r+(2r)+(r-3)+(r-3)=20

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Solution for r+(2r)+(r-3)+(r-3)=20 equation: 



r+(2r)+(r-3)+(r-3)=20

We move all terms to the left:

r+(2r)+(r-3)+(r-3)-(20)=0

We add all the numbers together, and all the variables

3r+(r-3)+(r-3)-20=0

We get rid of parentheses

3r+r+r-3-3-20=0

We add all the numbers together, and all the variables

5r-26=0

We move all terms containing r to the left, all other terms to the right

5r=26

r=26/5

r=5+1/5

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