How do you solve #abs(11+6x)<=47#?

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Answer 1 · 2018-08-02T23:54:39

#-29/3<=x<=6#

With this absolute value inequality, we have two scenarios:

#6x+11<=47# and #6x+11>=-47#

From both, let's subtract #11# from both sides to get

#6x<=36# and #6x>= -58#

Next, divide both sides by #6# to get

#x<=6# and #x>= -29/3#, which can be rewritten as

#-29/3<=x<=6#

Hope this helps!