How do you solve the system x^2+3y^2=16 and 3x^2+y^2=16 and y=-x?

1 Answer
Jul 10, 2016

{x,y} = {2,-2} or {x,y} = {-2,2}

Explanation:

Three equations and two incognitas. One of them must be redundant, to have a solution. Being a nonlinear system is not trivial to detect redundancy.

First we solve for x_2 = x^2 and y_2 = y^2 the system

{ (x_2 + 3 y_2 = 16), (3 x_2 + y_2 = 16) :}

giving

x_2 = x^2 = 4 and y_2 = y^2 = 4

testing the solutions x=pm2 and y=pm2 in the last equation

x+y=0

we follow with the solution which is

{x,y} = {2,-2} or {x,y} = {-2,2}

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