Respuesta :
divide both sides by 2
|x-3|>4 and |x-3|<-4
ok so |x-3|<-4 is false, since |x|≥0 always
so we have
|x-3|>4
now assume
x-3>4 and
x-3<-4
x-3>
add 3
x>7
x-3<-4
add 3
x<-1
so
-1>x and x>7
so basically it is all numbers from -∞ to +∞ except from -1 to 7
in interval notaion
(-∞,-1)U(7,∞)
S={x|x<-1 or x>7}
|x-3|>4 and |x-3|<-4
ok so |x-3|<-4 is false, since |x|≥0 always
so we have
|x-3|>4
now assume
x-3>4 and
x-3<-4
x-3>
add 3
x>7
x-3<-4
add 3
x<-1
so
-1>x and x>7
so basically it is all numbers from -∞ to +∞ except from -1 to 7
in interval notaion
(-∞,-1)U(7,∞)
S={x|x<-1 or x>7}