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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 \(a\)

 

 

Please help me.

 Jul 7, 2019
edited by KeyLimePi  Jul 7, 2019
 #1
avatar+8853 
+5

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
avatar+7763 
+3

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

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 Jul 8, 2019

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