+152 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 Add the two equations together to get
2a = 8b
a = 4b
Substitute this value in for a in the first given equation.
4b + (4b)b2 = 40b
4b + 4b3 = 40b
Subtract 40b from both sides of the equation.
4b3 - 36b = 0
Factor b out of the terms on the left side.
b(4b2 - 36) = 0
Factor 4b2 - 36 as a difference of squares.
b(2b + 6)(2b - 6) = 0
Set each factor equal to zero and solve for b
| b = 0 | ____or____ | 2b + 6 = 0 | ____or____ | 2b - 6 = 0 |
| 2b = -6 |
| 2b = 6 | ||
| b = -3 | b = 3 |
If b = 0 then a = 4b = 4(0) = 0 so (0, 0) is a solution.
If b = -3 then a = 4b = 4(-3) = -12 so (-12, -3) is a solution.
If b = 3 then a = 4b = 4(3) = 12 so (12, 3) is a solution.
