Find all values $a$ for which there exists an ordered pair (a,b) satisfying the following system of equations: a + ab^2 & = 40b, a - ab^2 & = -32b. List only the values for a.
+128475 a + ab^2 = 40b
a - ab^2 = -32b add these
2a = 8b divide by 2
a = 4b (1)
Sub (1) into either equation
(4b) + (4b)b^2 = 40b
4b^3 + 4b - 40b = 0
4b^3 - 36b = 0
4b ( b^2 - 9) = 0
4b ( b + 3) (b - 3) = 0 set each factor to 0 and solve for b and we have that
b = 0 b = - 3 and b = 3
And a = 4(0) = 0 4(-3) = - 12 4(3) = 12
So....the solutions are (a, b) = (0, 0) (-3, -12) and ( 3, 12)
