How do you solve #27=3c-3(6-2c)#?

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Answer 1 · 2018-04-16T03:13:43

#c = 5#

#27 = 3c - 3(6-2c)#

Distribute:
#27 = 3c - 18 + 6c#

Combine #3c# and #6c#:
#27 = 9c - 18#

Add #18# to both sides of the equation:
#27 quadcolor(red)(+quad18) = 9c - 18 quadcolor(red)(+quad18)#

#45 = 9c#

Divide both sides by #9#:
#45/color(red)9 = (9c)/color(red)9#

#5 = c#

Therefore, #c = 5#.

Hope this helps!