To solve this system of equations, we can start by isolating one variable in the first equation and then substituting it into the second equation. Let's solve for x in the first equation:

[x - 2y + 1 = 0]
[x = 2y - 1]

Now, substitute this expression for x into the second equation:

[\sqrt{2y - 1} + \sqrt{y} = 2]

Square both sides to eliminate the square roots:

[(\sqrt{2y - 1} + \sqrt{y})^2 = 2^2]
[2y - 1 + 2\sqrt{y(2y - 1)} + y = 4]
[2y - 1 + 2\sqrt{2y^2 - y} + y = 4]
[3y + 2\sqrt{2y^2 - y} = 5]
[2\sqrt{2y^2 - y} = 5 - 3y]
[4(2y^2 - y) = (5 - 3y)^2]
[8y^2 - 4y = 25 - 30y + 9y^2]
[8y^2 - 4y - 9y^2 + 30y - 25 = 0]
[-y^2 + 26y - 25 = 0]
[y^2 - 26y + 25 = 0]
[(y - 1)(y - 25) = 0]

Therefore, y = 1 or y = 25.

Using these values to solve for x using the equation x = 2y - 1, we get:

If y = 1, then x = 2(1) - 1 = 1
If y = 25, then x = 2(25) - 1 = 49

Therefore, the solutions to the system of equations are:
(1, 1) and (49, 25)