3(4-2(y+2))=2y-4(1+2(1+y))

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Solution for 3(4-2(y+2))=2y-4(1+2(1+y)) equation:


Simplifying
3(4 + -2(y + 2)) = 2y + -4(1 + 2(1 + y))

Reorder the terms:
3(4 + -2(2 + y)) = 2y + -4(1 + 2(1 + y))
3(4 + (2 * -2 + y * -2)) = 2y + -4(1 + 2(1 + y))
3(4 + (-4 + -2y)) = 2y + -4(1 + 2(1 + y))

Combine like terms: 4 + -4 = 0
3(0 + -2y) = 2y + -4(1 + 2(1 + y))
3(-2y) = 2y + -4(1 + 2(1 + y))

Remove parenthesis around (-2y)
3 * -2y = 2y + -4(1 + 2(1 + y))

Multiply 3 * -2
-6y = 2y + -4(1 + 2(1 + y))
-6y = 2y + -4(1 + (1 * 2 + y * 2))
-6y = 2y + -4(1 + (2 + 2y))

Combine like terms: 1 + 2 = 3
-6y = 2y + -4(3 + 2y)
-6y = 2y + (3 * -4 + 2y * -4)
-6y = 2y + (-12 + -8y)

Reorder the terms:
-6y = -12 + 2y + -8y

Combine like terms: 2y + -8y = -6y
-6y = -12 + -6y

Add '6y' to each side of the equation.
-6y + 6y = -12 + -6y + 6y

Combine like terms: -6y + 6y = 0
0 = -12 + -6y + 6y

Combine like terms: -6y + 6y = 0
0 = -12 + 0
0 = -12

Solving
0 = -12

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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