1/3(r-3)+2=1/3(r-9)

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Solution for 1/3(r-3)+2=1/3(r-9) equation: 



1/3(r-3)+2=1/3(r-9)

We move all terms to the left:

1/3(r-3)+2-(1/3(r-9))=0



Domain of the equation: 3(r-3)!=0

r∈R





Domain of the equation: 3(r-9))!=0

r∈R



We calculate fractions


(3rr/(3(r-3)*3(r-9)))+(-3rr/(3(r-3)*3(r-9)))+2=0



We calculate terms in parentheses: +(3rr/(3(r-3)*3(r-9))), so:

3rr/(3(r-3)*3(r-9))

We multiply all the terms by the denominator


3rr

Back to the equation:

+(3rr)





We calculate terms in parentheses: +(-3rr/(3(r-3)*3(r-9))), so:

-3rr/(3(r-3)*3(r-9))

We multiply all the terms by the denominator


-3rr

Back to the equation:

+(-3rr)



We get rid of parentheses

3rr-3rr+2=0

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

3rr-3rr=-2

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