document.write( "Question 36424: You have $24.05 with 210 coins in quarters, nickles, and dimes, such that twice the number of quarters is 13 more than the number of nickels. Find how many of each coin you have. \n" ); document.write( "
Algebra.Com's Answer #22420 by Paul(988)\"\" \"About 
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Let the dimes be x
\n" ); document.write( "Let the nickels be y
\n" ); document.write( "Let the quarters be z
\n" ); document.write( "Equation 1 :
\n" ); document.write( "\"twice the number of quarters is 13 more than the number of nickels\"
\n" ); document.write( "2(z)=13+y
\n" ); document.write( "y=2z-13 (subsitution 1)
\n" ); document.write( "Equation 2:
\n" ); document.write( "x+y+z=210 (subsitute for y)
\n" ); document.write( "x+(2z-13)+z=210
\n" ); document.write( "x=210+13-2z-z
\n" ); document.write( "x=223-3z (subsitution 2)
\n" ); document.write( "Equation 3:
\n" ); document.write( "25z+5y+10x=2405
\n" ); document.write( "Subsitute for y:
\n" ); document.write( "25z+5(2z-13)+10x=2405
\n" ); document.write( "subsitute for x:
\n" ); document.write( "25z+10z-65+10(223-3z)=2405
\n" ); document.write( "35z-30z=2405-2230+65
\n" ); document.write( "5z=240
\n" ); document.write( "z=48
\n" ); document.write( "(subsitution 1)-- y=2(48)-13
\n" ); document.write( "y=83
\n" ); document.write( "(subsitution 2)-- x=223-3(48)
\n" ); document.write( "x=79\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hence, there are 48 quarters, 79 dimes and 83 nickels.
\n" ); document.write( "Paul.\r
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