document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=163739>Question 163739</A>: <I><SPAN STYLE=\"word-break: break-all;\"> $1.65 is made up of pennies, nickels, and dimes. Half of the coins are nickels. List a set of coins that will solve the problem.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #120626 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=120626\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=number of pennies<BR>\n" );
document.write( "y=number of nickels<BR>\n" );
document.write( "And z=number of dimes<BR>\n" );
document.write( "(we'll deal in pennies)<BR>\n" );
document.write( "x+5y+10z=165----------------------------eq1<BR>\n" );
document.write( "y=x+z-----------------------------------------eq2<BR>\n" );
document.write( "subtract eq2 from eq1 and we get:\r<BR>\n" );
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document.write( "6y+9z=165  subtract 9z from each side<BR>\n" );
document.write( "6y=165-9z  divide each side by 6<BR>\n" );
document.write( "y=(165-9z)/6=(55-3z)/2-------------eq3  <BR>\n" );
document.write( "eq3 shows the relationship between number of nickels and number of dimes.  Now, we know the following about this problem:\r<BR>\n" );
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document.write( "(1)----We cannot have fractions of coins;  we must deal in whole numbers<BR>\n" );
document.write( "(2)----And we cannot have negative coins;  we must deal in positive numbers\r<BR>\n" );
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document.write( "By inspection, we see immediately that z, the number of dimes, cannot be greater than 18 lest we start having negative nickels.  In fact, z cannot be greater than 16 otherwise we break the bank. Actually, if half the coins are nickels, z needs to be much less than 16. Lets start assuming values for z, starting at z=1 and see what happens:\r<BR>\n" );
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document.write( "z=1-----------y=52/2=26 now we substitute y=26 and z=1 into eq2:<BR>\n" );
document.write( "26=x+1<BR>\n" );
document.write( "x=25<BR>\n" );
document.write( "CK<BR>\n" );
document.write( "26 nickels, 1 dime and 25 pennies---------------half are nickels<BR>\n" );
document.write( "and<BR>\n" );
document.write( "26*5+1*10+25*1=165<BR>\n" );
document.write( "130+10+25=165<BR>\n" );
document.write( "165=165\r<BR>\n" );
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document.write( "Hope this helps----ptaylor\r<BR>\n" );
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document.write( " </SPAN>\n" );
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