<s>[INST] Solving $x\sqrt{2}=1-x$ Solve for $x$ of the equation,
$$x\sqrt{2}=1-x$$
According to the solution, $x$ is supposed to be $\sqrt{2}-1$. The solutions to this test have proven wrong from case to case, so I wonder ...
If you solve it like this: $x\sqrt{2}+x = 1$ and pull out $x$ to get $x(\sqrt{2}+1) = 1$. Then, dividing by $\sqrt{2}+1$, we get 
$$x=\frac1{\sqrt{2}+1}$$
right? 
 [/INST] $\frac 1{\sqrt 2 + 1}=\frac 1{\sqrt 2+1}\frac {\sqrt 2 -1}{\sqrt 2 -1} = \frac {\sqrt 2 -1}{\sqrt 2^2 + \sqrt 2 - \sqrt 2 - 1} = \frac {\sqrt 2 -1}{2-1} =\frac {\sqrt 2-1}1 = \sqrt 2 -1$.
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