John has 14 coins that have a total of $2.30. the coins are nickels and quarters. which system of equations can be used to find the answer to how many of each coin john has?

Total coins = 14
n = # of nickels, .05n = amount of $ in nickels
q = # of quarters, .25q = amount of $ in quarters
So the following 2 equations can be used to solve this system:
1) .05n + .25q = 2.30
2) n + q = 14

I'm going to use the elimination method, choosing to eliminate the q, so I need a -.25q to cancel out the other +.25q:
-.25(n + q = 14)
-.25n - .25q = -3.5
+.05n + .25q = 2.30
-.20n = -1.2
-.20n/-.20 = -1.2/-.20
n = 6
n + q = 14
q = 14-n = 14-6 = 8

Therefore John has 6 nickels and 8 quarters !

Ithink the answer is c just trying to others