Question 23382
I'm assuming that you mean {{{x/(x+2) >-1}}}.


If so, then I would begin by setting the inequality to zero.  DO NOT MULTIPLY BOTH SIDES OF THE INEQUALITY BY (X+2)!!

{{{x/(x+2) +1>0}}}.


Next, find a common denominator, which is x+2:
{{{(x/(x+2))+1*((x+2)/(x+2)) >0}}}
{{{(x + x+2)/(x+2) >0}}}

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


Now, you have two choices for method of solving the inequality.  You can use some algebra explanations to solve it, or you can use a graphing calculator (or the algebra.com calculator!!).  I will choose the latter method.  You need to graph {{{y= (2x+2)/(x+2) }}}.   Before drawing the graph, notice that there is one place that the numerator equals zero, and that is at x= -1.  There is one value of x that would make the denominator zero, which is NOT allowed, and that would be x= -2.  This tells me that there is a ROOT at x = -1 and an ASYMPTOTE at x= -2.  Now, draw the graph with this in mind:

{{{graph(500,500, -10,10,-10,10, (2x+2)/(x+2) )}}}


Now, since the problem is to solve {{{( 2x + 2)/(x+2) >0 }}}, which is solving a problem that is "GREATER THAN", you need to find all values of x for which the graph is ABOVE the x-axis.  And do NOT include the endpoints.  
Notice that the graph is above the x-axis from -infinity to -2, and also from -1 to infinity.  
This means that the solution is x<-2 or x>1.  
In interval notation this will be:(-inf, -2) U (-1, inf).


R^2 at SCC