How do you solve for x in #2x + 3y = 6#?

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Answer 1 · 2018-05-24T12:02:58

See a solution process below:

First, subtract #color(red)(3y)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x + 3y - color(red)(3y) = 6 - color(red)(3y)#

#2x + 0 = 6 - 3y#

#2x = 6 - 3y#

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

#(2x)/color(red)(2) = (6 - 3y)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = (6 - 3y)/2#

#x = (6 - 3y)/2#

Or

#x = 6/2 - (3y)/2#

#x = 3 - 3/2y#