How do you solve and write the following in interval notation: #12<7/2a+(-2)#?

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Answer 1 · 2016-08-12T21:01:01

#(4,oo)#

As we have #12<7/2a+(-2)#, this is equivalent to

#12-(-2)<7/2a# or

#14<7/2a# or

#7/2a>14# or

#a>(14×2)/7# or

#a>4# which can be written in interval form as

#(4,oo)#