Find the complex number $z$ that satisfies \[(1 + i) z - 2 \overline{z} = -11 + 25i.\]

(1 + i) z  - 2z(conjugate)  =  -11 + 25i

(1 + i) (a + bi) - 2(a - bi) = -11 + 25i

a + ai + bi + bi^2 - 2a + 2bi  = -11 + 25i

-a +3bi + ai - b  = -11 + 25i 

-(a + b) + (a + 3b)i =  -11 + 25i

Equate real and complex parts and we have this system

- (a + b) = - 11   ⇒  a + b = 11    ⇒  b = 11 - a     (1)

a + 3b = 25      (2)

Sub (1) into (2)  and we have that

a + 3(11 - a) = 25

a + 33 - 3a = 25

-2a = - 8

a = 4

And using (1)    b = 11 - a  = 11 - 4  = 7

So.....the complex number z is

4 + 7i