SOLUTION: Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than te number

Question 394523: Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than te number of nickels, then how many of each does he have?

Answer by pie90010(44) (Show Source): You can put this solution on YOUR website!
the definite easiest way to solve this is to use matrices
so first, you need 3 equations to get the ball rolling for a 3 variable equation
"Bernard has 41 coins" tells us that

"they are worth a total of $4.00" tells us that

"the number of quarters is one more than te number of nickels" tells us

just get all variables on one side, and line them up and you get

Now that you have your 3 equations, make your matrix
[1.00 1.00 1.00] [N]
|.50 .100 0.25| |D|
[-1.0 1.00 1.00] [Q]
would be equal to
[41]
| 4|
[ 1|

now that you set that up, just take the inverse of the first matrix, and multiply it by the last one, as you would solve a regular equation
[A]^-1 [B]

your solution should be
20 nickels
15 dimes
6 quarters
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