
#1
Dec909, 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. 



#2
Dec909, 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 nexttolast to the last line by multiplying by (1). These steps let you avoid the alltocommon problem of multiplying both sides of an inequality by a variable term when you don't know whether it's positive or negative. 



#3
Dec909, 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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