Question

( |x+3| geq 10 )
( 8 b quad|x| geqslant a Rightarrow x geqslant a ) or ( x leqslant-a )
( therefore x+3 geqslant 10 )
or ( x+3 leq-10 )
( Rightarrow x geqslant 10-3 )
of ( x leq-10-3 )
( x geqslant 7 )
7
( -13 quad j quad 7 )
( therefore x in(-infty,-13] cup[7, infty) )

# | 1 |x+3210 then xe (a) (-13,7) (b) (-13, 7) (c) (-00,-13) [7,00) (d) (-0, -13] [7,)

Solution