I'm trying to solve the following system of equations: $$3x^2 - 3y = 0$$ $$-3x + 4y^3 = 0$$
I started out by rearranging the first equation to get x in terms of y: $$3x^2 - 3y = 0$$ $$x^2-y=0$$ $$x^2=y$$ $$x=\pm\sqrt{y}$$
Then, I plugged in the above result into the second equation and solved for y: $$-3x + 4y^3 = 0$$ $$-3(\pm\sqrt{y})+4y^3=0$$ $$4y^3=3(\pm\sqrt{y})$$ $$16y^6=9y$$ $$y^5=\frac{9}{16}$$ $$y=(\frac{3}{4})^\frac{2}{5}; y=0$$
Lastly, I plugged in the above result back into the result of the first equation to get the corresponding x values. The 3 answers are: $$(0,0),((\frac{3}{4})^\frac{1}{5},(\frac{3}{4})^\frac{2}{5}), (-(\frac{3}{4})^\frac{1}{5},(\frac{3}{4})^\frac{2}{5})$$
The issue is that the first two answers are correct, but the third one (with the negative) is not. It's easy to tell that it's incorrect if you plug the third solution back into the second equation. Why is it incorrect though? I made sure to include the plus/minus signs when taking the square root back when I was simplifying the first equation, but that's actually the reason why I'm getting that negative answer, which turns out to be incorrect. I can't figure out where I went wrong in the algebra.
May be, it could have been simpler to work the other way.
From the first equation, you have $y=x^2$. Then, the second equation becomes $$-3x+4x^6=x(4x^5-3)=0$$ from which the real solutions are $x=0$ and $x=\big(\frac 3 4\big)^{1/5}$ to which corresponf $y=0$ and $y=\big(\frac 3 4\big)^{2/5}$.
This avoids the problem of squaring which always introduces extra solutions.