How do you solve #\frac { 1} { 3} ( 9c - 18- 3c ) = \frac { 1} { 4} ( 12c - 28)#?

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Answer 1 · 2016-11-17T14:41:53

#c = 1#

Step 1) Expand the terms in parenthesis:

#(9c)/3 - 18/3 - (3c)/3 = (12c)/4 - 28/4#

#3c - 6 - c = 3c - 7#

Step 2) Combine like terms:

#3c - c - 6 = 3c - 7#

#(3 - 1)c - 6 = 3c - 7#

#2c - 6 = 3c - 7#

Step 3) Solve for #c# while keeping the equation balanced:

#2c - 6 - 2c + 7 = 3c - 7 - 2c + 7#

#-6 + 7 = 3c - 2c#

#1 = c# or #c = 1#