document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1010644>Question 1010644</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One number is 3 less than a second number.  Twice the second number is 12 more than 3 times the first.  Find the two numbers. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #626126 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=626126\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the two numbers, x and y.  Then we have<BR>\n" );
document.write( "x = y - 3  ad<BR>\n" );
document.write( "2y = 12 + 3x<BR>\n" );
document.write( "Now substitute the first into the second equation and we get<BR>\n" );
document.write( "2y = 12 + 3(y - 3)<BR>\n" );
document.write( "2y = 12 + 3y - 9<BR>\n" );
document.write( "2y = 3y + 3<BR>\n" );
document.write( "-y = 3<BR>\n" );
document.write( "y = -3<BR>\n" );
document.write( "x = -3 - 3 = -6<BR>\n" );
document.write( "(-6, -3) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );