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Solve the following systems of equations.

 

[1]  4x² + 8x + 5 - y = 1

     3x² - x + 3 = y + x + 8

 Apr 19, 2019
 #1
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[1]  4x² + 8x + 5 - y = 1

     3x² - x + 3 = y + x + 8

 

We can write

 

y  = 4x^2 + 8x + 4      and

y = 3x^2 -2x - 5

 

Set these equal

 

4x^2 + 8x + 4  =  3x^2 - 2x  - 5      rearrange as

 

x^2 + 10x + 9  = 0      factor

 

(x + 9) ( x + 1)  = 0           set both factors equal to 0  and solve  and we get that

 

x = -9    and y  = 3(-9)^2 -2(-9) - 5  =  256

 

x = -1   and y =3(-1)^2 -2(-1) - 5  =  0

 

So...the solutions are (-9, 256) and (-1, 0)

 

 

cool cool cool

 Apr 19, 2019

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