x+y=9
x²-3xy+2y²=0

Factor the left side of the second equation:

(x-y)(x-2y)=0

Use the zero-factor principle:

x-y = 0;  x-2y = 0
  x = y;     x = 2y

Now you have two systems to solve:

x + y = 9    |   x + y = 9
    x = y    |       x = 2y
             |
y + y = 9    |  2y + y = 9
   2y = 9    |      3y = 9
    y =     |       y = 3
    
    x = y    |       x = 2y
    x =     |       x = 2(3)
             |       x = 6

Two solutions:

(x,y) = (,)
(x,y) = (6,3)

Edwin