How do you solve #c+(4-3c)-2=0#?

Question

17107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-06T02:30:25

See a solution process below:

First, rewrite the left side of the equation and group and combine like terms:

#c + (4 - 3c) - 2 = 0#

#c + 4 - 3c - 2 = 0#

# - 3c + c + 4 - 2 = 0#

#-3c + 1c + 4 - 2 = 0#

#(-3 + 1)c + (4 - 2) = 0#

#-2c + 2 = 0#

Next, subtract #color(red)(2)# from each side of the equation to isolate the #c# term while keeping the equation balanced:

#-2c + 2 - color(red)(2) = 0 - color(red)(2)#

#-2c + 0 = -2#

#-2c = -2#

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

#(-2c)/color(red)(-2) = (-2)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))c)/cancel(color(red)(-2)) = 1#

#c = 1#