<s>[INST] Solving for $x$ in the absolute value equation $$|x-1|+|2x-3|=|3x-4|$$
I tried to solve the cases separately for the range of values of $x$, but it did not work. Can anyone please show me the way.
 [/INST] Twisted/Confusing way to solve this problem:
$|x-1|+|2x-3|=|3x-4|$...eq(I).  Let us call the three expressions as $A, B, C$
, so we have $A+B=C$
$\begin{matrix} \text{when} & \text{expression}\\ x\geq1&A=x-1 & i\\x<1& A=-(x-1) & ii\\ x\geq\frac{3}{2} & B=2x-3 & iii\\ x<\frac{3}{2} & B=-(2x-3) & iv\\ x\geq \frac{4}{3} & C=3x-4 & v\\ x<\frac{4}{3} & C=-(3x-4) & vi
\end{matrix}$


*

*When we say that $x\geq \frac{3}{2}$ we also say that $x\geq1$ and $x\geq \frac{4}{3}$; $(\frac{3}{2}-\frac{4}{3}=\frac{1}{6}\Rightarrow \frac{3}{2}>\frac{4}{3})$


LHS of eq (I) $=  i + iii = v$ RHS of eq (I)


*

*When we say that $x\leq 1$ we also say that $x\leq \frac{4}{3}$ and $x\leq \frac{3}{2}$


LHS of eq (I) $=  ii + iv = vi $ RHS of eq (I)
for all the values of $x \leq 1 $ and $ x \geq \frac{3}{2}$, eq (I) holds


*

*When $ 1 < x < \frac{4}{3} $, LHS of eq (I) $i + iv \neq vi  $ RHS of eq (I), eq (i) does not hold

*When $ \frac{4}{3} < x < \frac{3}{2}$, LHS of eq (I) $ i + iv \neq v$ RHS of eq (I) ,eq (I) does not hold 


So the solution is $x\leq 1$ and $x\geq \frac{3}{2}$
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