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My question is:

Solve: $|x-4|< a$, where $a$ belongs to the real numbers. Solve this by considering various cases depending upon whether $a$ is negative, positive or zero.

What I have tried so far: If $a>0$ then: $x < a+4$ and $x>4-a$, if $a=0$ then there is no solution.

My doubt is: Should I consider the case $a<0$ as again $|x-4|<a$ which is not possible as absolute value cannot be negative.

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Note that if $a \le 0$, there is no $x$ such that $|x-4|\lt a$, because $|w|$ is always $\ge 0$. That will com out if you use the general machinery described by robjohn, but you might as well handle it separately. –  André Nicolas Jun 2 '12 at 19:07
Probably the hint should be: depending on whether $x-4$ is negative, positive, or zero. –  GEdgar Jun 3 '12 at 23:29

1 Answer 1

up vote 4 down vote accepted

Hint: $|x-4|<a$ means that $x$ is closer than $a$ units to $4$.

Another Hint: $|x-4|<a$ means that $(x-4)<a$ and $-(x-4)<a$.

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I really dont understand these modulus based questions. plz can you try to show it on a number line? –  mgh Jun 2 '12 at 14:27
There is no modulus here; there is an absolute value. Try this: draw a number line from $0$ to $10$, then color the part of the number line that is closer than $2$ units to $4$. Try the same thing for the part closer than $3$ units to $4$. –  robjohn Jun 2 '12 at 14:33
@ robjohn ya I did that –  mgh Jun 2 '12 at 14:43
@ robjohn Plz can you tell me the solution as I am getting very confused. –  mgh Jun 2 '12 at 14:47
@user1396721: using the second hint, can you plot $(x-4)<a$ and $-(x-4)<a$? The answer is the intersection of these sets. –  robjohn Jun 2 '12 at 15:59

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