document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=979209>Question 979209</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a fraction twice the numerator in 2 more than the denominator. If 3 is added to the numerator and to the denominator the new fraction is 2/3. Find original fraction.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #600491 by </B><A HREF=/tutors/aboutme.mpl?userid=subhash+Sharma+><B>subhash Sharma (1)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=subhash+Sharma+><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=subhash+Sharma+><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=600491\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=the numerator and y= the denominator                                                                                                  Now we are told that<BR>\n" );
document.write( "2x=y+2----------equation (1)<BR>\n" );
document.write( "and (x+3)/(y+3)=2/3<BR>\n" );
document.write( "        3(x+3)=2(y+3)<BR>\n" );
document.write( "         3x+9=2y+6<BR>\n" );
document.write( "         3x-2y=-3------------equation (2)<BR>\n" );
document.write( "     Solve from eqn. (1) and (2) we get, <BR>\n" );
document.write( "        2x-y=2----------(1)×2<BR>\n" );
document.write( "        3x-2y=-3--------(2)×1<BR>\n" );
document.write( " Now eqn.(1)-(2) we get, <BR>\n" );
document.write( "  4x-2y=4 <BR>\n" );
document.write( "  3x-2y=-3<BR>\n" );
document.write( "                                               X=7<BR>\n" );
document.write( "   And from equation(1)     y=12 <BR>\n" );
document.write( "   Therefore, original fraction x/y=7/12.  </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );