document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=952031>Question 952031</A>: <I><SPAN STYLE=\"word-break: break-all;\"> -7m-2p=-13<BR>\n" );
document.write( "m-2p=11\r<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 #581417 by </B><A HREF=/tutors/aboutme.mpl?userid=diazabdulm><B>diazabdulm(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=diazabdulm><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=581417\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> So far, it looks like the best way to solve this system is by elimination of the 2p\r<BR>\n" );
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document.write( "Multiply either equation by -1, in this case I choose the first equation\r<BR>\n" );
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document.write( "-1(-7m-2p=-13)<BR>\n" );
document.write( "m-2p=11\r<BR>\n" );
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document.write( "that would equal\r<BR>\n" );
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document.write( "7m+2p=13<BR>\n" );
document.write( "m-2p=11\r<BR>\n" );
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document.write( "Add the two equations and get\r<BR>\n" );
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document.write( "8m=24\r<BR>\n" );
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document.write( "Divide by 8\r<BR>\n" );
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document.write( "m=3\r<BR>\n" );
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document.write( "Now after I know one number, I substitute that number into either equation to know the second number, I choose the second equation\r<BR>\n" );
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document.write( "m=3\r<BR>\n" );
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document.write( "m-2p=11\r<BR>\n" );
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document.write( "Substitue m=3\r<BR>\n" );
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document.write( "3-2p=11\r<BR>\n" );
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document.write( "Subtract 3\r<BR>\n" );
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document.write( "-2p=8\r<BR>\n" );
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document.write( "Divide by -2\r<BR>\n" );
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document.write( "p=-4\r<BR>\n" );
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document.write( "So\r<BR>\n" );
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document.write( "m=3 and p=-4 </SPAN>\n" );
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