3[(2r+3)-(r+1)-(r+2)]=0

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Solution for 3[(2r+3)-(r+1)-(r+2)]=0 equation:


Simplifying
3[(2r + 3) + -1(r + 1) + -1(r + 2)] = 0

Reorder the terms:
3[(3 + 2r) + -1(r + 1) + -1(r + 2)] = 0

Remove parenthesis around (3 + 2r)
3[3 + 2r + -1(r + 1) + -1(r + 2)] = 0

Reorder the terms:
3[3 + 2r + -1(1 + r) + -1(r + 2)] = 0
3[3 + 2r + (1 * -1 + r * -1) + -1(r + 2)] = 0
3[3 + 2r + (-1 + -1r) + -1(r + 2)] = 0

Reorder the terms:
3[3 + 2r + -1 + -1r + -1(2 + r)] = 0
3[3 + 2r + -1 + -1r + (2 * -1 + r * -1)] = 0
3[3 + 2r + -1 + -1r + (-2 + -1r)] = 0

Reorder the terms:
3[3 + -1 + -2 + 2r + -1r + -1r] = 0

Combine like terms: 3 + -1 = 2
3[2 + -2 + 2r + -1r + -1r] = 0

Combine like terms: 2 + -2 = 0
3[0 + 2r + -1r + -1r] = 0
3[2r + -1r + -1r] = 0

Combine like terms: 2r + -1r = 1r
3[1r + -1r] = 0

Combine like terms: 1r + -1r = 0
3[0] = 0

Multiply 3 * 0
0 = 0

Anything times zero is zero.
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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