4(3y-2)+28=4(5+2y)+4y

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Solution for 4(3y-2)+28=4(5+2y)+4y equation: 


Simplifying
4(3y + -2) + 28 = 4(5 + 2y) + 4y

Reorder the terms:
4(-2 + 3y) + 28 = 4(5 + 2y) + 4y
(-2 * 4 + 3y * 4) + 28 = 4(5 + 2y) + 4y
(-8 + 12y) + 28 = 4(5 + 2y) + 4y

Reorder the terms:
-8 + 28 + 12y = 4(5 + 2y) + 4y

Combine like terms: -8 + 28 = 20
20 + 12y = 4(5 + 2y) + 4y
20 + 12y = (5 * 4 + 2y * 4) + 4y
20 + 12y = (20 + 8y) + 4y

Combine like terms: 8y + 4y = 12y
20 + 12y = 20 + 12y

Add '-20' to each side of the equation.
20 + -20 + 12y = 20 + -20 + 12y

Combine like terms: 20 + -20 = 0
0 + 12y = 20 + -20 + 12y
12y = 20 + -20 + 12y

Combine like terms: 20 + -20 = 0
12y = 0 + 12y
12y = 12y

Add '-12y' to each side of the equation.
12y + -12y = 12y + -12y

Combine like terms: 12y + -12y = 0
0 = 12y + -12y

Combine like terms: 12y + -12y = 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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