Find all values \(a\) for which there exists an ordered pair \((a, b)\) satisfying the following system of equations:
\(\begin{align*}a + ab^2 & = 40b, \\a - ab^2 & = -32b.\end{align*}\)
+9481 a + ab2 = 40b
a - ab2 = -32b
The purple values are equal to each other and the blue values are equal to each other.
a - ab2 = -32b Add 40b to both sides of the equation.
a - ab2 + 40b = -32b + 40b Since a + ab2 = 40b we can substitute a + ab2 in for 40b
a - ab2 + a + ab2 = -32b + 40b The elimination method is really just like substitution 
a - ab2 + a + ab2 = -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 + ab2 = 40b
Substitute \(\frac14\)a in for b
a + a(\(\frac14\)a)2 = 40(\(\frac14\)a)
Simplify both sides of the equation.
a + \(\frac{1}{16}\)a3 = 10a
Multiply through by 16
16a + a3 = 160a
Subtract 16a from both sides and subtract a3 from both sides
0 = 144a - a3
Factor a out of both terms on the right side
0 = a( 144 - a2 )
Factor 144 - a2 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 |
+2863 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:
\(40-ab=-32+ab\)
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
+9481 Here is something taken from what I wrote in my notes on my phone a long time ago... 
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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.
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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 + ab2 must equal 40b but 1 + 362 ≠ 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 )
