For what x is 1 x + 3 > 2 x - 3 ?

Step 1
Note that ${\displaystyle x=-3}$ and ${\displaystyle x=3}$ are not solutions since one or the other side of the inequality is undefined for these values of x.
Split the remaining possibilities into cases:
Case ${\displaystyle x\in (-\infty ,-3)\cup (3,\infty )}$
${\displaystyle x+3}$ and ${\displaystyle x-3}$ are both negative or both positive.
In either case ${\displaystyle (x+3)(x-3)>0}$ so multiply both sides of the inequality by that to get:
${\displaystyle x-3>2x+6}$
Subtract ${\displaystyle x+6}$ from both sides to get:
${\displaystyle -9>x}$
Combining with the conditions of the case, that gives us the solution:
${\displaystyle x\in (-\infty ,-9)}$
Case ${\displaystyle x\in (-3,3)}$.
${\displaystyle x+3>0}$ and ${\displaystyle x-3<0}$, so ${\displaystyle (x+3)(x-3)<0}$, so multiply both sides of the inequality by this and reverse the inequality to get:
${\displaystyle x-3<2x+6}$
Step 2
Subtract ${\displaystyle x+6}$ from both sides to get:
${\displaystyle -9<x}$
Combining with the conditions of the case, that gives us the solution:
${\displaystyle x\in (-3,3)}$
Conclusion
Putting the two cases together, the original inequality holds for all
${\displaystyle x\in (-\infty ,-9)\cup (-3,3)}$