Does the final answer depend on the original expression or its simplified form? I came across this simple probability problem.

From first $50$ natural numbers, a number is selected at random. Find the probability that selected number satisfies $$\left|\frac{3}{x-4}\right|>1$$

I solved it in this way: 
Sample space is $^{50}C_1$ 
Case $1$: $\left(\frac{3}{x-4}\right)<-1$
$$\begin{array}{l|l}
\mathrm {When} \;x-4>0 &\mathrm{When}\;x-4<0\\
\implies x>4 & \implies x<4\\
\therefore 3<4-x & \therefore 3>4-x\\
\therefore x<1 & \therefore x>1\\
\mathrm {This \;is\;absurd} & \therefore x=2,3
\end{array}$$
Case  $2$: $\left(\frac{3}{x-4}\right)>1$
$$\begin{array}{l|l}
\mathrm {When} \;x-4>0 &\mathrm{When}\;x-4<0\\
\implies x>4 & \implies x<4\\
\therefore 3>x-4 & \therefore 3<x-4\\
\therefore x<7 & \therefore x>7\\
\therefore x=5,6 & \mathrm {This \;is\;absurd}
\end{array}$$
Thus, I got four values of $x$, namely, $2,3,5,6$ and hence, the probability is $4 \over 50$
Now, the book solved it in this way:
$$\left|\frac{3}{x-4}\right|>1 \implies |x-4|<3 \implies -3<x-4<3 \implies 1<x<7$$
$$\therefore x=2,3,4,5,6$$
Thus, probability is $5 \over 50$
What I wanna ask is, shouldn't $4$ be excluded? If we consider the original inequality, then putting $x=4$ doesn't satisfy it. However, it does satisfy when we put it in the simplified form, that is, $|x-4|<3$. 
Often, in limit problems, we simplify the original expression to get a well-defined answer. Otherwise, just substituting the limit in the original expression gives us an indeterminate form. Does the same apply here?

Community wiki answer so the question can be marked as answered:
As has been stated in various comments, you're right and the book is wrong. Substituting $4$ into the inequality results in an undefined expression that cannot be said to be satisfied or not satisfied.
