Rational Inequality...3

Section 2.7
Question 46



5/(x - 6) > 3/(x + 2)

5/(x - 6) - 3/(x + 2) > 0

We know that x - 6 = 0 leads to x = 6.
We also know that x + 2 leads to x = - 2.

At x = -2 and x = 6, we draw a HOLE on the number line to indicate where the fraction is undefined.

There are three intervals to check.

When x = -3, we get -5 > -27. True statement.
When x = 0, we get -5/6 > 3/2. False statement.
When x = 7, we get 15 > 1. True statement.

The original inequality is satisfied in the following intervals: (-infinity, -2) U (6, infinity).



I think one more should do it in terms of practice. I want to spend time in the Calculus Forum today.

 

Solutions:
x>6
-14<x<-2

Interval notation:
(6, ∞)
(-14, -2)

 

Ok. I am wrong again. Where did (x + 14) come from?

 

from here you dee that numerator is x+14, if equal to zero you got x=-14