3(z+4)-2=2(2z+5)-z

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Solution for 3(z+4)-2=2(2z+5)-z equation:


Simplifying
3(z + 4) + -2 = 2(2z + 5) + -1z

Reorder the terms:
3(4 + z) + -2 = 2(2z + 5) + -1z
(4 * 3 + z * 3) + -2 = 2(2z + 5) + -1z
(12 + 3z) + -2 = 2(2z + 5) + -1z

Reorder the terms:
12 + -2 + 3z = 2(2z + 5) + -1z

Combine like terms: 12 + -2 = 10
10 + 3z = 2(2z + 5) + -1z

Reorder the terms:
10 + 3z = 2(5 + 2z) + -1z
10 + 3z = (5 * 2 + 2z * 2) + -1z
10 + 3z = (10 + 4z) + -1z

Combine like terms: 4z + -1z = 3z
10 + 3z = 10 + 3z

Add '-10' to each side of the equation.
10 + -10 + 3z = 10 + -10 + 3z

Combine like terms: 10 + -10 = 0
0 + 3z = 10 + -10 + 3z
3z = 10 + -10 + 3z

Combine like terms: 10 + -10 = 0
3z = 0 + 3z
3z = 3z

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

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

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

Solving
0 = 0

Couldn't find a variable to solve for.

This equation is an identity, all real numbers are solutions.

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