I honestly cannot believe I am stuck on this.

[flyingpig]

Homework Statement

Solve

\left | \frac{5}{x + 2} \right | < 1

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

\left |5\right | &lt; \left|x + 2\right|
-|x + 2| &lt; -5
5 &lt; -(x + 2) &lt; -5
-5 &gt; x + 2 &gt; 5
-7 &gt; x &gt; 3

I am sorry lol

 

[Saitama]

I don't understand how you get the third step?

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

If you solve these two equations:-
5<(x+2)
and
5<-(x+2)
You will get the desired result.

 

[BloodyFrozen]

-7 &gt; x &gt; 3

?

x is greater than 3 and less than -7?

 

[flyingpig]

See I don't know what I am doing lol

 

[Dembadon]

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.

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]

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.

 

[Mark44]

Solve

\left | \frac{5}{x + 2} \right | &lt; 1

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

\left |5\right | &lt; \left|x + 2\right|

Your inequality is equivalent to
|x + 2| > 5 and x \neq -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]

Homework Statement

Solve

\left | \frac{5}{x + 2} \right | &lt; 1

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

\left |5\right | &lt; \left|x + 2\right|
-|x + 2| &lt; -5
5 &lt; -(x + 2) &lt; -5
-5 &gt; x + 2 &gt; 5
-7 &gt; x &gt; 3

I am sorry lol

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

RGV

 

[NascentOxygen]

A graph of the curve: http://fooplot.com/index.php?q0=abs(5/(x+2))"