document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=52688>Question 52688</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0918091624/algebrahomeworkh>Intoductory Algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> A metallurgist has one alloy cantaining 20% copper and another containing 70% copper.  How many pounds of each alloy must he use to make 50 pounds of a third alloy containing 50% copper?<BR>\n" );
document.write( "Find the equation(s) and solve. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #35167 by </B><A HREF=/tutors/aboutme.mpl?userid=funmath><B>funmath(2933)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=funmath><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=35167\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the amount of 20% copper=x<BR>\n" );
document.write( "Then the rest of the mixture would be 70%, let the amount of 70% mixture=50-x<BR>\n" );
document.write( "You multiply percent times amount to get the solutions.<BR>\n" );
document.write( ".20x+.70(50-x)=.50(50) To eliminate the decimals, multiply both sides of the equation by 100 or 10.<BR>\n" );
document.write( "10(.20x)+10(.70)(50-x)=10(.50)(50)<BR>\n" );
document.write( "2x+7(50-x)=5(50)<BR>\n" );
document.write( "2x+350-7x=250<BR>\n" );
document.write( "-5x+350=250<BR>\n" );
document.write( "-5x+350-350=250-350<BR>\n" );
document.write( "-5x=-100<BR>\n" );
document.write( "-5x/-5=-100/-5<BR>\n" );
document.write( "x=20<BR>\n" );
document.write( "The amount of 20% alloy is 20 pounds.<BR>\n" );
document.write( "The amount of 70% alloy is 50-20=30 pounds. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );