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Question 27750: This is a rational inequality problem, I need help with.
a/a+2 > 0, any suggestions, please.
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! 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
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