Question

( frac{|x+3|+x}{x+2}>1 )
( |x+3|+2>x+2 )
( |x+3|>2 )
( x+3>pm 2 )
( x+3>2 quad ) (0) ( quad x+3>-2 )
( x>-1 )
( x in(-5,-1) cup(-1,2) )

# > 1 is 1 x + 3+x Solution of ! x + 2 (A) x € (-5, -2] V (-1,0) (B) x € (-5, -1) U (-1,00) (C) x € (-5, -2) (-1,00) (D) x € (-5, -2] U (0,00)

Solution