Is this absolute value inequality correct?

Question

Given $ -3 - x \lt x - 4 \lt -3 + x $, I want to add absolute value bracket to $x - 4$, making it $ |x - 4| \lt a $.

$\because|x - 4| < a$ means $ -a \lt x - 4 \lt a . $

$ \because-(- 3- x) > (-3 + x) $

$ \therefore -3 -x \lt x - 4\lt -3 + x \lt 3+ x$

$ \therefore |x - 4| < 3+ x $

Is this correct?

Answer 1

We have

$$-3 - x \lt x - 4 \iff x> \frac12$$

$$x - 4 \lt -3 + x \iff -4<-3$$

therefore

$$|x - 4| \ge 0 $$

Answer 2

Given $$ -3 - x \lt x - 4 \lt -3 + x $$ you can add $3$ to all sides and get $$ - x \lt x - 1 \lt + x $$ hence $$|x-1| \lt x.$$

The solution is $x>1/2. $