How do you solve #6- 2y = 4y + 8#?

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Answer 1 · 2018-01-15T02:58:10

See a solution process below:

First, add #color(red)(2y)# and subtract #color(blue)(8)# from each side of the equation to isolate the #y# term while keeping the equation balanced:

#6 - color(blue)(8) - 2y + color(red)(2y) = 4y + color(red)(2y) + 8 - color(blue)(8)#

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

#-2 = 6y#

Now, divide each side of the equation by #color(red)(6)# to solve for #y# while keeping the equation balanced:

#-2/color(red)(6) = (6y)/color(red)(6)#

#-1/3 = (color(red)(cancel(color(black)(6)))y)/cancel(color(red)(6))#

#-1/3 = y#

#y = -1/3#

Answer 2 · 2018-01-15T02:58:15

#y = -1/3#

#6-2y = 4y + 8#

The first thing we need to do is put all the variables on one side of the equation and numbers on the other:
#-2 = 6y#

Now we divide both sides by #6# so that we can find #y#:
#-1/3 = y#

Put #y# on the left side of the equation so that it is clearer:
#y = -1/3#