I am trying to solve this (knowing that $w$ is bigger than zero and real):
$$|1-\frac{7}{6}w-\sqrt{19}|=|1-\frac{7}{6}w+\sqrt{19}|$$
I put it into wolframalpha and it spat out $w = 6/7$, but how is that possible when the $w$-part is the same on each side??
Interpret geometrically the absolute value of the difference: $\;\lvert x-a\rvert=\lvert x-b\rvert $ means $x$ is at the same distance from $a$ and $b$.
So precisely, it is having the same $1-\dfrac76 w\;$ which makes the equation easy to solve: this expression is equidistant from $\sqrt{19}$ and $-\sqrt{19}$, i.e. is $0$.
\begin{align*}\left|1-\frac76w-\sqrt{19}\right|=\left|1-\frac76w-\sqrt{19}\right|&\iff\left(1-\frac76w-\sqrt{19}\right)^2=\left(1-\frac76w+\sqrt{19}\right)^2\\&\iff-2\left(1-\frac76w\right)\sqrt{19}=2\left(1-\frac76w\right)\sqrt{19}\\&\iff1-\frac76w=0\\&\iff w=\frac67.\end{align*}
Setting $$1-\frac{7}{6}w=a$$ then you have to solve $$|a-\sqrt{19}|=|a+\sqrt{19}|$$ squaring this you will have $$a^2-2\sqrt{19}a+19=a^2+2\sqrt{19}a+19$$ therefore $a=0$
There are two possibilities $$1-\frac{7}{6}w-\sqrt{19}=1-\frac{7}{6}w+\sqrt{19}\lor 1-\frac{7}{6}w-\sqrt{19}=-\left(1-\frac{7}{6}w+\sqrt{19}\right)$$ The first is impossible, the second gives $$1-\frac{7}{6}w-\sqrt{19}=-1+\frac{7}{6}w-\sqrt{19}$$ that is $$\frac{7}{3} w=2\rightarrow w=\frac{6}{7}$$ This is much simpler than squaring both sides...
