document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166593>Question 1166593</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please explain in detail how to use substitution to solve linear system:\r<BR>\n" );
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document.write( "y=6x-11<BR>\n" );
document.write( "-2x-3y=-7 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #791101 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=791101\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y = 6x-11<BR>\n" );
document.write( "-2x-3y = -7<br>\r<BR>\n" );
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document.write( "Since the first equation is already expressed in terms of y, you can take the<BR>\n" );
document.write( "right hand side (RHS) of the first equation and substitute it for 'y' in the <BR>\n" );
document.write( "2nd equation:<br>\r<BR>\n" );
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document.write( "-2x - 3(6x-11) = -7<BR>\n" );
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document.write( "Now its just turning the crank to solve for x.  Once you have x, substitute that value into either of the two equations and solve it for y.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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