(c+3)-2c-(1+3c)=2

Solution for (c+3)-2c-(1+3c)=2 equation:

Simplifying
(c + 3) + -2c + -1(1 + 3c) = 2

Reorder the terms:
(3 + c) + -2c + -1(1 + 3c) = 2

Remove parenthesis around (3 + c)
3 + c + -2c + -1(1 + 3c) = 2
3 + c + -2c + (1 * -1 + 3c * -1) = 2
3 + c + -2c + (-1 + -3c) = 2

Reorder the terms:
3 + -1 + c + -2c + -3c = 2

Combine like terms: 3 + -1 = 2
2 + c + -2c + -3c = 2

Combine like terms: c + -2c = -1c
2 + -1c + -3c = 2

Combine like terms: -1c + -3c = -4c
2 + -4c = 2

Add '-2' to each side of the equation.
2 + -2 + -4c = 2 + -2

Combine like terms: 2 + -2 = 0
0 + -4c = 2 + -2
-4c = 2 + -2

Combine like terms: 2 + -2 = 0
-4c = 0

Solving
-4c = 0

Solving for variable 'c'.

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

Divide each side by '-4'.
c = 0

Simplifying
c = 0