Question 27750
a/a+2 > 0,
First of  all (a+2) the denominator cannot be zero as division by 0 is not  allowed(not defined in fact)
So (a+2) not 0 means a not (-2)
Therefore  either a>-2 or                          a<-2
That is either (a+2) >0 or (a+2)<0
Case 1. Let (a+2) >0
Given a/a+2 > 0
Multiplying both the sides of the inequality by (a+2)>0
We have  [a/(a+2)]X(a+2) > 0X(a+2)   (multiplication by a positive qty retains the inequalityand therefore > remains >)
This implies  a>0  ( as zero multiplied by anything is zero)
This situation a>0 has arisen out of the hypothesis a>(-2)
Judging these two the verdict is  
a>0   
(as asnything to the right of zero is definitely to the right of (-2)
Case 2. Let (a+2)<0
Given a/a+2 < 0
Multiplying both the sides of the inequality by (a+2)<0
We have  [a/(a+2)]X(a+2) < 0X(a+2)   (multiplication by a positive qty alters the inequality and therefore > becomes <)
This implies  a<0  ( as zero multiplied by anything is zero)
This situation a<0 has arisen out of the hypothesis a<(-2)
Judging these two the verdict is  
a<(-2)  
(as asnything to the left of (-2) is definitely to the left of 0
Answer: a<(-2) and a>0
That is  a   can take any  value to the right of (minus infinity) to the       left of (-2)  and a can take  any value to the right of zero
That is a  cannot be -2, a cannot be any value to the right  of -2 uptill zero(including  zero
Just draw a horizontal line and mark these things. You will know in a jiffy. I am yet to understand the art of decoding  graph on your answer page.If I try to transfer  my  graphs from my applicastions, nothing turns  out prop erly according to my liking. My phone and internet charges become  astronomical