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

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

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

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

Reorder the terms:
3 + 3z = 2(-3 + z) + z
3 + 3z = (-3 * 2 + z * 2) + z
3 + 3z = (-6 + 2z) + z

Combine like terms: 2z + z = 3z
3 + 3z = -6 + 3z

Add '-3z' to each side of the equation.
3 + 3z + -3z = -6 + 3z + -3z

Combine like terms: 3z + -3z = 0
3 + 0 = -6 + 3z + -3z
3 = -6 + 3z + -3z

Combine like terms: 3z + -3z = 0
3 = -6 + 0
3 = -6

Solving
3 = -6

Couldn't find a variable to solve for.

This equation is invalid, the left and right sides are not equal, therefore there is no solution.