document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=272377>Question 272377</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solving using the substitution method: \r<BR>\n" );
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document.write( "8x - 5y = -47<BR>\n" );
document.write( "4x + 3y = -7\r<BR>\n" );
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document.write( "I got some help from the site, but I dont know how it derived to that answer. When doing it long hand, I couldn't get the same results.\r<BR>\n" );
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document.write( "Expressing x as a function of y (from the first equation) x = (-47 +5y)/8<BR>\n" );
document.write( "Substituting this expression into the second equation:\r<BR>\n" );
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document.write( "4(-47-5y)/8+3y = -7\r<BR>\n" );
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document.write( "Simplifying this expression:\r<BR>\n" );
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document.write( "44y = 132 or y = 3\r<BR>\n" );
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document.write( "Now, substituting y = 3 back into the x expressed as a function of y:<BR>\n" );
document.write( "x = (-47+5y)/8 = -4<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 #199265 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=199265\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> take advantage of easy ratios of coefficients\r<BR>\n" );
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document.write( "4x + 3y = -7 ___ 4x = -3y - 7\r<BR>\n" );
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document.write( "2(4x) - 5y = -47\r<BR>\n" );
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document.write( "substituting ___ 2(-3y - 7) - 5y = -47\r<BR>\n" );
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document.write( "-6y - 14 - 5y = -47\r<BR>\n" );
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document.write( "-11y = -33 ___ y = 3\r<BR>\n" );
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document.write( "substituting ___ 4x = -3(3) - 7 ___ 4x = -16 ___ x = -4 </SPAN>\n" );
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