document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=121550>Question 121550</A>: <I><SPAN STYLE=\"word-break: break-all;\"> im not very good at algebra. Joe's dad s 2 times joe's age. 7 years ago, he was 3 times older than joe. find both's ages. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #89204 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=89204\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let D = Dad's present age and J = Joe's present age.<BR>\n" );
document.write( "From the problem description you can write:<BR>\n" );
document.write( "D = 2J  \"Dad is two times Joe's age\". Substitute this value of D into the second equation and solve for J.<BR>\n" );
document.write( "D-7 = 3(J-7) \"Seven years ago (that's -7), Dad was three times Joe's age\".<BR>\n" );
document.write( "2J-7 = 3(J-7) Simplify.<BR>\n" );
document.write( "2J-7 = 3J-21 Subtract 2J from both sides.<BR>\n" );
document.write( "-7 = J-21  Add 21 to both sides.<BR>\n" );
document.write( "14 = J<BR>\n" );
document.write( "Joe's age is 14<BR>\n" );
document.write( "D = 2J<BR>\n" );
document.write( "D = 2(14)<BR>\n" );
document.write( "D = 28<BR>\n" );
document.write( "Dad's age is 28 </SPAN>\n" );
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