Let's use algebra to solve this problem. Let's represent the number of dimes as "d" and the number of nickels as "n."
The value of dimes in cents is 10d.
The value of nickels in cents is 5n.
The total value of the coins in cents is 310 cents (given that $3.10 is equal to 310 cents).
According to the problem, Ennis has 14 more nickels than dimes, so we can set up the equation:
n = d + 14
Now, we can write the equation for the total value of the coins:
10d + 5n = 310
Substitute the value of "n" from the first equation into the second equation:
10d + 5(d + 14) = 310
Now, solve for "d":
10d + 5d + 70 = 310
15d = 240
d = 16
Now that we have the value of "d" (number of dimes), we can find the value of "n" (number of nickels) using the first equation:
n = d + 14
n = 16 + 14
n = 30
So, Ennis has 16 dimes and 30 nickels.