(8r+3+3r)+(4r+6+2r)=6

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Solution for (8r+3+3r)+(4r+6+2r)=6 equation:



(8r+3+3r)+(4r+6+2r)=6
We move all terms to the left:
(8r+3+3r)+(4r+6+2r)-(6)=0
We add all the numbers together, and all the variables
(11r+3)+(6r+6)-6=0
We get rid of parentheses
11r+6r+3+6-6=0
We add all the numbers together, and all the variables
17r+3=0
We move all terms containing r to the left, all other terms to the right
17r=-3
r=-3/17
r=-3/17

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