4(z+4)=3(z-2)+z

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Solution for 4(z+4)=3(z-2)+z equation: 


Simplifying
4(z + 4) = 3(z + -2) + z

Reorder the terms:
4(4 + z) = 3(z + -2) + z
(4 * 4 + z * 4) = 3(z + -2) + z
(16 + 4z) = 3(z + -2) + z

Reorder the terms:
16 + 4z = 3(-2 + z) + z
16 + 4z = (-2 * 3 + z * 3) + z
16 + 4z = (-6 + 3z) + z

Combine like terms: 3z + z = 4z
16 + 4z = -6 + 4z

Add '-4z' to each side of the equation.
16 + 4z + -4z = -6 + 4z + -4z

Combine like terms: 4z + -4z = 0
16 + 0 = -6 + 4z + -4z
16 = -6 + 4z + -4z

Combine like terms: 4z + -4z = 0
16 = -6 + 0
16 = -6

Solving
16 = -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.

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