How do you solve #2c+13=19#?

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Answer 1 · 2017-06-12T20:44:41

See a solution process below:

Step 1) Subtract #color(red)(13)# from each side of the equation to isolate the #c# term while keeping the equation balanced:

#2c + 13 - color(red)(13) = 19 - color(red)(13)#

#2c + 0 = 6#

#2c = 6#

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

#(2c)/color(red)(2) = 6/color(red)(2)#

#(color(red)(cancel(color(black)(2)))c)/cancel(color(red)(2)) = 3#

#c = 3#

Answer 2 · 2017-06-12T20:47:35

#c=3#

We need to move all terms that don't contain #c# to the right side and solve.

#2c + 13 = 19#

-subtract #13# from each side

#2c cancel(+13 - 13) =19 -13#

#2c=6#

-divide each side by #2#

#c=3#