document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1120930>Question 1120930</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A man has two daughters, one three times as old as the other. The man is five times as old as his older daughter and in 5 years he will be five times as old as the younger. Find their present ages. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #736623 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=736623\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> older daughter is x years now.  Man is 5x, younger daughter is 1/3 x.<BR>\n" );
document.write( "In 5 years, man will be 5x+5 and younger daughter 1/3 x+5<BR>\n" );
document.write( "so 5x+5=5((1/3 x +5)<BR>\n" );
document.write( "5x+5=(5/3)x+25<BR>\n" );
document.write( "(10/3)x=20, since 5x=15/3 x<BR>\n" );
document.write( "10x=60<BR>\n" );
document.write( "x=6 years older daughter<BR>\n" );
document.write( "(1/3)x=2 years younger daughter<BR>\n" );
document.write( "5x=30 years man.<BR>\n" );
document.write( "In 5 years, he will be 35 and younger daughter 7, which is 5 times as much. </SPAN>\n" );
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