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Use the elimination method to find all solutions of the system of equations. (Order your answers from smallest to largest x, then from smallest to largest y.)

 

2x^2 + 4y = 11

x^2-y^2=5/2

 Dec 9, 2019
 #1
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2x^2 + 4y = 11

x^2 -y^2  =  5/2

 

Multiply the second equation through by  -  2        and we get that

 

2x^2  + 4y  =  11

-2x^2 + 2y^2  = - 5        add these

 

2y^2  + 4y  = 6

 

2y^2  + 4y - 6  =  0          divide through by 2

 

y^2  + 2y  - 3   =  0       factor

 

(y + 3)  ( y - 1)  =  0

 

Setting each factor to 0   and solving for y   we get that   y   = -3  and  y  =  1

 

And using 2x^2 + 4y  =  11 we can find the  values for x

 

2x^2  + 4(-3)  = 11          2x^2  + 4(1) = 11

2x^2  - 12  = 11               2x^2  + 4  = 11

2x^2  =  23                     2x^2  = 7

x^2  =  23/2                     x^2 = 7/2

x = ±√[23/2]                     x = ±√[7/2]

 

So.....the solutions are

 

(x, y )   =    (-3, ±√23/2] )   and  ( 1 , ±√[7/2]  )

 

 

 

cool cool cool

 Dec 9, 2019

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