solve linear-quadratic system

Question

solve linear-quadratic system

 

^y=9x+16

^{y=4x^2-11x-40}

Answer 1

y = 9x+16

y=4x²-11x-40       Set the two equations equal to each other.

 

4x²-11x-40=9x+16

4x²-20x-56=0

x²-5x-14=0

(x-7)(x+2) = 0

x-7=0   or   x+2=0

x = 7     or   x =-2

 

y = 9(7) + 16      y = 9(-2) + 16

y = 63 +16           y = -18+16

y = 79                   y = -2

 

(7,79)     (-2,-2)

Answer 2

 Hello, Nick

What I am going to do is make them equal to each other since both are y=. So, this is what we get

 

9x+ 16 = 4x^2-11x-40.        Subtract 16 from both sides

      -16                  -16

 9x      = 4x^2-11x -56         Subtract 9x from both sides

 -9x                -9x

     0=  4x^2-20x-56             Divide by 4

     0= x^2-5x-14                  Factoring

  0= (x-7)( x+2)   

So, x=7   X= -2

 

I hope this helps.  Thanks for posting.

 

D.Y. T.

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