Erin has a total of 52 coins consisting of nickels, dimes, and quarters. the value of the coins is $7.80. there are six more nickels and dimes than quarters. find the number of nickels

Let n = number of nickels.
Let d = number of dimes.
Let q = number of quarters.

Number of coins:
n + d + q = 52

Value of coins:
0.05n + 0.1d + 0.25q = 7.8

Info about number of coins:
n + d = q + 6
n + d - q = 6

We have a system of three equations:
n + d + q = 52
0.05 + 0.1d + 0.25q = 7.8
n + d - q = 6

Subtract the third equation from the first equation:

       n + d + q = 52
-      n + d - q  = 6

                  2q = 46

q = 23

Now replace q in the first ans second equations with 23.

n + d + 23 = 52
n + d = 29             Eq.1

0.05n + 0.1d + 0.25(23) = 7.8
0.05n + 0.1d + 5.75 = 7.8
0.05n +0.1d = 2.05           Eq. 2

Now solve equations 1 and 2 as a system of two equations.

n + d = 29
0.05n + 0.1d = 2.05

d = 29 - n

0.05n + 0.1(29 - n) = 2.05

0.05n + 2.9 - 0.1n = 2.05

-0.05n = -0.85

n = 17

There are 17 nickels.

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