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 #1
avatar+8577 
+3

a + ab2   =   40b

a - ab2   =   -32b

 

The purple values are equal to each other and the blue values are equal to each other.

 

a - ab2   =   -32b                     Add  40b  to both sides of the equation.

 

a - ab2 + 40b   =   -32b + 40b        Since  a + ab2  =  40b   we can substitute  a + ab2  in for  40b

 

a - ab2 + a + ab2   =   -32b + 40b      The elimination method is really just like substitution  smiley

 

a - ab2 + a + ab2   =   -32b + 40b      Simplify both sides by combining like terms.

 

2a   =   8b          Divide both sides of the equation by  8

 

\(\frac14\)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  from both sides 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  
Jul 20, 2019
Jul 19, 2019
 #1
avatar+102320 
+2

Here's my best attempt....

 

x^2  + 7x - 44   can be factored as  (x + 11) ( x - 4)

 

So....we are looking for two "a's"   such  that   x^2 + 5x +  a  has  factors of 

(x + 11)  or ( x - 4)

 

So

                x  - 6

x + 11   [  x^2  + 5x  + a  ]

               x^2  + 11x

              _____________

                         -6x   + a

                         -6x  - 66

                       _________

                               a + 66    must be 0  ....so......a  =-66

Proof

 

x^2  + 5x  - 66         (x + 11) ( x - 6)             x  - 6

_____________  =   _____________   =    _______

x^2 + 7 x  -  44          (x + 11)(x - 4)               x -  4

 

Also

               x  +  9

x -  4   [   x^2  + 5x   + a  ]

               x^2  - 4x

              _____________

                         9x  + a

                         9x - 36

                        _______

                              a+ 36   must  = 0 ....so.......a  =-36

 

Second proof

 

x^2 + 5x  - 36         ( x - 4) ( x + 9)            x +  9

___________   =   _____________  =    _______

x^2 + 7x -  44          (x - 4) ( x + 11)           x  + 11

 

So   either   a  =  -66   or  a  =  - 36

 

And the sum of the possible  a's   =  - [ 66 + 36 ]  =   -  102

 

 

cool cool cool

Jul 19, 2019
 #1
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Jul 19, 2019

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