Question 859192
Given;
(1) (x + iy)*(2 + i) = 2x - i(y + 1) or
(2) 2x + ix + 2iy +i^2y = 2x - i(y + 1) 
Since i^2 = -1, we have
(3) 2x + ix + 2iy -y = 2x - i(y + 1) or
(4) 2x - y + i(x + 2y) = 2x - i(y + 1) 
In order for us to have equal complex numbers, both the real parts have to be equal AND the imaginary parts have to be equal. Therefore (4) gives us TWO equations
(5) 2x - y = 2x and
(6) x + 2y = -(y + 1)
The solution to (5) is
(7) y = 2x - 2x or
(8) y = 0 
The solution to (6) is, with y=0,
(9) x + 2*0 = -(0 + 1) or
(10) x = -1
The solution is (x,y) = (-1,0)
Let's check this pair in (1).
Is ((-1 + i*0)*(2 + i) = 2(-1) - i(0 + 1))?
Is ((-1)*(2 + i) = 2(-1) - i)?
Is (-2 - i = -2 - i)? Yes
Answer: The real pair of (x,y) that satisfy the given equation is (-1,0).