Question

|x+3|+x/x+2>1 solve the following inequality ​

Answer 1

Case-1

when x+3≥0ORx≥−3

|x+3|=x+3

so, x+3+xx+2>1

2x+3x+2−1>0

x+1x+2>0 OR (x+1)(x+2)>0 [∵abanda×bhavesamesign]

⇒xϵ(−∞,−2)∪(−1,∞)

but this solution is valid when x≥−3

means solution from case -1 is :

(x≥−3) and xϵ(−∞,−2)∪(−1,∞)

or xϵ[−3,−2)∪(−1,∞).......(1)

Case-2

when x+3<0ORx<−3

|x+3|=−(x+3)

so, −(x+3)+xx+2>1

on solving this we get ,

x+5x+2<0

or (x+5)(x+2)<0 [ ∵abanda×bhavesamesign ]

⇒xϵ(−5,−2)

but this solution is valid when x<−3

it means solution from case -2 is :

(x<−3) and xϵ(−5,−2)

or xϵ(−5,−3).......(2)

Hence , final answer is (Solution from 1st case) ∪ (Solution from 2nd case)

so take union of solution of 1st and 2nd equation

[−3,−2)∪(−1,∞)∪(−5,−3)

we get final answer as,

xϵ(−5,−2)∪(−1,∞)