4(c+3)+4+2c=10+6c+6

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Solution for 4(c+3)+4+2c=10+6c+6 equation: 


Simplifying
4(c + 3) + 4 + 2c = 10 + 6c + 6

Reorder the terms:
4(3 + c) + 4 + 2c = 10 + 6c + 6
(3 * 4 + c * 4) + 4 + 2c = 10 + 6c + 6
(12 + 4c) + 4 + 2c = 10 + 6c + 6

Reorder the terms:
12 + 4 + 4c + 2c = 10 + 6c + 6

Combine like terms: 12 + 4 = 16
16 + 4c + 2c = 10 + 6c + 6

Combine like terms: 4c + 2c = 6c
16 + 6c = 10 + 6c + 6

Reorder the terms:
16 + 6c = 10 + 6 + 6c

Combine like terms: 10 + 6 = 16
16 + 6c = 16 + 6c

Add '-16' to each side of the equation.
16 + -16 + 6c = 16 + -16 + 6c

Combine like terms: 16 + -16 = 0
0 + 6c = 16 + -16 + 6c
6c = 16 + -16 + 6c

Combine like terms: 16 + -16 = 0
6c = 0 + 6c
6c = 6c

Add '-6c' to each side of the equation.
6c + -6c = 6c + -6c

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

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

Solving
0 = 0

Couldn't find a variable to solve for.

This equation is an identity, all real numbers are solutions.

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