document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=37719>Question 37719</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you do y= 2x + 3  using substitution?<BR>\n" );
document.write( "             y= 3x + 1 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #23267 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=23267\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y=2x+3 now we substitute 2x+3 for y in equation #2 we get\r<BR>\n" );
document.write( "\n" );
document.write( "2x+3=3x+1 or -x=-2 or x=2  then y=2*2+3 or y=7 & y=2*3+1 or y=7\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );