Question 1012038
this inequality is true when x > 7/4 and x < 4.


this can also be expressed as x > 1.75 x < 4.


this can also be expressed as 1.75 < x < 4.


this can be seen easily in the following graph.


<img src = "http://theo.x10hosting.com/2016/010202.jpg" alt="$$$" </>


the equation is discontinuous at x = 4, and has a vertical asymptote at x = 4, as you can see on the graph.


as the graph approaches x = 4 from the left, (x-4)/3 - (3/(x-4) approaches positive infinity.


as the graph approaches x = 4 from the right, (x-4)/3 - (3/(x-4) approaches minus infinity.


at x = 7/4 the graph of (x-4)/3 - (3/(x-4) is equal in value to the graph of x/3.


the graph helps you tremendously to see what is happening.


if you don't have a good graphing software available, then get one.


it helps to visualize what is happening.


otherwise, you're working blind unless you can construct a graph manually.


the one i used is at http://www.desmos.com/calculator.


one of the ways you would solve this algebraically as follows:


start with:


{{{(x-4)/3 - 3/(x-4) > x/3}}}


subtract x/3 from both sides of the equation to get:


{{{(x-4)/3 - 3/(x-4) - x/3 > 0}}}


set the equation equal to 0 to get:


{{{(x-4)/3 - 3/(x-4) - x/3 = 0}}}


solve for x to get x = 7/4.


look at your equation to see that it has a vertical asymptote at x = 4.


you have 2 checkpoints.


you have y = 0 at x = 7/4 and you have y = undefined at x = 4.


the graph will be continuous up to x = 4 and will be continuous after x = 4.


you need to check 3 intervals.


the intervals are:


x < 7/4
x > 7/4 and < 4
x > 4


when x < 7/4, y is negative.
when x > 7/4 and less than 4, y is positive.
when x > 4, y is negative


this means the graph is positive between x = 7/4 and x = 4 but not including x = 7/4 and x = 4.


that's your solution, because:


if {{{(x-4)/3 - 3/(x-4) > x/3}}}, then {{{(x-4)/3 - 3/(x-4) - x/3 > 0}}}