5(r+1)+6=3(4+r)+(2r-1)

Solution for 5(r+1)+6=3(4+r)+(2r-1) equation:

Simplifying
5(r + 1) + 6 = 3(4 + r) + (2r + -1)

Reorder the terms:
5(1 + r) + 6 = 3(4 + r) + (2r + -1)
(1 * 5 + r * 5) + 6 = 3(4 + r) + (2r + -1)
(5 + 5r) + 6 = 3(4 + r) + (2r + -1)

Reorder the terms:
5 + 6 + 5r = 3(4 + r) + (2r + -1)

Combine like terms: 5 + 6 = 11
11 + 5r = 3(4 + r) + (2r + -1)
11 + 5r = (4 * 3 + r * 3) + (2r + -1)
11 + 5r = (12 + 3r) + (2r + -1)

Reorder the terms:
11 + 5r = 12 + 3r + (-1 + 2r)

Remove parenthesis around (-1 + 2r)
11 + 5r = 12 + 3r + -1 + 2r

Reorder the terms:
11 + 5r = 12 + -1 + 3r + 2r

Combine like terms: 12 + -1 = 11
11 + 5r = 11 + 3r + 2r

Combine like terms: 3r + 2r = 5r
11 + 5r = 11 + 5r

Add '-11' to each side of the equation.
11 + -11 + 5r = 11 + -11 + 5r

Combine like terms: 11 + -11 = 0
0 + 5r = 11 + -11 + 5r
5r = 11 + -11 + 5r

Combine like terms: 11 + -11 = 0
5r = 0 + 5r
5r = 5r

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

Combine like terms: 5r + -5r = 0
0 = 5r + -5r

Combine like terms: 5r + -5r = 0
0 = 0

Solving
0 = 0

Couldn't find a variable to solve for.

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