document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=792777>Question 792777</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Five yrs ago,a mother was twice as old as his son.in 6yrs time,d sum of their ages will b 82.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 #480134 by </B><A HREF=/tutors/aboutme.mpl?userid=psbhowmick><B>psbhowmick(878)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=psbhowmick><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=psbhowmick><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=480134\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the current age of son be x years and of mother be y years.\r<BR>\n" );
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document.write( "5 years ago:<BR>\n" );
document.write( "y-5=2(x-5)<BR>\n" );
document.write( "y=2x-5<BR>\n" );
document.write( "2x-y=5\r<BR>\n" );
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document.write( "6 years down:<BR>\n" );
document.write( "(x+6)+(y+6)=82<BR>\n" );
document.write( "x+y=70\r<BR>\n" );
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document.write( "x=25<BR>\n" );
document.write( "Y=45 </SPAN>\n" );
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