Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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: math.stackexchange.com/questions/152869/… –  TMM Jun 4 '12 at 11:07

3 Answers 3

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 $4$.

share|improve this answer
: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

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.

share|improve this answer
This is a followup to the exchange in math.stackexchange.com/questions/153644/… –  Ross Millikan Jun 4 '12 at 10:56
I don't see lasting value in keeping this content posted. I deleted my answer, but it was undeleted by moderator. –  Jonas Meyer Aug 7 '14 at 3:24

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.