SOLUTION: How do you solve for x and y as a system, in these types of equations? 4x^2 + 3y^2 = 12 5x^2 + 6y^2 = 30

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do you solve for x and y as a system, in these types of equations? 4x^2 + 3y^2 = 12 5x^2 + 6y^2 = 30      Log On


   

Question 612035: How do you solve for x and y as a system, in these types of equations?
4x^2 + 3y^2 = 12
5x^2 + 6y^2 = 30
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Use the elimination method, just like with linear equations.

Multiply your first equation by -2



Add the second equation to it





Which you should be able to see has no real solution. This should make sense to you because you can see that each is the equation of an ellipse, each being centered at the origin, but the second one has both major and minor axes that are larger than the first one. Ergo, they do not intersect and the system is inconsistent.

John

My calculator said it, I believe it, that settles it
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