How do you solve #4+ 2y - 11= 4y + 15#?

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Answer 1 · 2017-12-16T03:03:38

See a solution process below:

First, subtract #color(red)(2y)# and #color(blue)(15)# from each side of the equation to isolate the #y# term while keeping the equation balanced:

#4 + 2y - color(red)(2y) - 11 - color(blue)(15) = 4y - color(red)(2y) + 15 - color(blue)(15)#

#4 + 0 - 11 - 15 = (4 - color(red)(2))y + 0#

#-22 = 2y#

Now, divide each side of the equation by #color(red)(2)# to solve for #y# while keeping the equation balanced:

#-22/color(red)(2) = (2y)/color(red)(2)#

#-11 = (color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2))#

#-11 = y#

#y = -11#