3(d+1)+8=2(d+3)+d+5

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Solution for 3(d+1)+8=2(d+3)+d+5 equation: 


Simplifying
3(d + 1) + 8 = 2(d + 3) + d + 5

Reorder the terms:
3(1 + d) + 8 = 2(d + 3) + d + 5
(1 * 3 + d * 3) + 8 = 2(d + 3) + d + 5
(3 + 3d) + 8 = 2(d + 3) + d + 5

Reorder the terms:
3 + 8 + 3d = 2(d + 3) + d + 5

Combine like terms: 3 + 8 = 11
11 + 3d = 2(d + 3) + d + 5

Reorder the terms:
11 + 3d = 2(3 + d) + d + 5
11 + 3d = (3 * 2 + d * 2) + d + 5
11 + 3d = (6 + 2d) + d + 5

Reorder the terms:
11 + 3d = 6 + 5 + 2d + d

Combine like terms: 6 + 5 = 11
11 + 3d = 11 + 2d + d

Combine like terms: 2d + d = 3d
11 + 3d = 11 + 3d

Add '-11' to each side of the equation.
11 + -11 + 3d = 11 + -11 + 3d

Combine like terms: 11 + -11 = 0
0 + 3d = 11 + -11 + 3d
3d = 11 + -11 + 3d

Combine like terms: 11 + -11 = 0
3d = 0 + 3d
3d = 3d

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

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

Combine like terms: 3d + -3d = 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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