3(r+5)+3(3r-4)=27

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Solution for 3(r+5)+3(3r-4)=27 equation: 



3(r+5)+3(3r-4)=27

We move all terms to the left:

3(r+5)+3(3r-4)-(27)=0

We multiply parentheses

3r+9r+15-12-27=0

We add all the numbers together, and all the variables

12r-24=0

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

12r=24

r=24/12

r=2

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