How do you solve for b in #d = 3a + 3b #?

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Answer 1 · 2018-04-23T21:40:48

#=>b = d/3 - a #

We are given

#d = 3a + 3b#

Subtract #3a# from both sides

#d - 3a = 3a + 3b - 3a#

#d - 3a = cancel(3a)+3b cancel(-3a)#

#d - 3a = 3b#

Divide both sides by #3#

#(d-3a)/3 = (3b)/3#

#d/3 - (cancel(3)a)/cancel3 = (cancel(3)b)/cancel3#

#d/3 - a = b#