document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=938571>Question 938571</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can you help me with algebra 1 please I need help it I due tomorrow here is the problem <BR>\n" );
document.write( "M+n=23/12<BR>\n" );
document.write( "M-n=17/12\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 #571798 by </B><A HREF=/tutors/aboutme.mpl?userid=algebrahouse.com><B>algebrahouse.com(1659)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=algebrahouse.com><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=algebrahouse.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=571798\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> m + n = 23/12<BR>\n" );
document.write( "m - n = 17/12<BR>\n" );
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document.write( "2m = 40/12 {added the two equations}<BR>\n" );
document.write( "m = 5/3 {divided each side by 2}\r<BR>\n" );
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document.write( "m + n = 23/12 {top original equation}<BR>\n" );
document.write( "(5/3) + n = (23/12) {substituted (5/3), in for m, into top equation}<BR>\n" );
document.write( "n = 1/4 {subtracted 5/3 from each side}\r<BR>\n" );
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document.write( "m = 5/3 and n = 1/4<BR>\n" );
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