How do you solve and write the following in interval notation: #abs(2x+9) ≤ 3#?

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Answer 1 · 2017-02-22T16:05:53

The solution is #x in [-6,-9/2] uu[-9/2,-3]#

There are 2 solutions to this equation

#2x+9<=3# and #-2x-9<=3#

#2x<=-6# and #2x>=-12#

#x<=-3# and #x>=-6#

We need to solve

#2x+9=0#

#x=-9/2#

Therefore, the solutions are

#x in [-6,-9/2] uu[-9/2,-3]#