Jose has 6 more dimes than quarters. He has $1.65 total. How many coins of each type does he have?

number of quarters --- x

number of dimes ----- x + 6

25x + 10(x+6) = 165

solve for x, sub into my definition

its 1.65 not 165

ohh wait I got it now

To solve this problem, we can use a system of equations. Let's represent the number of quarters as "q" and the number of dimes as "d".

Given that Jose has 6 more dimes than quarters, we can write the equation: d = q + 6.

We also know that the total value of the coins is $1.65. The value of a quarter is $0.25, so the value of the quarters can be represented as 0.25q. Similarly, the value of a dime is $0.10, so the value of the dimes can be represented as 0.10d.

Therefore, we have the equation: 0.25q + 0.10d = 1.65.

Now we can solve this system of equations to find the values of q and d. Here's how:

1. Substitute the value of d from the first equation into the second equation:
0.25q + 0.10(q + 6) = 1.65.

2. Simplify and solve for q:
0.25q + 0.10q + 0.60 = 1.65.
0.35q = 1.65 - 0.60.
0.35q = 1.05.
q = 1.05 / 0.35.
q = 3.

So, Jose has 3 quarters.

3. Substitute the value of q into the first equation to find the value of d:
d = q + 6.
d = 3 + 6.
d = 9.

Therefore, Jose has 3 quarters and 9 dimes.