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a + ab²  =  40b

a - ab²  =  -32b

 

The purple values are equal and the blue values are equal.  purple + blue  =  purple + blue

 

(a + ab²) + (a - ab²)  =  40b + -32b

 

(a + ab²) + (a - ab²)  =  40b + -32b

 

2a  =  8b

 

$\frac14$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  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

 

Here is another answer for this question:  https://web2.0calc.com/questions/help-plz_7742

$\begin{cases} a + ab^2 = 40b \;\;\;\!\;\!--- (1)\\ a-ab^2 = -32b --- (2)\\ \end{cases}\\ (1) + (2) : 2a = 8b\\ b = \dfrac{a}{4}\\ \text{Substitute }b = \dfrac{a}4\text{ into (1),}\\ a + a \cdot \left(\dfrac{a^2}{16}\right) = 40\left(\dfrac{a}{4}\right)\\ a^3 -144a = 0\\ a(a+12)(a-12) = 0\\ a \in\{ 0, 12, -12\}$

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