document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=184814>Question 184814</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve Problems Using Linear Systems<BR>\n" );
document.write( "16. One type of fertilizer has 30% nitrogen and a second type has 15% nitrogen. If a farmer needs 600kg of fertilizer that is 20% nitrogen, how much of each type should the farmer mix together?<BR>\n" );
document.write( "Thank you very much. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #138785 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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=138785\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .30X+.15(600-X)=600*.20<BR>\n" );
document.write( ".30X+90-.15X=120<BR>\n" );
document.write( ".15X=120-90<BR>\n" );
document.write( ".15X=30<BR>\n" );
document.write( "X=30/.15<BR>\n" );
document.write( "X=2000 KG. OF 30% NITROGEN.<BR>\n" );
document.write( "600-200=400 KG. OF .15% NITROGEN.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( ".30*200+.15*400=120<BR>\n" );
document.write( "60+60=120 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );