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# Help

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Find all values $$a$$ for which there exists an ordered pair $$(a,b)$$ satisfying the following system of equations:

$$a+ab^2=40b$$

$$a-ab^2=-32b$$

List only the values for $$a$$

Jul 7, 2019
edited by KeyLimePi  Jul 7, 2019

#1
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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

Jul 7, 2019
#2
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$$\begin{cases} a + ab^2 = 40b \;\;\;\!\;\!--- (1)\\ a-ab^2 = -32b --- (2)\\ \end{cases}\\ (1) + (2) : 2a = 8b\\ b = \dfrac{a}{4}\\ \text{Substitute }b = \dfrac{a}4\text{ into (1),}\\ a + a \cdot \left(\dfrac{a^2}{16}\right) = 40\left(\dfrac{a}{4}\right)\\ a^3 -144a = 0\\ a(a+12)(a-12) = 0\\ a \in\{ 0, 12, -12\}$$

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Jul 8, 2019