Leroy has written down four numbers. The mean of these numbers is $2x + y$, the median is $x + y$ and the mode is $x - y$, where $x$ and $y$ are constants and $x > y$. Find the four numbers in terms of $x$ and $y$. Answer ... , ... , ... , ...

1. Determine the first two numbers using the mode and median.

Let the four numbers be $a \le b \le c \le d$.
The mode is $x - y$. Since $x > y$, the mode $x - y$ is strictly less than the median $x + y$.
For the median to be greater than the mode, the mode must appear exactly twice and be the two smallest numbers.
Therefore, $a = x - y$ and $b = x - y$.

2. Find the third number using the median.

The median of four numbers is the average of the 2nd and 3rd numbers.
$\frac{b + c}{2} = x + y$
$\frac{(x - y) + c}{2} = x + y$
$x - y + c = 2x + 2y$
$c = x + 3y$

3. Find the fourth number using the mean.

The mean of the four numbers is $2x + y$, so their sum is $4(2x + y) = 8x + 4y$.
$a + b + c + d = 8x + 4y$
$(x - y) + (x - y) + (x + 3y) + d = 8x + 4y$
$3x + y + d = 8x + 4y$
$d = 5x + 3y$

4. State the four numbers.

The four numbers are $x - y$, $x - y$, $x + 3y$, and $5x + 3y$.