A cashier has 25 coins consisting of nickels, dimes, and quarters with a value of $4.90. if the number of dimes is 1 less than twice the number of nickels, how many of each type of coin does she have?

1) x - number of nickels

2x - 1 - number of dimes

y - number of quarters

Altogether number of coins: x+2x-1+y = 25, or 3x - 1 + y = 25, 3x+y=26

2) 5x + 10 (2x - 1) + 25y = 490 (cents)

5x + 20x - 10 + 25y = 490

25x + 25y=500

x+y=20

3) Now, we need to solve a system of equations.

3x+y=26

x+y=20 → y=20 - x

3x+y=26 → 3x + 20 - x = 26 → 2x = 6 → x = 3 (nickels)

y=20 - x → y = 20 - 3 = 17 (quarters)

2x - 1 = 2*3 - 1 = 5 (dimes)

So, we have 3 (nickels), 17 (quarters), 5 (dimes).