You could try solving the first equation for y and substituting it int the second....that will give you an equation in terms of x only.
Or you could solve for x in the 2nd equation, which is quadratic in x, by using the quadratic formula.
[tex]yx^2 + 2y^2 x + y^3 -2 = 0[/tex]
You will of course get two equations for x.
You could expand (x+y)3 and then substitute the values for the two expressions that you have. That would be a lot simpler.
Notice that both equations only have terms of degree 3 in both variables. So one thing you could do is let r=x/y, and divide both equations by y^3. Each left-hand side will then depend only on r, and each right-hand side only on y^3. Eliminating y^3 will give you a cubic equation in r. Solve this by guessing a root, or by graphing to get a root. Once you know one root you can reduce it to a quadratic to get the other two.
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