3(2f+3)+1=3(f-4)

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Solution for 3(2f+3)+1=3(f-4) equation: 



3(2f+3)+1=3(f-4)

We move all terms to the left:

3(2f+3)+1-(3(f-4))=0

We multiply parentheses

6f-(3(f-4))+9+1=0



We calculate terms in parentheses: -(3(f-4)), so:

3(f-4)

We multiply parentheses

3f-12

Back to the equation:

-(3f-12)



We add all the numbers together, and all the variables

6f-(3f-12)+10=0

We get rid of parentheses

6f-3f+12+10=0

We add all the numbers together, and all the variables

3f+22=0

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

3f=-22

f=-22/3

f=-7+1/3

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