Solve : |x-4|>a if case1:a>0 and case2:a<0
I am getting answers which look similar in both cases. please i wish to know why it is so and how different both answers are when plotted on a number line
If $a \lt 0$, all $x$ will satisfy it as all absolute values are $ \ge 0$. If $a \gt 0$ you need the points more than $a$ from $4$.
The work in your comment was a good start. Another thing to notice is that if $a>0$, then $4-a<4+a$, so your solution consists of two disjoint "rays" corresponding to the inequalities you wrote. However, when $a<0$, then $4+a<4-a$, and in particular each $x$ such that $x\geq 4-a$ satisfies $x>4+a$. That is one way to see that your method leads to the same answer mentioned in Ross's answer. But Ross's method of observing that $|x-4|\geq 0>a$ for all $x$ if $a<0$ is a little easier.
To make things a little more concrete, consider what happens when $a=5$: Your method says that $x>9$ or $x<-1$, which is correct. Now when $a=-5$, your method says that $x>-1$ or $x<9$, which is true for all real numbers.
Consider multiple cases
Case 1: $a = 0$
Then, any number other than $4$ will satisfy your inequality.
Case 2: $a < 0$
Then, any $x$ will satisfy your inequality since absolute values are $\ge 0$
Case 3: $a > 0$
Then $|x - 4| > a$ if and only if $x$ is farther than $a$ units from $4$. Hence, $x - 4 > a$ or $x - 4 < -a$. So the set of all real numbers that satisfy your inequality is $$(-\infty, 4 - a) \cup (4+a, \infty)$$