To solve this system of equations, we can use the substitution method.

Let's solve the second equation for x:

x + 2y = 3
x = 3 - 2y

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

(3 - 2y)^2 - 4(3 - 2y)y + y = 25
Expand and simplify:

(9 - 12y + 4y^2) - 12y + 8y^2 + y = 25
9 - 12y + 4y^2 - 12y + 8y^2 + y = 25
Combine like terms:

9 - 2y + 12y^2 = 25
12y^2 - 2y - 16 = 0
Factor the quadratic equation:

(3y - 4)(4y + 4) = 0
Using the zero-product property:

3y - 4 = 0 or 4y + 4 = 0
y = 4/3 or y = -1

Now, substitute these values back into the second equation to solve for x:

if y = 4/3:
x + 2(4/3) = 3
x + 8/3 = 3
x = 3 - 8/3
x = 1/3

if y = -1:
x + 2(-1) = 3
x - 2 = 3
x = 3 + 2
x = 5

Therefore, the solutions to the system of equations are x = 1/3, y = 4/3 and x = 5, y = -1.