Starting system: 

In this example, we use the simpler "substitution" method.
Solve the top equation for "y", to get:
y = x + 1
Substitute "x + 1" for "y" in the 2nd equation, to get :
x + 1 = x^{2}  2x  9
or
x^{2}  3x  10 = 0.
Now factor the left side, to get :
(x  5)(x + 2) = 0
Thus, either x = 5 and hence y = 6
or x = 2 and hence y = 1