jmc

We can make our work easier by rewriting all fractions in the inequality so that they have a common denominator of $48$: $$\frac{10}{48}+\left|x-\frac{11}{48}\right|<\frac{15}{48}$$Then we subtract $\frac{10}{48}$ from both sides: $$\left|x-\frac{11}{48}\right|<\frac{5}{48}$$The expression on the left side is the positive difference between $x$ and $\frac{11}{48}$. So, the inequality says that $x$ is strictly between $\frac{11}{48}-\frac{5}{48}$ and $\frac{11}{48}+\frac{5}{48}$. Simplifying these expressions and writing our answer in interval notation, we have $x\in \boxed{\left(\frac{1}{8},\frac{1}{3}\right)}$.