How do you solve #12x=2y-6# for #y#?

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Answer 1 · 2017-04-07T01:20:47

See the entire solution process below:

First, add #color(red)(6)# to each side of the equation to isolate the #y# term while keeping the equation balanced:

#12x + color(red)(6) = 2y - 6 + color(red)(6)#

#12x + 6 = 2y - 0#

#12x + 6 = 2y#

Next, multiply each side of the equation by #color(red)(1/2)# to solve for #y# while keeping the equation balanced:

#color(red)(1/2)(12x + 6) = color(red)(1/2) xx 2y#

#(color(red)(1/2) xx 12x) + (color(red)(1/2) xx 6) = color(red)(1/color(black)(cancel(color(red)(2)))) xx color(red)(cancel(color(black)(2)))y#

#6x + 3 = y#

#y = 6x + 3#