Suppose a and b are integers. How many solutions are there to the equation ab = 2a + 3b?
a*b=2a + 3b
a = -3 and b = 1
a = 0 and b = 0
a = 1 and b = -1
a = 2 and b = -4
a = 4 and b = 8
a = 5 and b = 5
a = 6 and b = 4
a = 9 and b = 3
+130081 ab = 2a + 3b
Square both sides
a^2b^2 = 4a^2 + 12ab + 9b^2 rearrange as
[ 4-b^2]a^2 + [ 12b]a + 9b^2 = 0
Setting this up as a quadratic and solving for a, we have that
a = -12b ±√ [ (-12b)^2 - 4 (4 - b^2)( 9b^2) ]
________________________________
2 [ 4 - b^2]
a = -12b ±√ [ 144b^2 - 144b^2 + 36b^4 ]
_______________________________
2 [ 4 - b^2 ]
a = -12b ± 6b^2
___________
2 [ 4 - b^2 ]
a = 6b [ -2 ± b ]
___________
2 [ 4 - b^2 ]
a = 3b [ -2 ± b ]
_____________
(2 + b) (2 - b)
Choosing a = 3b [ -2 - b] -3b [ 2 + b] -3b
__________ = _____________ = ___
(2 + b) (2 -b) (2 + b) (2 -b) 2 -b
Will produce all the possible integer answers
Simplifying, we have that
a = -3b 3b
_____ = ____
2 - b b - 2
Dividing this fraction produces :
a = 3 + 6
____
b - 2
And when b = 8 5 4 3 1 0 -1 -4
a = 4 5 6 9 -3 0 1 2
So.........( a, b) = (2, -4), (1, -1), (0, 0), (-3, 1), (9, 3), (6,4), (5,5) and (4,8)
a =
