I honestly cannot believe I am stuck on this.

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Homework Help Overview

The discussion revolves around solving the inequality involving an absolute value: \(\left | \frac{5}{x + 2} \right | < 1\). Participants are exploring the implications of the inequality and the discontinuity at \(x = -2\).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods to approach the inequality, including test points and double inequalities. Some express confusion about the steps involved, particularly regarding the interpretation of absolute values and the conditions under which the inequality holds.

Discussion Status

The discussion is ongoing, with participants providing different perspectives on how to set up the problem. Some have suggested visualizing the functions involved, while others have pointed out potential misunderstandings in the application of absolute value properties. There is no clear consensus yet on the correct approach.

Contextual Notes

Participants note the discontinuity at \(x = -2\) and the need to consider this when determining the solution set. There is also mention of the computer-generated solution, which raises questions about its derivation.

flyingpig
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Homework Statement




Solve

[tex]\left | \frac{5}{x + 2} \right | < 1[/tex]

The Attempt at a Solution



All I know is that x= -2 is discontinuous. I thought about doing test points, but it seemed rather useless seeing I will probably end up with (-2, infinity) or (-inf, -2)

Computer says the solution is x < -7, x >3

I then thought about doing this

[tex]\left |5\right | < \left|x + 2\right|[/tex]
[tex]-|x + 2| < -5[/tex]
[tex]5 < -(x + 2) < -5[/tex]
[tex]-5 > x + 2 > 5[/tex]
[tex]-7 > x > 3[/tex]

I am sorry lol
 
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I don't understand how you get the third step? :confused:

You should go by taking two cases.
What do get when you solve |x+2|? Definitely (x+2) and -(x+2). :smile:

If you solve these two equations:-
5<(x+2)
and
5<-(x+2)
You will get the desired result. :smile:
 
[tex]-7 > x > 3[/tex]

?

x is greater than 3 and less than -7?
 
See I don't know what I am doing lol
 
flyingpig said:
See I don't know what I am doing lol

You can set up two inequalities (see Pranav-Anora's post). Or you can set up a double inequality since this problem is of the "less-than" variety. You attempted to do the double inequality method, but you shouldn't have a negative on your absolute value expression. I think you mixed two different methods into one. :wink:

Here's a general example of how you should've setup the double inequality:

Given |expression| < k, set up the double inequality, -k < |expression| < k and solve.
 
flyingpig said:
See I don't know what I am doing lol

Why don't you try making a sketch of y= |5/(x+2)| and y=1 on the same set of axes.
 
flyingpig said:
Solve

[tex]\left | \frac{5}{x + 2} \right | < 1[/tex]

The Attempt at a Solution



All I know is that x= -2 is discontinuous.
No, the rational expression is undefined at x = -2.
flyingpig said:
I then thought about doing this

[tex]\left |5\right | < \left|x + 2\right|[/tex]
Your inequality is equivalent to
|x + 2| > 5 and x [itex]\neq[/itex] -2

In general, if you have an inequality of the form |x + a| > b, you can get rid of the absolute values by rewriting this as
x + a > b or x + a < - b

For your problem, since x cannot be -2, any interval you end up with cannot include -2.
 
flyingpig said:

Homework Statement




Solve

[tex]\left | \frac{5}{x + 2} \right | < 1[/tex]

The Attempt at a Solution



All I know is that x= -2 is discontinuous. I thought about doing test points, but it seemed rather useless seeing I will probably end up with (-2, infinity) or (-inf, -2)

Computer says the solution is x < -7, x >3

I then thought about doing this

[tex]\left |5\right | < \left|x + 2\right|[/tex]
[tex]-|x + 2| < -5[/tex]
[tex]5 < -(x + 2) < -5[/tex]
[tex]-5 > x + 2 > 5[/tex]
[tex]-7 > x > 3[/tex]

I am sorry lol

For real A, the inequality |A| < 1 is the same as -1 < A < 1.

RGV
 

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