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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.
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.
