How do you solve #6x + 2= x + 27#?

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Answer 1 · 2017-07-22T01:28:38

See a solution process below:

Step 1) Subtract #color(red)(2)# and #color(blue)(x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#6x - color(blue)(x) + 2 - color(red)(2) = x - color(blue)(x) + 27 - color(red)(2)#

#6x - color(blue)(1x) + 0 = 0 + 25#

#(6 - color(blue)(1))x) = 25#

#5x = 25#

Step 2) Divide each side of the equation by #color(red)(5)# to solve for #x# while keeping the equation balanced:

#(5x)/color(red)(5) = 25/color(red)(5)#

#(color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) = 5#

#x = 5#