How do you solve #10 >29 - 3b#?

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Answer 1 · 2016-06-22T16:12:48

#19/3 < b# or #6 1/3 < b#

To solve #10>29-3b#

Begin by isolating the term with the variable by additive inverse

#10-29>cancel(29) cancel(-29)-3b#

#-19>(-3b)#

Isolate the variable by multiplicative inverse

#-19/-3 > (cancel(-3)b)/cancel(-3)#

When you divide by the negative you must flip the inequality sign

#19/3 < b# or #6 1/3 < b#