document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1106566>Question 1106566</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Triple one number decreased by the second is equal to eight. The sum of the numbers is 12. Find the numbers. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #721564 by </B><A HREF=/tutors/aboutme.mpl?userid=davidb52><B>davidb52(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=davidb52><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=721564\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the first number and y be the second number.<BR>\n" );
document.write( "You'd get two equations:\r<BR>\n" );
document.write( "\n" );
document.write( "3x - y = 8<BR>\n" );
document.write( "x + y = 12\r<BR>\n" );
document.write( "\n" );
document.write( "The second equation can be used to define y in terms of x, (or x in terms of y, whichever you prefer), and then subbed into the first equation to solve for x (or y.) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );