document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=216315>Question 216315</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much of a 90% gasoline mixture and a 75% gasoline mixture would be needed for 600 liters of an 85% gasoline mixture? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #163368 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=163368\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 0.90x+.75(600-x)=.85*600<BR>\n" );
document.write( ".90x+450=.75x=510<BR>\n" );
document.write( ".90x-.75x=510-450<BR>\n" );
document.write( ".15x=60<BR>\n" );
document.write( "x=60/.15<BR>\n" );
document.write( "x=400 gallons of 90% gas is used.<BR>\n" );
document.write( "600-400=200 gallons of 75% gas is used.<BR>\n" );
document.write( "Proof:<BR>\n" );
document.write( ".90*400+.75*200=.85*600<BR>\n" );
document.write( "360+150=510<BR>\n" );
document.write( "510=510 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );