|3-4x|(x+5)(3x-2)>0--thank y
Note that $\displaystyle |3-4x| \geq 0$ always.
If $\displaystyle x\not = \tfrac{3}{4}$ then $\displaystyle |3-4x| \not = 0$ and so $\displaystyle |3-4x| > 0$.
We know that $\displaystyle x=\tfrac{3}{4}$ cannot be such a number which makes this true - thus we will look at all other numbers to solve this.
Divide by $\displaystyle |3-4x|$ to get $\displaystyle (x+5)(3x-2)>0$.
Case 1: Both $\displaystyle (x+5),(3x-2) > 0$. This means that $\displaystyle x>-5 \text{ and }x>\tfrac{2}{3}$ this combines into just saying $\displaystyle x>\tfrac{2}{3}$.
Case 2: Both $\displaystyle (x+5),(3x-2) < 0$. This means that $\displaystyle x<-5 \text{ and }x< \tfrac{2}{3}$ this combines into just saying $\displaystyle x < -5$.
This gives us an answer that $\displaystyle x \in (-\infty, -5) \cup (\tfrac{2}{3},\infty)$.
But that is not quite right because remember we said $\displaystyle x\not = \tfrac{3}{4}$. Thus we need to take that point out of this solution set. Thus, the correct answer is $\displaystyle x\in (-\infty,5)\cup (\tfrac{2}{3},\tfrac{3}{4}) \cup (\tfrac{3}{4},\infty)$.