An inequality with absolute values

[theself]

Homework Statement

Solve Ix+3I>2
*I is used for absolute value notation

The Attempt at a Solution

Considering both
a) Ix+3I > 0 then Ix+3I= x+3
b) Ix+3I < 0 then Ix+3I= -(x+3)

when solved this would yield to;

a) x>-3 and x>-1
b) x<-5 and x<-3

from my general reasoning i think the answer should be x>-1 and x<-5. Why are the solutions x>-3 and x<-3 omitted? Is it because the other two include a broader range?

thanks for your help.

 

[ehild]

Hi theself, welcome to PF.

Homework Statement

Solve Ix+3I>2
*I is used for absolute value notation

The Attempt at a Solution

Considering both
a) Ix+3I > 0 then Ix+3I= x+3
b) Ix+3I < 0 then Ix+3I= -(x+3)

when solved this would yield to;

a) x>-3 and x>-1
b) x<-5 and x<-3

from my general reasoning i think the answer should be x>-1 and x<-5. Why are the solutions x>-3 and x<-3 omitted? Is it because the other two include a broader range?

thanks for your help.

You can find an absolute value key (vertical line) on the keyboard.

So the problem is: Solve |x+3|>2

The absolute value is never negative. The two cases are x+3≥0 (x>-3) and x+3<0 (x<-3).

You wrote correctly that |x+3|=x+3 in the first case and |x+3|=-(x+3) in the second case.

In the first case, you got x>-1 if x>-3 is true. x>-1 is more strict requirement than x>-3. So x>-1 is solution.

In the second case, you got x<-5 if x<-3. x<-5 is more strict condition than x<-3 so x<-5 is solution.

"Why are the solutions x>-3 and x<-3 omitted?"

They are not solutions. Substitute x=-2 for example: (it is greater then -3). |x+3|=1 less then 2. Choose x=-4: (it is less then -3) |-4+3|=|-1|=1.

ehild

 

[Mark44]

Homework Statement

Solve Ix+3I>2
*I is used for absolute value notation

Why? Use the | character instead.

The Attempt at a Solution

Considering both
a) Ix+3I > 0 then Ix+3I= x+3
b) Ix+3I < 0 then Ix+3I= -(x+3)

when solved this would yield to;

a) x>-3 and x>-1
b) x<-5 and x<-3

How did you get what you have just above?
|x + 3| > 2
<==> x + 3 > 2 OR x + 3 < -2
Can you take it from here?

from my general reasoning i think the answer should be x>-1 and x<-5. Why are the solutions x>-3 and x<-3 omitted? Is it because the other two include a broader range?

thanks for your help.

 

[HallsofIvy]

The simplest way to solve a general inequality is to first solve the corresponding equation. If |x+ 3|= 2 then either x+ 3= 2 or x+ 3= -2. In the first case x= -1 and in the second x= -5.

The key point here is that, since absolute value is continuous, we can go from "<" to ">", and vice-versa, only at "=". That is, the two points, -5 and -1, divide the real numbers into three intervals and on each interval only one of "<" or ">" can apply. And we need only check one point in each interval to see which.

For example, -6< -5 and |-6+3|= |-3|= 3 which is larger than 2, not less. The original inequality is false at x= -6 and so false for all x less than -5. -4 lies between -5 and -1 and |-4+3|= |-1|= 1 which is less than 2. The original inequality is true at x= -4 and so true for all x between -5 and -1. Finally, 0 is larger than -1 and |0+ 3|= |3|= 3 which is larger than 2, not less. The original inequality is false at x= 0 and so false for all x larger than -1.

By the way, every keyboard I have ever seen has a "|" just above the return key. Are you using a keyboard that doesn't?

 

[theself]

In the first case, you got x>-1 if x>-3 is true. x>-1 is more strict requirement than x>-3. So x>-1 is solution.

In the second case, you got x<-5 if x<-3. x<-5 is more strict condition than x<-3 so x<-5 is solution.

ehild

Is the if condition there because we define it like that at the beginning ?

 

[theself]

How did you get what you have just above?
|x + 3| > 2
<==> x + 3 > 2 OR x + 3 < -2
Can you take it from here?

Well, I kind of did it in a long way;
If |x+3|> 0 then |x+3|= x+3, and substituted the term (x+3) to where |x+3| is found
this gives
if x> -3, x > -1
and did the same for the other condition

 

[theself]

The simplest way to solve a general inequality is to first solve the corresponding equation. If |x+ 3|= 2 then either x+ 3= 2 or x+ 3= -2. In the first case x= -1 and in the second x= -5.

Wouldn't using the equation create a problem as you can not do everything you do to an equation to an inequality?
For eg: dividing both sides by - 1

* Also I apologize for my clumsiness for not finding the key for "|"

 

[ehild]

Is the if condition there because we define it like that at the beginning ?

I think the answer is "yes"
The result x>-1 came from the condition that x>-3. x>-3 and x>-1 are not two solutions. Those x values for which -3<x<-1 do not satisfy the original inequality. In similar way, when we supposed that x<-3 we got the result x<-5. Again, the original inequality is not true for -5<x<-3.

ehild