3(x+y+z)=3x+3y+3z

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Solution for 3(x+y+z)=3x+3y+3z equation:


Simplifying
3(x + y + z) = 3x + 3y + 3z
(x * 3 + y * 3 + z * 3) = 3x + 3y + 3z
(3x + 3y + 3z) = 3x + 3y + 3z

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

Reorder the terms:
3x + -3x + 3y + 3z = 3x + 3y + -3x + 3z

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

Reorder the terms:
3y + 3z = 3x + -3x + 3y + 3z

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

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

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

Combine like terms: 3y + -3y = 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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