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

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


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

Reorder the terms:
5(-4 + z) + -3z = 4(z + -5) + -2z
(-4 * 5 + z * 5) + -3z = 4(z + -5) + -2z
(-20 + 5z) + -3z = 4(z + -5) + -2z

Combine like terms: 5z + -3z = 2z
-20 + 2z = 4(z + -5) + -2z

Reorder the terms:
-20 + 2z = 4(-5 + z) + -2z
-20 + 2z = (-5 * 4 + z * 4) + -2z
-20 + 2z = (-20 + 4z) + -2z

Combine like terms: 4z + -2z = 2z
-20 + 2z = -20 + 2z

Add '20' to each side of the equation.
-20 + 20 + 2z = -20 + 20 + 2z

Combine like terms: -20 + 20 = 0
0 + 2z = -20 + 20 + 2z
2z = -20 + 20 + 2z

Combine like terms: -20 + 20 = 0
2z = 0 + 2z
2z = 2z

Add '-2z' to each side of the equation.
2z + -2z = 2z + -2z

Combine like terms: 2z + -2z = 0
0 = 2z + -2z

Combine like terms: 2z + -2z = 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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