How do you solve and write the following in interval notation: #-5(2x+3) >3#?

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Answer 1 · 2016-07-20T02:52:33

#x < -9/5#

#(-oo,-9/5)#

Distribute the #-5# across the parenthesis.

#-5(2x) + -5(3) > 3#

#-10x - 15 > 3#

Add #15# to both sides and isolate #x#.

#-10x cancel(-15+15) > 3+15#

#-10x > 18#

Divide by #-10# and remember to switch the sign.

#(cancel(-10)x)/(cancel(-10)) < 18/-10#

#x < -18/10#

Simplify #-18/10#.

#x < -9/5#

Now write in interval notation:

#(-oo,-9/5)#