2n+[3(n+1)]=4(n-1)

Solution for 2n+[3(n+1)]=4(n-1) equation:

Simplifying
2n + [3(n + 1)] = 4(n + -1)

Reorder the terms:
2n + [3(1 + n)] = 4(n + -1)
2n + [(1 * 3 + n * 3)] = 4(n + -1)
2n + [(3 + 3n)] = 4(n + -1)
2n + [3 + 3n] = 4(n + -1)

Remove brackets around [3 + 3n]
2n + 3 + 3n = 4(n + -1)

Reorder the terms:
3 + 2n + 3n = 4(n + -1)

Combine like terms: 2n + 3n = 5n
3 + 5n = 4(n + -1)

Reorder the terms:
3 + 5n = 4(-1 + n)
3 + 5n = (-1 * 4 + n * 4)
3 + 5n = (-4 + 4n)

Solving
3 + 5n = -4 + 4n

Solving for variable 'n'.

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

Add '-4n' to each side of the equation.
3 + 5n + -4n = -4 + 4n + -4n

Combine like terms: 5n + -4n = 1n
3 + 1n = -4 + 4n + -4n

Combine like terms: 4n + -4n = 0
3 + 1n = -4 + 0
3 + 1n = -4

Add '-3' to each side of the equation.
3 + -3 + 1n = -4 + -3

Combine like terms: 3 + -3 = 0
0 + 1n = -4 + -3
1n = -4 + -3

Combine like terms: -4 + -3 = -7
1n = -7

Divide each side by '1'.
n = -7

Simplifying
n = -7