document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=246084>Question 246084</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Jason is now three times as old as helston. three years from now jason will be six times as old as helston one year ago find their present ages<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #179727 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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=179727\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let J = Jason's present age and H = Helston's present age.<BR>\n" );
document.write( "1) J = 3*H \"Jason is three times as old as Helston\"<BR>\n" );
document.write( "2) J+3 = 6(H-1) \"Three years from now Jason will be six times as old as Helston (was) one year ago\"<BR>\n" );
document.write( "Substitute J from equation 1) into equation 2) and simplify.<BR>\n" );
document.write( "(3*H)+3 = 6H-6 Subtract 3H from both sides.<BR>\n" );
document.write( "3 = 3H-6 Add 6 to both sides.<BR>\n" );
document.write( "9 = 3H Divide both sides by 3.<BR>\n" );
document.write( "3 = H and...<BR>\n" );
document.write( "J = 3*H = 3*3 = 9<BR>\n" );
document.write( "Jason is 9 and Helston is 3<BR>\n" );
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