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I really need help solving this system:


\(\begin{align*} x + y &= 8 \\ x^2 + y^2 &= 38 \end{align*}\)

 Feb 15, 2022
 #1
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x = 8-y

 

(8-y)^2 + y^2 = 38

 

y^2 + 64 + y^2 - 16y = 38

 

2y^2 - 16y + 26 = 0

 

y^2 - 8y + 13 = 0                 y = \( {8 \pm \sqrt{64-52} \over 2a} = {8 \pm \sqrt{12} \over 2a}\) = (8+-2√3)/2 , which is 4+-√3. 

 

(x, y) = (4-√3, 4+√3), (4+√3, 4-√3)

 Feb 15, 2022

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