How do you solve #9y - 27= 14y + 3#?

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Answer 1 · 2017-05-25T23:41:07

See a solution process below:

First, subtract #color(red)(9y)# and #color(blue)(3)# from each side of the equation to isolate the #y# term while keeping the equation balanced:

#-color(red)(9y) + 9y - 27 - color(blue)(3) = -color(red)(9y) + 14y + 3 - color(blue)(3)#

#0 - 30 = (-color(red)(9) + 14)y + 0#

#-30 = 5y#

Now, divide each side of the equation by #color(red)(5)# to solve for #y# while keeping the equation balanced:

#-30/color(red)(5) = (5y)/color(red)(5)#

#-6 = (color(red)(cancel(color(black)(5)))y)/cancel(color(red)(5))#

#-6 = y#

#y = -6#