Question 1095251

Hi! I really find this problem difficult. Hoping someone could help me. Thanks!
(X)/(x-3) is greater than (2x-11)/(2x)
rational inequality
<pre>{{{x/(x - 3) > (2x - 11)/(2x)}}}
Based on the DENOMINATORS, {{{matrix(1,3, x <> 3, and, x <> 0)}}}
Therefore, we already have 2 critical values: 3 and 0
{{{2x^2 > (2x - 11)(x - 3)}}} ------- Multiplying each side by LCD, (x - 3)(2x)
{{{2x^2 > 2x^2 - 17x + 33}}}
{{{2x^2 - 2x^2 + 17x > 33}}}
17x > 33 
{{{matrix(1,3, x > 33/17, or, x > 1&16/17)}}}
We now have 3 CRITICAL VALUES: {{{matrix(1,6, "3,", "0,", and, 33/17, or, 1&16/17)}}}
There now exists the following 4 (FOUR) INTERVALS that need to be tested.  
==========<pre><b>Interval 1: x < 0, or x = - 1</b>
{{{x/(x - 3) > (2x - 11)/(2x)}}}
{{{(- 1)/(- 1 - 3) > (2(- 1) - 11)/(2(- 1))}}} ------- Substituting - 1 for x
{{{(- 1)/(- 4) > (- 13)/(- 2)}}}
Is {{{1/4 > 13/2}}}? No!! Therefore, a solution DOES NOT exist in this interval.
================</pre><pre><b>Interval 2: {{{0 < x < 1&16/17}}}, or x = 1</b>
{{{x/(x - 3) > (2x - 11)/(2x)}}}
{{{(1)/(1 - 3) > (2(1) - 11)/(2(1))}}} ------- Substituting 1 for x
{{{1/(- 2) > (- 9)/2}}}
Is {{{- 1/2 > - 9/2}}}? Yes!! Therefore, a solution DOES exist in this interval.
===============</pre><pre><b>Interval 3: {{{1&16/17 < x < 3}}}, or x = 2</b>
{{{x/(x - 3) > (2x - 11)/(2x)}}}
{{{2/(2 - 3) > (2(2) - 11)/(2(2))}}} ------- Substituting 2 for x
{{{2/(- 1) > (- 7)/4}}}
Is {{{- 2 > - 7/4}}}? No!! Therefore, a solution DOES NOT exist in this interval.
============</pre><pre><b>Interval 4: x > 3, or x = 4</b>
{{{x/(x - 3) > (2x - 11)/(2x)}}}
{{{4/(4 - 3) > (2(4) - 11)/(2(4))}}} ------- Substituting 4 for x
{{{4/1 > (- 3)/8}}}
Is {{{4 > (- 3/8)}}}? Yes!! Therefore, a solution DOES exist in this interval.</pre>The solution, in INTERVAL NOTATION is: *[Tex \Large (0,\frac{33}{17}) \cup (3,\infty)]