Sketch on the same coordinate axes , the graphs for $\displaystyle y=\frac{1}{|x|} $ and $\displaystyle 4y=3|x|-4$

Hence , find the set of values of x such that $\displaystyle 3|x|-4>\frac{1}{|x|}$

I managed to sketch $\displaystyle y=\frac{1}{|x|} $ and $\displaystyle y=\frac{3}{4}|x|-1$

Is $\displaystyle \frac{3}{4}|x|-1>\frac{1}{|x|}$ the same as $\displaystyle 3|x|-4>\frac{1}{|x|}$ ??

Zz i am confused . Thanks !