Finding the Solution Set for an Inequality


by Learning_Math
Tags: inequality, solution
Learning_Math
Learning_Math is offline
#1
Dec9-09, 12:52 PM
P: 7
1. I need to find the solution set for |(3x+2)/(x+3)|>3.

3. When I solve the inequality (3x+2)/(x+3)>3, I get 2>9 which is clearly false. When I solve the inequality (3x+2)/(x+3)> -3, I come with the solution set (-inf. -11/6). My teacher is saying that there the solutotion set is (-inf. -3)U(-3, -11/6).

I just can't figure out how to get to that solution. I can't figure out where that -3 is coming from. In his sparse notes on my assignment, he says there are two subcases for each of the two cases in number 1. Those are when x < -3 and when x > -3. I just cant' figure out how to use these cases.
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statdad
statdad is offline
#2
Dec9-09, 01:28 PM
HW Helper
P: 1,344
The first step for solving [tex] |X| > a [/tex], for any expression X and number a, is to eliminate the absolute values with this:

[tex]
X < -a \text{ or } X > a
[/tex]

If you need to solve an inequality like (this is entirely made up for illustration)

[tex]
\frac x {x+1} > 5
[/tex]

your first steps should be

[tex]
\begin{align*}
\frac x {x+1} - 5 & > 0 \\
\frac x {x+1} - \frac{5(x+1)}{x+1} & > 0 \\
\frac{x - (5x+5)}{x+1} & > 0 \\
\frac{-4x - 5}{x+1} & > 0 \\
\frac{(-1)(4x+5)}{x+1} & > 0\\
\frac{4x+5}{x+1} & < 0
\end{align*}
[/tex]

I passed from the next-to-last to the last line by multiplying by (-1).

These steps let you avoid the all-to-common problem of multiplying both sides of an inequality by a variable term when you don't know whether it's positive or negative.
Learning_Math
Learning_Math is offline
#3
Dec9-09, 01:46 PM
P: 7
Thanks for that setup. I will remember it for future use. I have also been completely overlooking that (3x+2)/(x+3) is undefined when x = -3. So that is how I get (-inf. -3)U(-3, -11/6) instead of just (-inf. -11/6).


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