Finding the Solution Set for an Inequality

[Learning_Math]

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.

 

[statdad]

The first step for solving |X| &gt; a, for any expression X and number a, is to eliminate the absolute values with this:

<br /> X &lt; -a \text{ or } X &gt; a<br />

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

<br /> \frac x {x+1} &gt; 5<br />

your first steps should be

<br /> \begin{align*}<br /> \frac x {x+1} - 5 &amp; &gt; 0 \\<br /> \frac x {x+1} - \frac{5(x+1)}{x+1} &amp; &gt; 0 \\<br /> \frac{x - (5x+5)}{x+1} &amp; &gt; 0 \\<br /> \frac{-4x - 5}{x+1} &amp; &gt; 0 \\<br /> \frac{(-1)(4x+5)}{x+1} &amp; &gt; 0\\<br /> \frac{4x+5}{x+1} &amp; &lt; 0<br /> \end{align*}<br />

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]

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).