Find all values \(a\) for which there exists an ordered pair \((a, b)\) satisfying the following system of equations:
\(\begin{align*}a + ab^2 & = 40b, \\a  ab^2 & = 32b.\end{align*}\)
a + ab^{2} = 40b
a  ab^{2} = 32b
The purple values are equal to each other and the blue values are equal to each other.
a  ab^{2} = 32b Add 40b to both sides of the equation.
a  ab^{2} + 40b = 32b + 40b Since a + ab^{2} = 40b we can substitute a + ab^{2} in for 40b
a  ab^{2} + a + ab^{2} = 32b + 40b The elimination method is really just like substitution
a  ab^{2} + a + ab^{2} = 32b + 40b Simplify both sides by combining like terms.
2a = 8b Divide both sides of the equation by 8
\(\frac14\)a = b
Now we can substitute this value for b into one of the original equations.
a + ab^{2} = 40b
Substitute \(\frac14\)a in for b
a + a(\(\frac14\)a)^{2} = 40(\(\frac14\)a)
Simplify both sides of the equation.
a + \(\frac{1}{16}\)a^{3} = 10a
Multiply through by 16
16a + a^{3} = 160a
Subtract 16a from both sides and subtract a^{3} from both sides
0 = 144a  a^{3}
Factor a out of both terms on the right side
0 = a( 144  a^{2} )
Factor 144  a^{2} as a difference of squares
0 = a( 12  a )( 12 + a )
Set each factor equal to 0 and solve for a
0 = a  ___ or ___  12  a = 0  ___ or ___  12 + a = 0 

a = 0  a = 12  a = 12 
EDIT: The following is wrong, but why?
1. Divide both sides by b
\(\frac{a}{b}+ab=40\)
\(\frac{a}{b}ab=32\)
2. Substitute out a/b so we get:
\(40ab=32+ab\)
3. Solving for ab we get
\(ab=36\)
So the values of a can be \((36, 18, 12 9, 6, 4, 3, 2, 1, 1, 2, 3, 4, 6, 9, 12, 18, 36)\)
Assuming if the problem asks for integer solutions
Here is something taken from what I wrote in my notes on my phone a long time ago...

a + b = 10a + b Divide both sides by a
a/a + b/a = 10a/a + b/a
1 + b/a = 10 + b/a Subtract b/a from both sides
1 = 10
So you'd be like, "NO SOLUTION!!!" right??? WRONG!! because a = 0 IS a solution to the original equation, but when you divided both sides by a it made it look like there was no solution. So that is why you gotta always say a ≠ 0 whenever you divide by an unknown number. So then the final answer would be basically saying "No solution when a ≠ 0. So a = 0 might be a solution might not be I dunno." Then you gotta check if a = 0 is a solution.

So that is why 0 is missing from your list.
When you divided both sides by b, you didn't note that the following equation holds only when b ≠ 0
Otherwise, it is true that ab = 36
There are three solutions to this system of equations (see here) and in both cases where b ≠ 0, ab = 36
But ab = 36 is not the full story! There are more restrictions...
Notice that (1, 36) can't be a solution for (a, b) because
a + ab^{2} must equal 40b but 1 + 36^{2} ≠ 40(36)
So (1, 36) can't be a solution.
There are more restrictions on a and b besides just ab = 36 ( when b ≠ 0 )