document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1101094>Question 1101094</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Carol has some dimes and quarters if she has 25 coins worth a total of $3.70 how many of each type of coin does she have? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #715665 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=715665\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br>Let me offer a couple of alternative methods for solving this kind of problem.  The other tutor uses a single variable and substitution, which is fine.  But I personally prefer either of these other methods.<br><BR>\n" );
document.write( "Look at all three methods of solution and try each one, to find which one \"works\" best for you.<br><BR>\n" );
document.write( "1. using two variables and elimination<br><BR>\n" );
document.write( "let d = the number of dimes<BR>\n" );
document.write( "and q = the number of quarters<br><BR>\n" );
document.write( "Then<BR>\n" );
document.write( "(a) <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d%2Bq+=+25 ALIGN=MIDDLE  ALT=\"d%2Bq+=+25\"   DISABLED_style= \"max-width:90%; height:auto;\" >  (there are 25 coins total)<BR>\n" );
document.write( "(b) <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10d%2B25q+=+370 ALIGN=MIDDLE  ALT=\"10d%2B25q+=+370\"   DISABLED_style= \"max-width:90%; height:auto;\" >  (the total value of the coins is $3.70, or 370 cents)<br><BR>\n" );
document.write( "Multiply the first equation by 10 (to get \"10d\"), then subtract the resulting equation from the second equation; this eliminates variable d, leaving an equation you can solve for q:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10d%2B10q+=+250 ALIGN=MIDDLE  ALT=\"10d%2B10q+=+250\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=15q+=+120 ALIGN=MIDDLE  ALT=\"15q+=+120\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=q+=+8 ALIGN=MIDDLE  ALT=\"q+=+8\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Then substitue q=8 in either original equation to find d:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8%2Bq+=+25 ALIGN=MIDDLE  ALT=\"8%2Bq+=+25\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=q+=+17 ALIGN=MIDDLE  ALT=\"q+=+17\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><br><BR>\n" );
document.write( "2. using logical analysis -- which ends up doing basically the same calculations as the elimination method above.<br><BR>\n" );
document.write( "If all 25 coins were dimes, the total value would be 250 cents; that is 120 cents less than the actual total of 370 cents.<BR>\n" );
document.write( "Each quarter is worth 15 cents more than each dime; so each time we trade in a dime for a quarter, we keep the same number of coins but increase the total by 15 cents.<BR>\n" );
document.write( "To increase the total by 120 cents to get the actual total of 370 cents, the number of quarters has to be 120/15 = 8.<BR>\n" );
document.write( "And then the number of dimes is 25-8 = 17. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );