<s>[INST] Hints on solving the equation $3\sqrt{x-1}+\sqrt{3x+1}=2$ I recently stumbled upon the problem $3\sqrt{x-1}+\sqrt{3x+1}=2$, where I am supposed to solve the equation for x. My problem with this equation though, is that I do not know where to start in order to be able to solve it. Could you please give me a hint (or two) on what I should try first in order to solve this equation?
Note that I only want hints.
Thanks for the help!
 [/INST] Let $A:=\sqrt{3x+1}$ and $B:=\sqrt{3x-3}$.  Then, $A+\sqrt{3}B=2$ and $A^2-B^2=4$.  That is,
$$(A+\sqrt{3}B)^2=A^2-B^2\,.$$

  Suppose that we are solving over the reals.  Thus, $\left(2\sqrt{3}A+4B\right)\,B=0$.  Since $A$ and $B$ are nonnegative and cannot simultaneously be $0$, we conclude that $B=0$.  


Alternatively, we note that $x\geq 1$ so that $\sqrt{x-1}$ is a real number.  Thus, $$\sqrt{3x+1}\geq \sqrt{3\cdot 1+1}=2\,.$$  For the required equality to hold, we must then have $\sqrt{3x+1}=2$.  Therefore, ....
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