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

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

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

Reorder the terms:
3(-2 + y) + 1 = 2(y + -3) + (y + 1)
(-2 * 3 + y * 3) + 1 = 2(y + -3) + (y + 1)
(-6 + 3y) + 1 = 2(y + -3) + (y + 1)

Reorder the terms:
-6 + 1 + 3y = 2(y + -3) + (y + 1)

Combine like terms: -6 + 1 = -5
-5 + 3y = 2(y + -3) + (y + 1)

Reorder the terms:
-5 + 3y = 2(-3 + y) + (y + 1)
-5 + 3y = (-3 * 2 + y * 2) + (y + 1)
-5 + 3y = (-6 + 2y) + (y + 1)

Reorder the terms:
-5 + 3y = -6 + 2y + (1 + y)

Remove parenthesis around (1 + y)
-5 + 3y = -6 + 2y + 1 + y

Reorder the terms:
-5 + 3y = -6 + 1 + 2y + y

Combine like terms: -6 + 1 = -5
-5 + 3y = -5 + 2y + y

Combine like terms: 2y + y = 3y
-5 + 3y = -5 + 3y

Add '5' to each side of the equation.
-5 + 5 + 3y = -5 + 5 + 3y

Combine like terms: -5 + 5 = 0
0 + 3y = -5 + 5 + 3y
3y = -5 + 5 + 3y

Combine like terms: -5 + 5 = 0
3y = 0 + 3y
3y = 3y

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

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

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

Solving
0 = 0

Couldn't find a variable to solve for.

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