28-(3c+4)=2(c+12)+c

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Solution for 28-(3c+4)=2(c+12)+c equation:


Simplifying
28 + -1(3c + 4) = 2(c + 12) + c

Reorder the terms:
28 + -1(4 + 3c) = 2(c + 12) + c
28 + (4 * -1 + 3c * -1) = 2(c + 12) + c
28 + (-4 + -3c) = 2(c + 12) + c

Combine like terms: 28 + -4 = 24
24 + -3c = 2(c + 12) + c

Reorder the terms:
24 + -3c = 2(12 + c) + c
24 + -3c = (12 * 2 + c * 2) + c
24 + -3c = (24 + 2c) + c

Combine like terms: 2c + c = 3c
24 + -3c = 24 + 3c

Add '-24' to each side of the equation.
24 + -24 + -3c = 24 + -24 + 3c

Combine like terms: 24 + -24 = 0
0 + -3c = 24 + -24 + 3c
-3c = 24 + -24 + 3c

Combine like terms: 24 + -24 = 0
-3c = 0 + 3c
-3c = 3c

Solving
-3c = 3c

Solving for variable 'c'.

Move all terms containing c to the left, all other terms to the right.

Add '-3c' to each side of the equation.
-3c + -3c = 3c + -3c

Combine like terms: -3c + -3c = -6c
-6c = 3c + -3c

Combine like terms: 3c + -3c = 0
-6c = 0

Divide each side by '-6'.
c = 0

Simplifying
c = 0

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