What are all ordered pairs of real numbers (x, y) for which:

Patrick
Algebra 2
Teacher

p^q = 1 in three cases

1. p = 1; in this case q can be any number
   
   
2. p = -1; in thus case q must be even
   
   
3. q = 0; in this case p can be any number except 0.

With your problem the three cases are:

1. y = 1; the second equation then gives x = 5. Thus (5,1) is one pair.
   
   
2. y = -1; the second equation gives x = 7 which is odd. Thus there is no pair with y = -1.
   
   
3. x² - 7x + 12 = (x - 4)(x - 3)= 0; Thus x = 3 or x = 4. In these cases the second equation then gives (3,3) and (4,2) as valid pairs.