How do you solve #3(2-b)<10-3(b-6)#?

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Answer 1 · 2017-09-02T11:43:14

The inequality is satisfied for all real b.
#{ b in RR }#

After rearranging:

#3( 2 - b ) < 10 - 3( b - 6 ) => -22 - 3b < -3b#

Adding #3b# to both sides :

# -22 < -3b + 3b#

for any real #b#

#-3b + 3b = 0#

# -22 < 0 # is always true.

Hope this helps.