document.write( "Question 829677: The number of nickels in a coin box is 8 more than the number of quarters and the number of dimes is 3 times the number of nickels. Find the number of each kind of coin if their total value is $7.00 \n" ); document.write( "
Algebra.Com's Answer #500019 by LinnW(1048)\"\" \"About 
You can put this solution on YOUR website!
Set n = no of nickles
\n" ); document.write( " d = no of dimes
\n" ); document.write( " q = no of quarters
\n" ); document.write( "n = q + 8
\n" ); document.write( "for the equation above solve for q.
\n" ); document.write( "subtract 8 from each side
\n" ); document.write( "n - 8 = q or
\n" ); document.write( "q = n - 8
\n" ); document.write( "d = 3n
\n" ); document.write( "We know that
\n" ); document.write( "(no of dimes)(0.10) + (no of nickles)(0.05) + (no of quarters)(0.25) = 7.00
\n" ); document.write( "d*(0.10) + n*(0.05) + q(0.25) = 7.00
\n" ); document.write( "usually written
\n" ); document.write( "0.10d + 0.05n + 0.25q = 7.00
\n" ); document.write( "Use the two equations q = n - 8 , d = 3n
\n" ); document.write( "and substitute (n-8) for q , and (3n) for d
\n" ); document.write( "0.10(3n) + 0.05n + 0.25(n - 8) = 7.00
\n" ); document.write( "0.30n + 0.05n + 0.25n - 2.00 = 7.00
\n" ); document.write( "add 2.00 to each side
\n" ); document.write( "0.30n + 0.05n + 0.25n = 9.00
\n" ); document.write( "0.60n = 9.00
\n" ); document.write( "multiply each side by 100
\n" ); document.write( "60n = 900
\n" ); document.write( "divide each side by 60
\n" ); document.write( "n = 15
\n" ); document.write( "d = 3n = 3(15) = 45
\n" ); document.write( "q = n - 8 = 15 - 8 = 7
\n" ); document.write( "Let's check
\n" ); document.write( "15*(0.05) + 45(0.10) + 7(0.25) ?= 7.00
\n" ); document.write( "0.75 + 4.50 + 1.75 = 7.00 which checks out\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );