Let d = number of dimes

and n = number of nickels

.

From:"Joe has a collection of nickels and dimes that is worth $6.05."

.10d + .05n = 6.05 (equation 1)

.

From:"If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85"

.10(2d) + .05(n-10) = 9.85 

.20d + .05n-.5 = 9.85

.20d + .05n = 10.35 (equation 2)

Solve equation 1 for n:

.10d + .05n = 6.05

.05n = 6.05 - .10d

n = 121 - 2d

.

Plug the above into equation 2 and solve for d:

.20d + .05n = 10.35

.20d + .05(121 - 2d) = 10.35

.20d + 6.05 - .1d = 10.35

.10d + 6.05 = 10.35