document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=896368>Question 896368</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Coin question<BR>\n" );
document.write( "Natalie has some nickels, Dirk has some dimes and Quincy has some quarters. Dirk has five more dimes than Quincy has quarters. If Natalie gives Dirk a nickel, Dirk gives Quincy a dime, and Quincy gives Natalie a quarter; they will all have the same amount of money. How many coins did each have originally?<BR>\n" );
document.write( "I set up an equation to show how much money Dirk has: .10d + .05 - .10 where d is the number of dimes. For Quincy: .25q + .10 - .25 where q is the number of quarters. I also set up 5 + d = q to show that Dirk has 5 more coins than Quincy then I solved for d (d = q - 5) and substituted that in Dirks formula, but this isn't working. I know intuitively that they each end up with $.85 and that means Natalie has 13 coins, Dirk has 9 and Quincy has 4 at the beginning, but I can't figure out how to write this as an equation. I appreciate the help, Judy </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #543447 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=543447\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Dirk has (x+5).10<BR>\n" );
document.write( "Quincy has x(.25)<BR>\n" );
document.write( "Natalie has (y-1).05<BR>\n" );
document.write( "Dirk has .05 + (x+5).10<BR>\n" );
document.write( "Dirk has .05 + (x+4).10<BR>\n" );
document.write( "Quincy has x(.25) +.10<BR>\n" );
document.write( "Quincy has (x-1).25 +.10<BR>\n" );
document.write( "Natalie has (y-1).05 +.25<BR>\n" );
document.write( "now we can write the final equations <BR>\n" );
document.write( "Dirk = Natalie = Quincy<BR>\n" );
document.write( ".05 + (x+4).10 = (y-1).05 +.25 = (x-1).25 +.10<BR>\n" );
document.write( ".05 +.10x +.40 = .05y -.05 +.25 = .25x -.25 +.10<BR>\n" );
document.write( ".10x +.45 = .05y +.20 = .25x -.15<BR>\n" );
document.write( "first solve for x<BR>\n" );
document.write( ".10x +.45 = .25x - .15<BR>\n" );
document.write( "-.15x = -.60<BR>\n" );
document.write( "x = 4<BR>\n" );
document.write( "now we can solve for y<BR>\n" );
document.write( ".05 + (4+4).10 = (y-1).05 +.25<BR>\n" );
document.write( ".85 = .05y -.05 + .25<BR>\n" );
document.write( ".85 = .05y +.20<BR>\n" );
document.write( ".05y = .65<BR>\n" );
document.write( "y = 13<BR>\n" );
document.write( "Dirk had 9 dimes, Quincy had 4 quarters and Natalie had 13 nickels </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );