document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1012894>Question 1012894</A>: <I><SPAN STYLE=\"word-break: break-all;\"> twice a number minus a second number is -1. twice the second number added to three times the first number is 9. 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 #628978 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=628978\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the numbers x and y.  Then we have<BR>\n" );
document.write( "2x - y = -1  and<BR>\n" );
document.write( "2y + 3x = 9<BR>\n" );
document.write( "Let's rearrange the first and get<BR>\n" );
document.write( "y = 2x + 1<BR>\n" );
document.write( "and substitute that into the second equation...we get<BR>\n" );
document.write( "2(2x+1) + 3x = 9<BR>\n" );
document.write( "4x + 2 + 3x = 9<BR>\n" );
document.write( "7x + 2 = 9<BR>\n" );
document.write( "7x = 7<BR>\n" );
document.write( "x = 1<BR>\n" );
document.write( "so that<BR>\n" );
document.write( "y = 2(1) + 1 = 3 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );