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Solve the following systems of equations

 

[1]  y = x^2 + 1    

      

      2x^2 + 3y = 28

 Apr 19, 2019
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Since    y =x^2 + 1    let's substitute this into the second equation to solve for 'x'

 

2 x^2 + 3 (x^2+1) = 28

2 x^2 + 3x^2 + 3 = 28 

5 x^2 - 25 = 0      divide by 5

x^2 - 5 = 0

x^2 = 5

x= +- sqrt 5                 then   y = 5 + 1 = 6        soooooo    = sqrt 5  , 6    and    -sqrt 5 , 6

 Apr 19, 2019
edited by ElectricPavlov  Apr 19, 2019

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