Complete set of values of x satisfying inequality $\left|\right|\mathrm{x}-1|-5|<2\mathrm{x}-5$ is

We have, $\left|\right|\mathrm{x}-1|-5|<2\mathrm{x}-5$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{x}-5>0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}>\frac{5}{2} \left(1\right)\end{array}$

Now from (1),

$\begin{array}{l} -2\mathrm{x}+5<|\mathrm{x}-1|-5<2\mathrm{x}-5\\ \Rightarrow -2\mathrm{x}+10<|\mathrm{x}-1|<2\mathrm{x}\end{array}$

Case I : $\mathrm{x}\ge 1$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-2\mathrm{x}+10<\mathrm{x}-1<2\mathrm{x}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{x}>11\text{ and }\mathrm{x}>-1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}>11/3 \left(2\right)\end{array}$

Case II : x < 1

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-2\mathrm{x}+10<1-\mathrm{x}<2\mathrm{x}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}>9\text{ and }\mathrm{x}>1/3\end{array}$

$\Rightarrow \mathrm{x}>9$, not possible as x < 1

Therefore, $\mathrm{x}\in (11/3,\mathrm{\infty })$