR>\n" );
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document.write( "1) When you take x kg of one alloy you must take (60-x)kg because we need 60 kg when we add x+(60-x) we will get 60 which is our requirement\r<BR>\n" );
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document.write( "2)We are adding both alloys but you subtract it!!!!\r<BR>\n" );
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document.write( "Now see what will be the equation\r<BR>\n" );
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document.write( "0.45x+0.15(60-x)=.32(60)\r<BR>\n" );
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document.write( "I hope you will solve the rest.<BR>\n" );
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