document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204507>Question 1204507</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Rachelle has  <BR>\n" );
document.write( "$<BR>\n" );
document.write( "4.45<BR>\n" );
document.write( "  in dimes and nickles in her car. The number of nickles is twenty more than the number of dimes. How many of each type of coin does she have?<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 #840794 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=840794\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Using formal algebra....<br><BR>\n" );
document.write( "x = # of dimes<BR>\n" );
document.write( "x+20 = # of nickels<br><BR>\n" );
document.write( "The total value is $4.45, or 445 cents:<br><BR>\n" );
document.write( "10(x)+5(x+20) = 445<BR>\n" );
document.write( "10x+5x+100 = 445<BR>\n" );
document.write( "15x+100 = 445<BR>\n" );
document.write( "15x = 345<BR>\n" );
document.write( "x = 345/15 = 23<br><BR>\n" );
document.write( "ANSWER: x=23 dimes and x+20=43 nickels<br><BR>\n" );
document.write( "CHECK: 23(10)+43(5)=230+215=445<br><BR>\n" );
document.write( "You can get good mental exercise by solving the problem informally, which uses essentially the same calculations as the formal algebraic solution.<br><BR>\n" );
document.write( "The \"extra\" 20 nickels have a value of 5(20) = 100 cents; the remaining coins then have a total value of 445-100 = 345 cents.<br><BR>\n" );
document.write( "The remaining coins are equal numbers of dimes and nickels.  The value of one dime and one nickel is 15 cents.  So the number of nickels and dimes remaining is 345/15 = 23.<br><BR>\n" );
document.write( "And of course again we find there are 23 dimes and 23+20 = 43 nickels.<br><BR>\n" );
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