3c-[c+(3c+1)+(c+3)]=6

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Solution for 3c-[c+(3c+1)+(c+3)]=6 equation:


Simplifying
3c + -1[c + (3c + 1) + (c + 3)] = 6

Reorder the terms:
3c + -1[c + (1 + 3c) + (c + 3)] = 6

Remove parenthesis around (1 + 3c)
3c + -1[c + 1 + 3c + (c + 3)] = 6

Reorder the terms:
3c + -1[c + 1 + 3c + (3 + c)] = 6

Remove parenthesis around (3 + c)
3c + -1[c + 1 + 3c + 3 + c] = 6

Reorder the terms:
3c + -1[1 + 3 + c + 3c + c] = 6

Combine like terms: 1 + 3 = 4
3c + -1[4 + c + 3c + c] = 6

Combine like terms: c + 3c = 4c
3c + -1[4 + 4c + c] = 6

Combine like terms: 4c + c = 5c
3c + -1[4 + 5c] = 6
3c + [4 * -1 + 5c * -1] = 6
3c + [-4 + -5c] = 6

Reorder the terms:
-4 + 3c + -5c = 6

Combine like terms: 3c + -5c = -2c
-4 + -2c = 6

Solving
-4 + -2c = 6

Solving for variable 'c'.

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

Add '4' to each side of the equation.
-4 + 4 + -2c = 6 + 4

Combine like terms: -4 + 4 = 0
0 + -2c = 6 + 4
-2c = 6 + 4

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

Divide each side by '-2'.
c = -5

Simplifying
c = -5

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