SOLUTION: Please help me solve this system of equations for all intersections.
(4/x^2)+ (6/y^4) = 7/2
(1/x^2)+ (2/y^4) = 0
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-> SOLUTION: Please help me solve this system of equations for all intersections.
(4/x^2)+ (6/y^4) = 7/2
(1/x^2)+ (2/y^4) = 0
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Question 745739: Please help me solve this system of equations for all intersections.
(4/x^2)+ (6/y^4) = 7/2
(1/x^2)+ (2/y^4) = 0 Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! (4/x^2)+ (6/y^4) = 7/2
(1/x^2)+ (2/y^4) = 0
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r = 1/x^2
s = 1/y^4
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4r + 6s = 7/2
r + 2s = 0 --> r = -2s
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4r + 6s = 7/2
-8s + 6s = 7/2
-2s = 7/2
s = -7/4
--> x^2 = -4/7
No real number intersections.
