document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=189274>Question 189274</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Word problem:<BR>\n" );
document.write( "A chainsaw runs on fuel that is 10% oil and 90% gas.  Howmuch oil will have to be added to 3 gallons of fuel that is 5% oil and 95% gas in order to use the chainsaw? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #142011 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=142011\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=amount of oil that has to be added<BR>\n" );
document.write( "The amount of pure gas before the oil is added (0.95*3) has to equal the amount of pure gas after the oil is added 0.90(3+x), so:<BR>\n" );
document.write( "0.95*3=0.90(3+x)<BR>\n" );
document.write( "2.85=2.70+0.90x  subtract 2.70 from each side<BR>\n" );
document.write( "2.85-2.70=2.70-2.70+0.90x  collect like terms<BR>\n" );
document.write( "0.15=0.90x  divide both sides by 0.90<BR>\n" );
document.write( "x=0.16667 gal ------------------ans<BR>\n" );
document.write( "CK<BR>\n" );
document.write( "3*0.95=0.90*(3.16667)<BR>\n" );
document.write( "2.85=2.85\r<BR>\n" );
document.write( "\n" );
document.write( "Hope this helps---ptaylor </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );