document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=64575>Question 64575</A>: <I><SPAN STYLE=\"word-break: break-all;\"> To solve systems of equations by the substitution method.  Please help my son to solve this homework questions:\r<BR>\n" );
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document.write( "x – y = 3<BR>\n" );
document.write( "6x + 4y = 13<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #45205 by </B><A HREF=/tutors/aboutme.mpl?userid=303795><B>303795(602)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=303795><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=45205\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> In the substitution method one equation is written with one variable on its own on one side of the equation.<BR>\n" );
document.write( "Rewrite the first equation by adding y to each side of the equation to get<BR>\n" );
document.write( "x – y + y = 3 + y<BR>\n" );
document.write( "x = y + 3<BR>\n" );
document.write( "Substitute for x in the second equation.<BR>\n" );
document.write( "6(y + 3) + 4y = 13<BR>\n" );
document.write( "6y + 18 + 4y = 13<BR>\n" );
document.write( "10y + 18 = 13<BR>\n" );
document.write( "Subtract 18 from each side of the equation<BR>\n" );
document.write( "10y + 18 - 18 = 13 - 18<BR>\n" );
document.write( "10y = -5<BR>\n" );
document.write( "Divide each side by 10<BR>\n" );
document.write( "y = -0.5<BR>\n" );
document.write( "Substitute into the first equation (or its modified form)<BR>\n" );
document.write( "x = y + 3<BR>\n" );
document.write( "x = -0.5 + 3<BR>\n" );
document.write( "x = 2.5 </SPAN>\n" );
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