document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=892535>Question 892535</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have to a system of two equations in two unknowns that I have to solve:\r<BR>\n" );
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document.write( "1 = C + D<BR>\n" );
document.write( "1 = ((1+sqrt(5))/2)*C + ((1-sqrt(5))/2)*D \r<BR>\n" );
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document.write( "Hope you can help... :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #540579 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=540579\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> it's a little complicated in the execution, but you follow the same rules for solving systems of equations as you would for a more normal looking problem.\r<BR>\n" );
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document.write( "your solutions are:\r<BR>\n" );
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document.write( "d = (5-sqrt(5))/10\r<BR>\n" );
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document.write( "c = (5+sqrt(5))/10\r<BR>\n" );
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document.write( "the execution could get messy so i assigned a = (1+sqrt(5))/2 and i assigned b = (1-sqrt(5))/2\r<BR>\n" );
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document.write( "replacing them in the original equation, then the original equation becomes:\r<BR>\n" );
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document.write( "1 = c + d<BR>\n" );
document.write( "1 = ac + bd\r<BR>\n" );
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document.write( "this is a lot easier on the eyes, at least for a while.\r<BR>\n" );
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document.write( "using the first equation, i solved for c to get c = 1 - d\r<BR>\n" );
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document.write( "i then used 1-d to replace c in the second equation to get:\r<BR>\n" );
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document.write( "1 = a(1-d) + bd\r<BR>\n" );
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document.write( "simplifying this equation got me:\r<BR>\n" );
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document.write( "1 = a - ad + bd\r<BR>\n" );
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document.write( "i factored out the d to get:\r<BR>\n" );
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document.write( "1 = a - d(a-b)\r<BR>\n" );
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document.write( "i subtracted 1 from both sides of this euation and i added d(a-b) to both sides of this equaiton to get:\r<BR>\n" );
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document.write( "d(a-b) = a-1\r<BR>\n" );
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document.write( "i then divided both sides of this equation by (a-b) to get:\r<BR>\n" );
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document.write( "d = (a-1) / (a-b)\r<BR>\n" );
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document.write( "this is where i substituted (1+sqrt(5)/2 for a and (1-sqrt(5) for b to get me back to my original expressions.\r<BR>\n" );
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document.write( "the equation became.\r<BR>\n" );
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document.write( "d = (1+sqrt(5)/2 - 1) / ((1+sqrt(5)/2) + (1-sqrt(5)/2))\r<BR>\n" );
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document.write( "the rest is just brute calculations which i will show you in the picture at the bottom.\r<BR>\n" );
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document.write( "the end of it all was that i got c = (5+sqrt(5)/10 and i got d = (5-sqrt(5))/10\r<BR>\n" );
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document.write( "i verified that these solutions were good by checking the original equation using my calculator and replacing c and d with their values.\r<BR>\n" );
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document.write( "all is good as far as i can tell.\r<BR>\n" );
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document.write( "once i found d, finding c was easy because c + d = 1 and i could solve for c to get c = 1 - d.\r<BR>\n" );
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document.write( "your solutions are what i showed you up top.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2014/081302.jpg\" alt=\"$$$\" </L\r<BR>\n" );
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document.write( " </SPAN>\n" );
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