What are all ordered pairs of real numbers (x, y) for which:
yx2 - 7x + 12 = 1 and x + y = 6?
pq = 1 in three cases
- p = 1; in this case q can be any number
- p = -1; in thus case q must be even
- q = 0; in this case p can be any number except 0.
With your problem the three cases are:
- y = 1; the second equation then gives x = 5. Thus (5,1) is one pair.
- y = -1; the second equation gives x = 7 which is odd. Thus there is no pair with y = -1.
- 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.
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