How do you solve the compound inequalities $6b < 42$ or $4b + 12 > 8$ ?

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Answer 1 · 2015-05-22T13:10:16

$6b<42$
$rarr b<7$

$4b+12>8$
$rarr 4b> -4$
$rarr b> -1$

Since all Real numbers are either $<7$ or $> -1$
the solution to the given compound inequalities is
all Real numbers

How do you solve the compound inequalities 6b < 426b < 42 or 4b + 12 > 84b + 12 > 8?