# I honestly cannot believe I am stuck on this.

1. Aug 28, 2011

### flyingpig

1. The problem statement, all variables and given/known data

Solve

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

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

$$\left |5\right | < \left|x + 2\right|$$
$$-|x + 2| < -5$$
$$5 < -(x + 2) < -5$$
$$-5 > x + 2 > 5$$
$$-7 > x > 3$$

I am sorry lol

2. Aug 28, 2011

### 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.

3. Aug 28, 2011

### BloodyFrozen

$$-7 > x > 3$$

?????????????????????????????????????????

x is greater than 3 and less than -7?

4. Aug 28, 2011

### flyingpig

See I don't know what I am doing lol

5. Aug 28, 2011

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.

6. Aug 28, 2011

### uart

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

7. Aug 28, 2011

### Staff: Mentor

No, the rational expression is undefined at x = -2.
|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.

8. Aug 29, 2011

### Ray Vickson

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

RGV

9. Aug 30, 2011

### Staff: Mentor

Last edited by a moderator: Apr 26, 2017