document.write( "Question 620375: $2.77 in a coin jar- all pennies(x), nickels(y), and dimes(z). If there are 53 coins in all, and 4 more nickels than dimes, how many of each coin do you have. \n" ); document.write( "
Algebra.Com's Answer #390096 by lwsshak3(11628)\"\" \"About 
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2.77 in a coin jar- all pennies(x), nickels(y), and dimes(z). If there are 53 coins in all, and 4 more nickels than dimes, how many of each coin do you have.
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\n" ); document.write( "x=number of pennies
\n" ); document.write( "y=number of nickels
\n" ); document.write( "z=number of dimes
\n" ); document.write( "..
\n" ); document.write( "z+4=number of nickels
\n" ); document.write( "z+(z+4)+x=53
\n" ); document.write( "2z+4+x=53
\n" ); document.write( "x=49-2z=number of pennies
\n" ); document.write( "..
\n" ); document.write( "pennies+nickels+dimes=$2.77
\n" ); document.write( ".01x+.05y+.1z=2.77
\n" ); document.write( ".01(49-2z)+.05(z+4)+(.1z)=2.77
\n" ); document.write( ".49-.02z+.05z+.2+.1z=2.77
\n" ); document.write( ".13z+.69=2.77
\n" ); document.write( ".13z=2.77-.69
\n" ); document.write( ".13z=2.08
\n" ); document.write( "z=2.08/.13
\n" ); document.write( "z=16
\n" ); document.write( "z+4=20
\n" ); document.write( "49-2z=17
\n" ); document.write( "..
\n" ); document.write( "number of pennies=17
\n" ); document.write( "number of nickels=20
\n" ); document.write( "number of dimes=16
\n" ); document.write( " \n" ); document.write( "
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