How can we algebraically solve <mrow class="MJX-TeXAtom-ORD"> | </mrow> x &#x2212;

Step 1
$x-<!--\; -\; -->1\ge <!--\; \ge \; -->0;x-<!--\; -\; -->2\ge <!--\; \ge \; -->0$ . Thus $x\ge <!--\; \ge \; -->1$ and $x\ge <!--\; \ge \; -->2$ .
This is the case that $x\ge <!--\; \ge \; -->2$
Okay: $|x-<!--\; -\; -->1|+|x-<!--\; -\; -->2|>1$ so $(x-<!--\; -\; -->1)+(x-<!--\; -\; -->2)>1$ so $2x-<!--\; -\; -->3>1$ so $2x>4$ and $x>2$ . And we restrict this to $x\ge <!--\; \ge \; -->2$ to get
$x>2$ and $x\ge <!--\; \ge \; -->2$ so
Conclusion $x>2$ .
Step 2
$(x-<!--\; -\; -->1)\ge <!--\; \ge \; -->0$ and $(x-<!--\; -\; -->2)<0$ . That is $x\ge <!--\; \ge \; -->1$ and $x<2$ so this is the case that $1\le <!--\; \le \; -->x<2$ .
We get $(x-<!--\; -\; -->1)-<!--\; -\; -->(x-<!--\; -\; -->2)>1$ so $1>1$ . This is never the case so there are no solutions where $1\le <!--\; \le \; -->x<2$
If we want to be thurough we would say.
We must restrict to where $1>1$ and $1\le <!--\; \le \; -->x<2$ . There are no cases where both are true.
Step 3
$(x-<!--\; -\; -->1)<0$ and $x-<!--\; -\; -->2\ge <!--\; \ge \; -->0$ . This means $x<1$ and $x\ge <!--\; \ge \; -->2$ . This is impossible. There are no such x and so no such x can be a solution (as there are no such x !).
If we want to be thorough (which we don't but let's pretend we do) we would solve
$-<!--\; -\; -->(x-<!--\; -\; -->1)+(x-<!--\; -\; -->2)>1$ so $-<!--\; -\; -->1>1$ and or solution occurs when $-<!--\; -\; -->1>1$ and $x<1$ and $x\ge <!--\; \ge \; -->2$ . As those three conditions are never concurrently true we have no solution in this interval which doesn't exist in the first place.
Step 4
$(x-<!--\; -\; -->1)<0$ and $(x-<!--\; -\; -->2)<0$ . This means $x<1$ and $x<2$ so is the case when $x<1$ .
So $-<!--\; -\; -->(x-<!--\; -\; -->1)-<!--\; -\; -->(x-<!--\; -\; -->2)>1$ so $-<!--\; -\; -->2x+3>1$ so $-<!--\; -\; -->2x>-<!--\; -\; -->2$ so $x<2$ .
So these solutions occur when $x<2$ and $x<1$
Conclusion: so these solutions occur whenever $x<1$
Combining Case 1, and Case 4 (and 2 and 3 although those had no result) we have final solution
$|x-<!--\; -\; -->1|+|x-<!--\; -\; -->2|>1$ if
$x>2$ or $x<1$ or $x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)\cup <!--\; \cup \; -->(2,\mathrm{\infty <!--\; \infty \; -->})$ .
If we want to be thorough (which be now you should know we don't)
We could so we have solutions when:
$x>2$ or $1<1$ or ( $x<1$ and $x\ge <!--\; \ge \; -->2$ ) or $x<1$ or $x\in <!--\; \in \; -->(2,\mathrm{\infty <!--\; \infty \; -->})\cup <!--\; \cup \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}\cup <!--\; \cup \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}\cup <!--\; \cup \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)=$
$(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)\cup <!--\; \cup \; -->(2,\mathrm{\infty <!--\; \infty \; -->})$
Step 5
Familiarity and common sense and we can allow ourselve to consider then intervals $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1],[1,2],$ and $[2,\mathrm{\infty <!--\; \infty \; -->})$ .
If $x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}1]$ then $(x-<!--\; -\; -->1)\le <!--\; \le \; -->0;x-<!--\; -\; -->2<0$ so $|x-<!--\; -\; -->1|+|x-<!--\; -\; -->2|=-<!--\; -\; -->(x-<!--\; -\; -->1)-<!--\; -\; -->(x-<!--\; -\; -->2)=-<!--\; -\; -->2x+3>1$ so $x<1$
If $x\in <!--\; \in \; -->[1,2]$ then $x-<!--\; -\; -->1\ge <!--\; \ge \; -->0$ and $x-<!--\; -\; -->2\le <!--\; \le \; -->0$ so $|x-<!--\; -\; -->1|+|x-<!--\; -\; -->2|=(x-<!--\; -\; -->1)-<!--\; -\; -->(x-<!--\; -\; -->2)=1>1$ which is impossible.
If $x\in <!--\; \in \; -->[2,\mathrm{\infty <!--\; \infty \; -->})$ then $x-<!--\; -\; -->1>0$ and $x-<!--\; -\; -->2\ge <!--\; \ge \; -->0$ so $|x-<!--\; -\; -->1|+|x-<!--\; -\; -->2|=x-<!--\; -\; -->1+x-<!--\; -\; -->2=2x-<!--\; -\; -->3>1$ so $x>2$
So $x<1$ or $x>2$ and $x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)\cup <!--\; \cup \; -->(2,\mathrm{\infty <!--\; \infty \; -->})$
this way we know $x-<!--\; -\; -->1<0$ while $x-<!--\; -\; -->2\ge <!--\; \ge \; -->0$ was absurd from the start and never needed to be considered in the first place.

Step 1
The best way to "try to avoid" errors is to consider the following intervals
$x<1⟹<!--\; ⟹\; -->|x-1|+|x-2|>1⟺<!--\; ⟺\; -->1-<!--\; -\; -->x+2-<!--\; -\; -->x>1⟺<!--\; ⟺\; -->2x<2⟺<!--\; ⟺\; -->x<1$
$1\le <!--\; \le \; -->x<2⟹<!--\; ⟹\; -->x-1+2-<!--\; -\; -->x>1⟺<!--\; ⟺\; -->1>1$
$x\ge <!--\; \ge \; -->2⟹<!--\; ⟹\; -->|x-1|+|x-2|>1⟺<!--\; ⟺\; -->x-<!--\; -\; -->1+x-<!--\; -\; -->2>1⟺<!--\; ⟺\; -->x>2$