To solve the system of equations given by:

\[ x^2 - 2y^2 = 2x + y \]
\[ y^2 - 2x^2 = 2y + x \]

we can start by rearranging each equation to isolate the quadratic terms on one side and the linear terms on the other side.

1. Rearrange the first equation:
\[ x^2 - 2y^2 - 2x - y = 0 \]

2. Rearrange the second equation:
\[ y^2 - 2x^2 - 2y - x = 0 \]

Now, let's try to find a relationship between \(x\) and \(y\) by subtracting the second equation from the first:

\[ (x^2 - 2y^2 - 2x - y) - (y^2 - 2x^2 - 2y - x) = 0 \]

Simplify the left-hand side:

\[ x^2 - 2y^2 - 2x - y - y^2 + 2x^2 + 2y + x = 0 \]
\[ x^2 + 2x^2 - 2y^2 - y^2 - 2x + x + 2y - y = 0 \]
\[ 3x^2 - 3y^2 - x + y = 0 \]

Factor out the common terms:

\[ 3(x^2 - y^2) - x + y = 0 \]

This simplifies to:

\[ 3(x^2 - y^2) = x - y \]

Now, let's solve for \(x\) and \(y\). We can try substituting \(x = y\) to see if it satisfies the equations:

If \(x = y\):

Substitute \(x = y\) into the first equation:
\[ x^2 - 2x^2 = 2x + x \]
\[ -x^2 = 3x \]
\[ x(x + 3) = 0 \]

So, \(x = 0\) or \(x = -3\).

If \(x = 0\), then \(y = 0\).

If \(x = -3\), then \(y = -3\).

Let's check these solutions in the second equation:

For \(x = 0\) and \(y = 0\):
\[ 0^2 - 2 \cdot 0^2 = 2 \cdot 0 + 0 \]
\[ 0 = 0 \]
This is true.

For \(x = -3\) and \(y = -3\):
\[ (-3)^2 - 2(-3)^2 = 2(-3) + (-3) \]
\[ 9 - 2 \cdot 9 = -6 - 3 \]
\[ 9 - 18 = -9 \]
\[ -9 = -9 \]
This is also true.

Thus, the solutions to the system of equations are:
\[ (x, y) = (0, 0) \]
\[ (x, y) = (-3, -3) \]