document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=42722>Question 42722</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Winston works faster. Winston can mow his dad’s lawn in 1 hour less than it takes his brother Willie. If they take 2 hours to mow it when working together, then how long would it take Winston working alone? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #27829 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=27829\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The working equation for this kind of problem is<BR>\n" );
document.write( "t/A + t/B = 1<BR>\n" );
document.write( "where t is the time worked together and A and B are individual times...<BR>\n" );
document.write( "Here t = 2...if we let A be Winston's time, then Willie's time is A + 1...we have<BR>\n" );
document.write( "2/A + 2/(A+1) = 1<BR>\n" );
document.write( "Now multiply through by the LCD and solve...we get<BR>\n" );
document.write( "2(A + 1) + 2A = A(A + 1)<BR>\n" );
document.write( "2A + 2 + 2A = A^2 + A<BR>\n" );
document.write( "A^2 - 3A - 2 = 0<BR>\n" );
document.write( "This can't be factored so use the quadratic...<BR>\n" );
document.write( "A = (3 ± sqrt(9 + 8)) / 2<BR>\n" );
document.write( "A = (3 ± sqrt(17)) / 2<BR>\n" );
document.write( "Since time can't be negative, Winston's time must be<BR>\n" );
document.write( "(3 + sqrt(17)) / 2 or about 3.7 hours... </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );