(r/9)+((r-27)/3))=3

Solution for (r/9)+((r-27)/3))=3 equation:

(r/9)+((r-27)/3))=3

We move all terms to the left:

(r/9)+((r-27)/3))-(3)=0

We add all the numbers together, and all the variables

(+r/9)+((r-27)/3))-3=0

We get rid of parentheses

r/9+((r-27)/3))-3=0

We calculate fractions


3r^2/()+(((r-27)*9)/()=0


We calculate terms in parentheses: +(((r-27)*9)/(), so:

((r-27)*9)/(

We multiply all the terms by the denominator


((r-27)*9)


We calculate terms in parentheses: +((r-27)*9), so:

(r-27)*9

We multiply parentheses

9r-243

Back to the equation:

+(9r-243)


We get rid of parentheses

9r-243

Back to the equation:

+(9r-243)


We get rid of parentheses

3r^2/()+9r-243=0

We multiply all the terms by the denominator


3r^2+9r*()-243*()=0

We add all the numbers together, and all the variables

3r^2+9r*()=0