How do you solve #2x+7y=z# for #y#?

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Answer 1 · 2017-05-14T03:44:44

See a solution process below:

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

#-color(red)(2x) + 2x + 7y = z - color(red)(2x)#

#0 + 7y = z - 2x#

#7y = z - 2x#

Now, divide each side of the equation by #color(red)(7)# to solve for #y# while keeping the equation balanced:

#(7y)/color(red)(7) = (z - 2x)/color(red)(7)#

#(color(red)(cancel(color(black)(7)))y)/cancel(color(red)(7)) = (z - 2x)/7#

#y = (z - 2x)/7#

Or

#y = z/7 - (2x)/7#