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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.

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  • $\begingroup$ You didn't do anything wrong. It just turns out that the first equation has two solutions for $y=(\frac{3}{4})^\frac{2}{5}$, but the second only has one. So of course you'll need to plug it into both to get all of the constraints taken care of. $\endgroup$ – got it--thanks Sep 25 '15 at 23:28
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    $\begingroup$ When you squared $4y^3=3(\pm\sqrt{y})$ to get $16y^6=9y$, you introduced a new solution. A simpler example, from $x=3$, you would get the equation $x^2=9$ by squaring. But this last equation has two solutions: $x=3$ and $x=-3$. This last solution was not contained in the original equation. Hence the only way to know which values in the set you found are really solutions, you must try every pair in the original equation. $\endgroup$ – Bernard Masse Sep 26 '15 at 0:09
  • $\begingroup$ That's not true, @BernardMasse: $4y^3=3(\pm\sqrt{y})$ and $16y^6=9y$ have the same solution set (check for yourself). The thing going on here is that any solutions obtained from one equation must be checked in the other so that each equation can constrain the solutions. In general this is the only way to tell which solutions actually hold in the entire system. If OP had a system of $9$ equations, after finding solutions in one of the equations, all of those solutions must then be checked in the other $8$ equations. $\endgroup$ – got it--thanks Sep 26 '15 at 2:52
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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.

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