Solve inequality help plz!!

**NeedHelp18** · September 10th 2008, 05:10 AM

|3-4x|(x+5)(3x-2)>0--thank y

**ThePerfectHacker** · September 10th 2008, 05:20 AM

Originally Posted by NeedHelp18

|3-4x|(x+5)(3x-2)>0--thank y

Note that $|3-4x| \geq 0$ always.
If $x\not = \tfrac{3}{4}$ then $|3-4x| \not = 0$ and so $|3-4x| > 0$ .
We know that $x=\tfrac{3}{4}$ cannot be such a number which makes this true - thus we will look at all other numbers to solve this.

Divide by $|3-4x|$ to get $(x+5)(3x-2)>0$ .

Case 1: Both $(x+5),(3x-2) > 0$ . This means that $x>-5 \text{ and }x>\tfrac{2}{3}$ this combines into just saying $x>\tfrac{2}{3}$ .

Case 2: Both $(x+5),(3x-2) < 0$ . This means that $x<-5 \text{ and }x< \tfrac{2}{3}$ this combines into just saying $x < -5$ .

This gives us an answer that $x \in (-\infty, -5) \cup (\tfrac{2}{3},\infty)$ .

But that is not quite right because remember we said $x\not = \tfrac{3}{4}$ . Thus we need to take that point out of this solution set. Thus, the correct answer is $x\in (-\infty,5)\cup (\tfrac{2}{3},\tfrac{3}{4}) \cup (\tfrac{3}{4},\infty)$ .

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