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.
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.
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).