document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=360319>Question 360319</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A piggy bank contains $2 made from 24 coins which are nickels, dimes and quarters. What combinations of coins are possible? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #257121 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=257121\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2 dollars = 200 cents.<BR>\n" );
document.write( "1 dime = 10 cents<BR>\n" );
document.write( "1 nickel = 5 cents<BR>\n" );
document.write( "1 quarter = 25 cents.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "let:\r<BR>\n" );
document.write( "\n" );
document.write( "d = number of dimes<BR>\n" );
document.write( "n = number of nickels<BR>\n" );
document.write( "q = number of quarters\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you will wind up with 2 equations that need to be solved simultaneously.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n + d + q = 24 (number of coins)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "5n + 10d + 25q = 200 (number of coins times the value of each coin)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "multiply the first equation by 5 to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "5n + 5d + 5q = 120 (first equation multiplied by 5)<BR>\n" );
document.write( "5n + 10d + 25q = 200 (second equation)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "subtract first equation from second equation to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "5d + 20q = 80\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "subtract 20q from both sides of the equation to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "5d = 80 - 20q\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "divide both sides of the equation by 5 to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "d = 16 - 4q\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "go back to your original equations and substitute for d.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "original equations are:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n + d + q = 24 (number of coins)<BR>\n" );
document.write( "5n + 10d + 25q = 200 (number of coins times the value of each coin)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "substitute for d to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n + (16-4q) + q = 24<BR>\n" );
document.write( "5n + 10*(16-4q) + 25q = 200\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "simplify to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n + 16 - 4q + q = 24<BR>\n" );
document.write( "5n + 160 - 40q + 25q = 200\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "combine like terms to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n + 16 - 3q = 24<BR>\n" );
document.write( "5n + 160 - 15q = 200\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "subtract 16 from both sides of the first equation, and subtract 160 from both sides of the second equation, to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n - 3q = 8<BR>\n" );
document.write( "5n - 15q = 40\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "multiply both sides of first equation by 5 to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "5n - 15q = 40<BR>\n" );
document.write( "5n - 15q = 40\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "subtract second equation from first equation to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "0 = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since this is true, then there appears to be any number of solutions to this equation as long as d = 16 - 4q provides us with a valid answer.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if we let q = 0, then we get d = 16<BR>\n" );
document.write( "if we let q = 1, then we get d = 12<BR>\n" );
document.write( "if we let q = 2, then we get d = 8<BR>\n" );
document.write( "if we let q = 3, then we get d = 4<BR>\n" );
document.write( "if we let q = 4, then we get d = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "those are the only valid solutions because (d or q) can't be negative.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "d is the number of dimes and q is the number of quarters.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since n + d + q = 24 from our first equation, then we get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "1 = 0, d = 16, n = 8, and 25q + 10d + 5n = 0 + 160 + 40 = 200<BR>\n" );
document.write( "q = 1, d = 12, n = 11, and 25q + 10d + 5n = 25 + 120 + 55 = 200<BR>\n" );
document.write( "q = 2, d = 8, n = 14, and we get 25q + 10d + 5n = 50 + 80 + 70 = 200<BR>\n" );
document.write( "q = 3, d = 4, n = 17, and we get 25q + 10d + 5n = 75 + 40 + 85 = 200<BR>\n" );
document.write( "q = 4, d = 0, n = 20, and we get 25q + 10d + 5n = 100 + 100 = 200\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "there are other combinations possible to get to 200 cents, but those combinations do not get us 24 coins in total.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "for example:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "1 quarter and 1 dime and 33 nickels add up to 5 + 10 + 165 = 200 cents but the number of coins add up to 35.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "while this satisfies the value equation (has to add up to 200), it doesn't satisfy the number of coins equation (has to add up to 24).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "value = 200 (200 is satisfied), number of coins = 35 (24 is not satisfied).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "both equations have to be satisfied, not just one.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the fact that your final 2 equations resolved to 0 = 0 meant that you have any number of possible solutions as long as d = 16 - q provides you with a valid answer, meaning that the number of dimes has to be positive and the number of quarters has to be positive.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that's because 0 = 0 is true.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if those last 2 equations resolved to something like 0 = 15, then you would have had no no possible number of solutions because 0 = 15 is false (not a true statement).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the fact that you had 2 equations in 3 unknowns means that you had to get the value of one of the variables in terms of the other variables and substitute so you could wind up with 2 equations in 2 unknowns.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );