3[3c-1]=4[3c+2]

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Solution for 3[3c-1]=4[3c+2] equation:


Simplifying
3[3c + -1] = 4[3c + 2]

Reorder the terms:
3[-1 + 3c] = 4[3c + 2]
[-1 * 3 + 3c * 3] = 4[3c + 2]
[-3 + 9c] = 4[3c + 2]

Reorder the terms:
-3 + 9c = 4[2 + 3c]
-3 + 9c = [2 * 4 + 3c * 4]
-3 + 9c = [8 + 12c]

Solving
-3 + 9c = 8 + 12c

Solving for variable 'c'.

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

Add '-12c' to each side of the equation.
-3 + 9c + -12c = 8 + 12c + -12c

Combine like terms: 9c + -12c = -3c
-3 + -3c = 8 + 12c + -12c

Combine like terms: 12c + -12c = 0
-3 + -3c = 8 + 0
-3 + -3c = 8

Add '3' to each side of the equation.
-3 + 3 + -3c = 8 + 3

Combine like terms: -3 + 3 = 0
0 + -3c = 8 + 3
-3c = 8 + 3

Combine like terms: 8 + 3 = 11
-3c = 11

Divide each side by '-3'.
c = -3.666666667

Simplifying
c = -3.666666667

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