Nick has two more than twice as many quarters as dimes.. one fewer than three times as many nickels as dimes, and eight more pennies than dimes.. he has $7.37. how many of each coin does he have?

you so much : )

Let's say we have x dimes.
This would mean we have:
(2x + 2) quarters,
(3x - 1) nickels and
(x + 8) pennies
The total value is 737 cents ($7.37)
We can thus say that if we multiply the number of each type of coin by the value of that coin and add it all, we will get 737, so we can write an equation:
10x + 25(2x + 2) + 5(3x - 1) + x+8 = 737
And if we tidy this up by multiplying out the brackets and collecting like terms, we get:
76x + 53 = 737
Now, solve for x:
76x = 684
x = 9
So by subbing into the expressions constructed at the start, we find how many of each coin we have:
9 dimes,
(2(9) + 2) quarters,
(3(9) - 1) nickels and
((9) + 8) pennies.
so... 9 dimes, 20 quarters, 26 nickels and 17 pennies.