document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=55821>Question 55821</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2a-5b+c=5<BR>\n" );
document.write( "3a+2b-c=17<BR>\n" );
document.write( "4a+3b+2c=17 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #37919 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=37919\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2a - 5b + c = 5<BR>\n" );
document.write( "3a + 2b - c = 17<BR>\n" );
document.write( "4a + 3b + 2c = 17<BR>\n" );
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document.write( "If we mult the 1st eq by -1 we can eliminate c from all three equation:<BR>\n" );
document.write( "-2a + 5b - c = -5<BR>\n" );
document.write( " 3a + 2b - c = 17<BR>\n" );
document.write( " 4a + 3b + 2c = 17<BR>\n" );
document.write( "---------------------add<BR>\n" );
document.write( " 5a + 10b = 29<BR>\n" );
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document.write( "Adding the 1st and 2nd equations eliminates c also:<BR>\n" );
document.write( "2a - 5b + c = 5<BR>\n" );
document.write( "3a + 2b - c = 17<BR>\n" );
document.write( "-------------------add<BR>\n" );
document.write( "5a - 3b = 22<BR>\n" );
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document.write( "WE have two equations with two unknowns a & b<BR>\n" );
document.write( "5a + 10b = 29<BR>\n" );
document.write( "5a -  3b = 22<BR>\n" );
document.write( "--------------- subtract<BR>\n" );
document.write( "     13b = 7<BR>\n" );
document.write( "       b = 7/13<BR>\n" );
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document.write( "Find a using the equation 5a - 10b = 29, substitute 5/13 for b<BR>\n" );
document.write( "5a + 10(7/13) = 29<BR>\n" );
document.write( "5a +  70/13 = 29<BR>\n" );
document.write( "        5a = 377/13 - 70/13, (converted 29 to 13th)<BR>\n" );
document.write( "        5a = 307/13<BR>\n" );
document.write( "         a = (307/13) / 5<BR>\n" );
document.write( "         a = 307/65<BR>\n" );
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document.write( "Find c using the 1st equation, 2a - 5b + c = 5, substitute for a & b<BR>\n" );
document.write( "2(307/65) - 5(7/13) + c = 5<BR>\n" );
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document.write( "614/65 - 5(35/65) + c = 5; convert 7/13 to 65th<BR>\n" );
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document.write( "614/65 - 175/65 + c = 5<BR>\n" );
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document.write( "439/65 + c = 5<BR>\n" );
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document.write( "c = 325/65 - 439/65; converted 5 to 65th<BR>\n" );
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document.write( "c = -114/65<BR>\n" );
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document.write( "So we end up with these nasty fractions (the guy who dreamed this up must be a sadist)<BR>\n" );
document.write( "a = 307/65; b = 7/13; c = -114/65<BR>\n" );
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document.write( "Because of the tremendous opportunity for error here, I check these solutions using the matrix feature on the TI*83 and got the decimal equivalent for each one of these fractions. \r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );