I honestly cannot believe I am stuck on this.

  • Thread starter flyingpig
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Broken] gives the picture of what is happening. The graph is undefined at x = -2, and as you approach x = -2 from the left or the right, the graph approaches infinity.In summary, the given inequality is equivalent to the double inequality -1 < 5/(x+2) < 1, which can be rewritten as -1 < x+2 < 1, giving the solution -7 < x < 3, with the restriction that x cannot equal -2. You attempted to solve the inequality by setting up a double inequality with a negative on the absolute value expression, which is incorrect.
  • #1
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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  • #2
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:
 
  • #3
[tex]-7 > x > 3[/tex]

?

x is greater than 3 and less than -7?
 
  • #4
See I don't know what I am doing lol
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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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