SOLUTION: Kevin and Randy Muise have a jar containing 79 coins, all of which are either quarters or nickels. The total value of the coins in the jar is $10.35. How many of each type of coins

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Question 924214: Kevin and Randy Muise have a jar containing 79 coins, all of which are either quarters or nickels. The total value of the coins in the jar is $10.35. How many of each type of coins do they have?
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Kevin and Randy Muise have a jar containing 79 coins, all of which are either
quarters or nickels. The total value of the coins in the jar is $10.35. How many
of each type of coins do they have?
Let the number of nuckels be N
Let the number of dimes be Q   
                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
NICKELS      N        $0.05    $0.05N 
QUARTERS     Q        $0.25    $0.25Q
-------------------------------------------
TOTALS      79        -----    $10.35

 The equations comes from the "Number of coins" column 
and the "Value of ALL coins" column:

%28matrix%283%2C1%2CNumber%2Cof%2Cnickels%29%29%22%22%2B%22%22%28matrix%283%2C1%2CNumber%2Cof%2Cquarters%29%29 %22%22=%22%22 %28matrix%283%2C1%2CNumber%2Cof%2Ccoins%29%29


%28matrix%284%2C1%2CValue%2Cof%2CALL%2Cnickels%29%29%22%22%2B%22%22%28matrix%284%2C1%2CValue%2Cof%2CALL%2Cquarters%29%29%22%22=%22%22%28matrix%284%2C1%2CTotal%2Cvalue%2Cof%2Ccoins%29%29

system%28D%2BQ=79%2C%0D%0A%0D%0A0.05N+%2B+0.25Q+=+10.35%29

Get rid of decimals in the second by multiplying 
every term by 100

      5N + 25Q = 1035

system%28N%2BQ=79%2C%0D%0A%0D%0A5D+%2B+25Q+=+1035%29

Solve the first for N

N+Q=79
  N=79-Q

Substitute in

       5N + 25Q = 1035
  5(79-N) + 25Q = 1035
   395-5Q + 25Q = 1035
        395+20Q = 1035
            20Q =  640 
              Q =   32 quarters

              N = 79-Q = 79-32 = 47 nickels

Edwin