How do you solve and write the following in interval notation: #(2/3)x – (4/3) ≥ (1/4)#?

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Answer 1 · 2016-08-19T19:41:20

#x>=9/4 -> [9/4,oo)#

[ means inclusive or closed; whilst ) means exclusive or open
So the interval includes #9/4# but excludes #oo#

Lets get rid of the fractions!

Change the denominators to 12

#(2/3xx4/4)x-(4/3xx4/4)>=(1/4xx3/3)#

#8/12x-16/12>=3/12#

It is also true that

#8x-16>=3#

Add 16 to both sides

#8x>=18#

divide both sides by 8

#x>=18/8 -=9/4#