Question 1074120
<pre><b>

The tutor above points out something that I have long known.
It is virtually impossible to successfully teach students to write
mathematics containing fractions correctly all on one line.

You wrote this:

4/x+2 > 2

which means this {{{4/x+2>2}}}, which is the way the tutor took
it. 

However I reason that since you are taking a higher level of
algebra, your teacher would not be testing you to see if you are
able to do a simple thing like the first step would require, to
subtract 2 from both sides to get {{{4/x>0}}}

Therefore I assume that you meant this: 

{{{4/(x+2) > 2}}}

instead, which when written on one line would require parentheses
around the denominator; otherwise the +2 will not be taken as part 
of the denominator but will be taken as a separate term.  So it 
should be written this way:

4/(x+2) > 2 

------------------------------

{{{4/(x+2) - 2 > 0}}}

{{{4/(x+2) - 2(x+2)/(x+2) > 0}}}

{{{(4 - 2(x+2))/(x+2) > 0}}}

{{{(4 - 2x-4)/(x+2) > 0}}}

{{{-(2x)/(x+2) > 0}}}

We find the critical numbers by setting
numerator = 0 and denominator = 0

2x = 0;   x+2 = 0
 x = 0;     x = -2

[Note: These critical numbers are NOT permissible values of x]

We place the critical numbers on a number line:

-------o-----o------------
-4 -3 -2 -1  0  1  2  3  4

We test a value for x left of -2, say -3 in this:

{{{-(2x)/(x+2) > 0}}}
{{{(-2(-3))/(-3)+2) > 0}}}
{{{6/(-1) > 0}}}
{{{-6 > 0}}}

That's false so we do not shade left of -2

We test a value for x between -2 and 0, say -1 in this:

{{{-(2x)/(x+2) > 0}}}
{{{(-2(-1))/(-1)+2) > 0}}}
{{{6/(1) > 0}}}
{{{6 > 0}}}

That's true so we shade between -2 and 0.

-------o=====o------------
-4 -3 -2 -1  0  1  2  3  4

We test a value for x right of 0, say 1 in this:

{{{-(2x)/(x+2) > 0}}}
{{{(-2(1))/(1)+2) > 0}}}
{{{(-2)/(3) > 0}}}
{{{-2/3 > 0}}}

This is false so we do not shade to the right of 0.

Answer {x | -2 < x < 0} in set-builder notation 
       or (-2,0) in interval notation.

Edwin</pre>