How do you solve # abs(8 - 6c ) = 1#?

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Answer 1 · 2017-02-19T11:42:46

#7/6<=c<=3/2#

The given equation is equivalent to:

#-1<=8-6c<=1#

Let's subtract 8 and get:

#-9<=-6c<=-7#

then change signs and inequality:

#7<=6c<=9#

and divide all terms by 6:

#7/6<=c<=cancel9^3/cancel6^2#

#7/6<=c<=3/2#