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


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.

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can u show your work? – Bhargav Jun 4 '12 at 10:12
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 answer's meaning is different i am unable to find out how the points included in both cases are different – mgh Jun 4 '12 at 10:14
Related:… – TMM Jun 4 '12 at 11:07

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 $4$.

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:i was having problem to solve:|x-2|+|x-5|=3 . I was trying to post this question but could not(i was being said that it does not meet our quality standards).So posted it here. – mgh Jun 4 '12 at 15:43
@meg_1997: when you have two absolute value signs, it is easiest to consider each region of $x$ and resolve the signs. So for $x\le 2$ both expressions are negative and need to be inverted. You are left with $7-2x=3$ AND $x \le 2$. You solve the equality and see if it meets the inequality. Then there are two more sections of the real line to consider the same way. – Ross Millikan Jun 4 '12 at 21:15

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

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