document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=157328>Question 157328</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve the problem. Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9.90. How many dimes does he have? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #115999 by </B><A HREF=/tutors/aboutme.mpl?userid=gonzo><B>gonzo(654)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=gonzo><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=115999\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let n = original number of nickels and d = original number of dimes.<BR>\n" );
document.write( "a nickel is worth .05 cents so value of n nickels is .05*n.<BR>\n" );
document.write( "a dime is worth .10 cents so value of d dimes is .10*n.<BR>\n" );
document.write( "first equation is .05*n + .10*d = 6.00 dollars.<BR>\n" );
document.write( "second equation says if you add 6 nickels and double the number of dimes you get $9.90.<BR>\n" );
document.write( "second equation would then be .05*(n+6) + .10*(2*d) = 9.90<BR>\n" );
document.write( "this becomes .05*n + .05*6 + .2*d = 9.90.<BR>\n" );
document.write( "this becomes .05*n + .30 * .2*d = 9.90.<BR>\n" );
document.write( "this becomes .05*n + .2*d = 9.90 - .30<BR>\n" );
document.write( "this finally becomes .05*n + .2*d = 9.60.<BR>\n" );
document.write( "now we have 2 equations that need to be solved simultaneously.<BR>\n" );
document.write( ".05*n + .1*d = 6.00<BR>\n" );
document.write( ".05*n + .2*d = 9.60<BR>\n" );
document.write( "since if we subtract the top equation from the bottom equation, the n will cancel out, we don't have to do anything to manipulate these equations in order to eliminate one of the unknowns.<BR>\n" );
document.write( "subtracting the top equation from the bottom equation yields .1*d = 3.60.<BR>\n" );
document.write( "d must then = 36 because 36 * (.1) = 3.60.<BR>\n" );
document.write( "going back to the first equation, we have (.05)*n + (.1)*36 = 6.00<BR>\n" );
document.write( "which becomes (.05)*n + 3.60 = 6.00<BR>\n" );
document.write( "which becomes (.05)*n = 2.40<BR>\n" );
document.write( "which becomes n = 2.40/.05<BR>\n" );
document.write( "which becomes n = 48.<BR>\n" );
document.write( "so number of nickels is 48 and number of dimes is 36.<BR>\n" );
document.write( "plugging these numbers in the second equation we get<BR>\n" );
document.write( "(.05)*48 + (.2)*36 = 2.4 + 7.2 = 9.6 which confirms that the number of nickels is 48 and the number of dimes is 36.<BR>\n" );
document.write( "in the bottom equation, (.2)d is really (.1)*2*d where d is the original number of dimes and .01 is the value of each dime and 2 is because we doubled the number of dimes. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );