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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