smc

We can do casework upon the value of $x$. First, consider the case where both absolute values are positive, which is when $x\ge 1$. In this case, the equation becomes $2x+1<5$. This turns into $x<2$. Combining this with our original assumption, we get the solutions $1\le x<2$. Next, consider the case when both absolute values are negative, which is when $x<-2$. This yields $-1-2x<5$, or $x>-3$. Combining this with our original assumption, we get $-3<x<-2$ The next case is when the first absolute value is positive and the second is negative. This occurs when $x\ge 1$ and $x<-2$. Obviously, this has no solutions, since the inequalities do not overlap. The final case is when the first absolute value is negative and the second is positive. This occurs when $x<1$ and $x\ge -2$. This yields $3<5$, which is always true. Therefore, we also get the solutions $-2\le x<1$. Therefore, after combining all of our solutions, we get the range $-3<x<2$, which is $\boxed{x\in (-3,2)}$.