How do you solve for y in # -x + 2y = 6#?

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Answer 1 · 2018-03-04T14:32:23

See a solution process below:

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

#color(red)(x) - x + 2y = color(red)(x) + 6#

#0 + 2y = x + 6#

#2y = x + 6#

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

#(2y)/color(red)(2) = (x + 6)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = x/color(red)(2) + 6/color(red)(2)#

#y = 1/2x + 3#