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SlipEternal $|x-1| = \begin{cases} x-1 & x\ge 1 \\ 1-x & x \le 1\end{cases}$
$|2x-3| = \begin{cases} 2x-3 & x\ge \dfrac{3}{2} \\ 3-2x & x \le \dfrac{3}{2}\end{cases}$
$|3x-4| = \begin{cases} 3x-4 & x\ge \dfrac{4}{3} \\ 4-3x & x \le \dfrac{4}{3}\end{cases}$
So, if $x \ge \dfrac{3}{2}$ then we have $(x-1)+(2x-3) = (3x-4)$ which is true for any $x \ge \dfrac{3}{2}$.
If $\dfrac{4}{3} \le x \le \dfrac{3}{2}$, then we have:
$(x-1) + (3-2x) = 2-x = 3x-4$ Solving for $x$ we get $4x=6 \Longrightarrow x = \dfrac{3}{2}$
If $1\le x \le \dfrac{4}{3}$, then we have:
$(x-1) + (3-2x) = 4-3x$
$x = 1$
Finally, for $x\le 1$ we have:
$(1-x) + (3-2x) = (4-3x)$ which is true for all $x \le 1$
If we combine all of the information we got for each interval, we have $x \le 1$ or $x \ge \dfrac{3}{2}$.