How do you solve #|8r - 3| <67#?

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Answer 1 · 2016-12-06T23:36:17

The solution to this problem is the interval:

#-8 < r < 35/4#

Because this is an absolute value inequality we must convert this to an inequality set and solve the inequality set while keeping the set balanced:

#-67 < 8r - 3 < 67#

#-67 + 3 < 8r - 3 + 3 < 67 + 3#

#-64 < 8r < 70#

#-64/8 < (8r)/8 < 70/8#

#-8 < r < (2/2) (35/4)#

#-8 < r < 35/4#