Picking the correct method for solving this first degree equation I have trouble solving this first degree equation:
$ 2(x-1) = \sqrt{2} (x + 1) - 1$
I can't see a way to factorize this, so using a distributive aproach I get this:
$ 2x  = x \sqrt{2} + \sqrt{2} + 1$
Now my issue is that, how do I get rid of $ \sqrt{2} $ terms ?
I tried to squared the whole equation this way:
$ 2x² = (x \sqrt{2})² + 3$
But now I'm stuck, the expected result is (a non-detailed results is given, hence I'm looking for how to continue solving):
$ x = \frac{4 + 3 \sqrt{2}}{2} $
Am I not using the correct way to solve this, or I'm mistaking in the evaluation of:
$(x \sqrt{2})²$ ?
Or should I even start by squaring both sides to this  way:
$ 2(x-1)² = (x + 1) - 1$
Thanks.
 A: You don't need to get rid of the $\sqrt2$ term.
You're looking to solve this in $x$, so all you need to do is group all $x$ terms together.
You have:
$$2x=x\sqrt2+\sqrt2+1$$
Which yields:
\begin{align}
\iff & 2x-x\sqrt2=1+\sqrt2\\
\iff &x(2-\sqrt2)=1+\sqrt2
\end{align}
Now you can divide both sides by $(2-\sqrt2)$, because it is nonzero, to obtain
\begin{align}
\iff &x=\frac{1+\sqrt2}{2-\sqrt2}
\end{align}
At this point, we usually try and rationalize the denominator.
In other words, we multiply numerator and denominator both by the same number $c$, so that the value of $x$ is not changed.
We choose $c$ such that the denominator (after multiplication) simplifies to a rational number.
This case is particularly easy, and we take advantage of the identity $(a-b)(a+b)=a^2-b^2$.
Our choice of constant $c$ is hence simply $2+\sqrt2$.
We get
\begin{align}
\iff &x=\frac{1+\sqrt2}{2-\sqrt2}\cdot\frac{2+\sqrt2}{2+\sqrt2}\\
\iff &x=\frac{(1+\sqrt2)(2+\sqrt2)}{2^2-{(\sqrt2)}^2}\\
\iff &x=\frac{2+\sqrt2+2\sqrt2+2}{4-2}\\
\iff &x=\frac{4+3\sqrt2}{2}
\end{align}
which is the answer sought.
A: Collect $x$ terms on the LHS and constants on the RHS:
$$\big(2-\sqrt{2}\big)x=\sqrt 2+1\Longrightarrow x=\frac{\sqrt 2+1}{2-\sqrt 2}=\frac{3\sqrt 2+4}{2}.$$
