How do you solve #|4y-4|>=11#?

1 Answer
Alan P.
Jun 18, 2018

#x>=15/4# or #x <=-7/4#

Explanation:

Consider the two cases for #(4y-4)# separately:
#{: ("if " (4x-4) >= 0,color(white)("xxx"),"if " (4x-4) < 0), ("then " abs(4x-4)=4x-4,,"then " abs(4x-4) =4-4x), ("and "abs(4x-4) >= 11,,"and " abs(4x-4) >= 11), (" "rarr 4x-4 >= 11,," "rarr 4-4x >= 11), (" "rarr 4x >= 15,," "rarr -7 >= 4x), (" "rarr x >=15/4,," "rarr -7/4 >=x) :}#

So we have
either #x>=15/4# or #x <=-7/4#

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