An inequality with absolute value and a parameter: $|x-4|>a$

Question

Solve : $|x-4|>a$.
Case 1: $a>0$; Case 2: $a<0$

Progress

I am getting answers which look similar in both cases:

• Let $a>0$ so $x>4+a$ or $x<4-a$ ,
• Let $a<0$ so $x>4+a$ or $x<4-a$ .

Though I know that both answers' meaning is different I am unable to find out how the points included in both cases are different

I wish to know why it is so and how different both answers are when plotted on a number line.

Answer 1

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 $4$.

Answer 2

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 $$(-\infty, 4 - a) \cup (4+a, \infty)$$