document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=244847>Question 244847</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have to use substitution to solve the equation below & don't know where or how to begin:\r<BR>\n" );
document.write( "\n" );
document.write( "3x-5y=11<BR>\n" );
document.write( "x-3y=1 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #179019 by </B><A HREF=/tutors/aboutme.mpl?userid=dabanfield><B>dabanfield(803)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dabanfield><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=179019\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First take the second equation X-(3*Y)=1 and rewrite it as X=(3*Y)+1. <BR>\n" );
document.write( "Now substitute (3*Y)+1 into the first equation as follows:\r<BR>\n" );
document.write( "\n" );
document.write( "(3*X)-(5*Y) = 11 <BR>\n" );
document.write( "3*((3*Y)+1))-(5*Y) = 11<BR>\n" );
document.write( "(9*Y)+3 - (5*Y) = 11<BR>\n" );
document.write( "Collecting terms we have:\r<BR>\n" );
document.write( "\n" );
document.write( "4*Y + 3 = 11<BR>\n" );
document.write( "4*Y = 8<BR>\n" );
document.write( "Y = 2\r<BR>\n" );
document.write( "\n" );
document.write( "Now substituting 2 for Y in the rewritten second equation we have:<BR>\n" );
document.write( "X = (3*2) + 1<BR>\n" );
document.write( "X = 7\r<BR>\n" );
document.write( "\n" );
document.write( "The solution then is x=7 and Y = 2 or (7,2) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );