Matt has twice as many quarters as dimes in his coin collection. If the dimes and quarters together total $9.00 how many of each kind of coin does he have?

.10*(1/2)x+.25x = 9.00

x(.10*(1/2)+.25)= 9.00
x= 9.00
___(.10*(1/2)+.25)
x = 30
Since you times the dimes by 1/2, you take half of x.
So Dimes = 15, and Quarters = 30
Check, .10(15)+ .25(30) = 9.00

To solve this problem, we can set up a system of equations based on the given information. Let's use the following variables:

- Let "q" represent the number of quarters.
- Let "d" represent the number of dimes.

According to the problem, Matt has twice as many quarters as dimes. This can be expressed as an equation: q = 2d.

We also know that the total value of the coins is $9.00. Quarters have a value of $0.25 each, so the total value of the quarters is 0.25q. Dimes have a value of $0.10 each, so the total value of the dimes is 0.10d. We can set up another equation based on this: 0.25q + 0.10d = 9.00.

Now we have a system of equations:
q = 2d
0.25q + 0.10d = 9.00

To solve this system, we can substitute the value of q from the first equation into the second equation:
0.25(2d) + 0.10d = 9.00
0.50d + 0.10d = 9.00
0.60d = 9.00
d = 9.00 / 0.60
d = 15

Now that we know the value of d, we can substitute it back into the first equation to find q:
q = 2d
q = 2 * 15
q = 30

Therefore, Matt has 30 quarters and 15 dimes.