document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=736981>Question 736981</A>: <I><SPAN STYLE=\"word-break: break-all;\"> You have $15.25 in quarters and dimes. You have 103 coins total. How many quarters do you have and how many dimes do you have. Show an equation  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #450074 by </B><A HREF=/tutors/aboutme.mpl?userid=mbogner><B>mbogner(1)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mbogner><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=450074\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = number of quarters<BR>\n" );
document.write( "Let y = number of dimes\r<BR>\n" );
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document.write( "x + y = 103 coins<BR>\n" );
document.write( "Also, $.25x + $.10y = $15.25\r<BR>\n" );
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document.write( "So, from the first equation y = 103 - x, substituting in the second equation,\r<BR>\n" );
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document.write( "$.25x + $.10 * (103 - x) = $15.25 <BR>\n" );
document.write( "or, $.25x + ($.10 * 103) -.10x = $15.25   <BR>\n" );
document.write( "or, $.25x - $.10x + $10.30 = $15.25<BR>\n" );
document.write( "or,   ($.25x - $.10x) + ($10.30 - $10.30) = ($15.25 -$10.30)<BR>\n" );
document.write( "or  $.15x = $4.95, x = $4.95/$.15 = 33 quarters<BR>\n" );
document.write( "hence 33 + y = 103, y = 103 - 33 =  70 dimes\r<BR>\n" );
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document.write( "Error checking is a must:<BR>\n" );
document.write( "($.25 * 33) + ($.10 * 70) = $8.25 + $7.00 = 15.25, which is correct.\r<BR>\n" );
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document.write( "Alternatively, one could start with x = 103 - y, and substitute for x.  This works, too.\r<BR>\n" );
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document.write( "Alternatively, one could start with $.25x + $.10y = $15.25 and solve for x (x= {15.25-.10y})/.25), or y (y={15.25-.25x}/.10), and substitute these into the first equation x + y - 103.   Although perfectly good ways, it can be seen that these add complexity and more opportunities for error. \r<BR>\n" );
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document.write( "Alternatively, one could use a guessing method and adjust.  Say 50 quarter and 53 dimes.  This gives 17.8, need fewer quarters.  One could use the midpoint method and try halfway, (0+50)/2 = 25 quarters and (103-25 = 78 dimes), 14.05.  Need between 25 and 50 quarters.  Eventually one would get 33 quarters and 70 dimes.      \r<BR>\n" );
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document.write( "Alternatively, one could load a spreadsheet from 1 to 102 in the first column (put 1 in cell a1, = 1 + a1 in cell a2, and copy cell a2 from cell a3 to cell a102).  Put = 103 - a1 into cell b1 and copy this formula from cell b2 to cell b102.  Tthe first row would be 1 (a1), 102 (b1), the second row would be 2 (a1), 101 (b1),  until the last row 102 (a102), 1 (b102)).  In the third column set up a formula in c1,=(a1*.25) + (b1*.10),giving 10.45.  copy this formula from c2 to c102.  Look for 15.25 in this column.  Row 33 has 33 (a33, quarters) and 70 (b33, dimes) has 15.25 in c33.    \r<BR>\n" );
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document.write( " </SPAN>\n" );
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