How do you solve #3b(9b-27)=0#?

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Answer 1 · 2017-04-18T00:07:40

See the entire solution process below:

We need to solve each term for #0# to determine the complete solution of the problem:

Solution 1)

#3b = 0#

#(3b)/color(red)(3) = 0/color(red)(3)#

#(color(red)(cancel(color(black)(3)))b)/cancel(color(red)(3)) = 0#

#b = 0#

Solution 2)

#9b - 27 = 0#

#9b - 27 + color(red)(27) = 0 + color(red)(27)#

#9b - 0 = 27#

#9b = 27#

#(9b)/color(red)(9) = 27/color(red)(9)#

#(color(red)(cancel(color(black)(9)))b)/cancel(color(red)(9)) = 3#

#b = 3#

The solution is: #b = 0# and #b = 3#