How do you solve and write the following in interval notation: #7 ≤ x + 4# or #x - 1 ≥ 5#?

1 Answer
NickTheTurtle
Apr 25, 2017

First, solve each inequality separately:

#7<=x+4#
#:.7-4<=x+4-4#
#:.3<=x#
#:.x>=3#
#x# can be any value between #3# (including) and #oo#. Therefore, #x in [3,oo)#. Brackets mean that it is included, while parentheses mean that it is not included. #oo# is never included for interval notations.

#x-1>=5#
#:.x-1+1>=5+1#
#:.x>=6#
#x# can be any value between #6# (including) and #oo#. Therefore, #x in [6,oo)#.

We can combine these two to get #x in [3,oo) nn [6,oo)#. This is equivalent to #x in [6,oo)#.

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