3(2f+5)=27

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Solution for 3(2f+5)=27 equation: 



3(2f+5)=27

We move all terms to the left:

3(2f+5)-(27)=0

We multiply parentheses

6f+15-27=0

We add all the numbers together, and all the variables

6f-12=0

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

6f=12

f=12/6

f=2

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