Find all values for which there exists an ordered pair satisfying the following system of equations:
a+ab^2=40b
a-ab^2=-32b
List only the values for a
The solutions (a,b) are (12,3) and (-12,-3), so the possible values of a are 12 and -12.
+128407 a + ab^2 = 40b (1)
a - ab^2 = 32b add these equations
2a = 72b
a = 36b (2)
Sub (2) into (1)
36b + (36b)b^2 = 40b
36b^3 = 4b
36b^3 - 4b = 0
4b ( 9b^2 - 1) = 0
4b ( 3b -1) (3b + 1) = 0
4b = 0 3b -1 = 0 3b + 1 = 0
b = 0 b =1/3 b = -1/3
a = 36 (0) = 0
a = 36 (1/3) = 12
a = 36 (-1/3) = -12
