Com10atorics wrote:
Solve the equation
$|x+1|-|x-1|=2$.
$\iff$
Either $x\ge 1$ and $(x+1)-(x-1)=2$ (always true)
Either $1\ge x\ge -1$ and $(x+1)+(x-1)=2$ (and so $x=1$)
Either $-1\ge x$ and $-(x+1)+(x-1)=2$ (never true)
Hence answer $\boxed{x\in[1,+\infty)}$