solve $\displaystyle \frac{2x+3}{x-5}<the modulus of \mid \frac{4x+12}{x+1}\mid$
2. $\displaystyle \frac{2x+3}{x-5}<\left| \frac{4x+12}{x+1} \right|$ implies that $\displaystyle \frac{4x+12}{x+1}>\frac{2x+3}{x-5}$ or $\displaystyle \frac{4x+12}{x+1}<-\frac{2x+3}{x-5}.$ Do each one of them. Solution set is the union of solution for the first inequality and the second one.
3. thanks i got $\displaystyle \frac{2\sqrt{17}}{4}<x<\frac{\sqrt{58}+1}{3}$ is this the right answer and what you mean by union? thanks