a + ab^{2} = 40b
a  ab^{2} = 32b
The purple values are equal and the blue values are equal. purple + blue = purple + blue
(a + ab^{2}) + (a  ab^{2}) = 40b + 32b
(a + ab^{2}) + (a  ab^{2}) = 40b + 32b
2a = 8b
\(\frac14\)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 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 
Here is another answer for this question: https://web2.0calc.com/questions/helpplz_7742
\(\begin{cases} a + ab^2 = 40b \;\;\;\!\;\! (1)\\ aab^2 = 32b  (2)\\ \end{cases}\\ (1) + (2) : 2a = 8b\\ b = \dfrac{a}{4}\\ \text{Substitute }b = \dfrac{a}4\text{ into (1),}\\ a + a \cdot \left(\dfrac{a^2}{16}\right) = 40\left(\dfrac{a}{4}\right)\\ a^3 144a = 0\\ a(a+12)(a12) = 0\\ a \in\{ 0, 12, 12\}\)
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