3(s+33)=4(s+12)

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Solution for 3(s+33)=4(s+12) equation: 



3(s+33)=4(s+12)

We move all terms to the left:

3(s+33)-(4(s+12))=0

We multiply parentheses

3s-(4(s+12))+99=0



We calculate terms in parentheses: -(4(s+12)), so:

4(s+12)

We multiply parentheses

4s+48

Back to the equation:

-(4s+48)



We get rid of parentheses

3s-4s-48+99=0

We add all the numbers together, and all the variables

-1s+51=0

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

-s=-51

s=-51/-1

s=+51

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