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I honestly cannot believe I am stuck on this.

  • Thread starter flyingpig
  • Start date
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1. The problem statement, all variables and given/known data


Solve

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

3. 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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92
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?
 
2,556
1
See I don't know what I am doing lol
 

Dembadon

Gold Member
607
89
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.
 

uart

Science Advisor
2,776
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32,520
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Solve

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

3. The attempt at a solution

All I know is that x= -2 is discontinuous.
No, the rational expression is undefined at x = -2.
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.
 

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
10,705
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1. The problem statement, all variables and given/known data


Solve

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

3. 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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