(5x-9)/9)+(9x-2)/3)=9

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Solution for (5x-9)/9)+(9x-2)/3)=9 equation: 



(5x-9)/9)+(9x-2)/3)=9

We move all terms to the left:

(5x-9)/9)+(9x-2)/3)-(9)=0



Domain of the equation: 9)+(9x!=0

x∈R



We calculate fractions


((5x-9)*3)-9)/27x^2+(-2)*9)+9x/27x^2=0



We calculate terms in parentheses: +((5x-9)*3), so:

(5x-9)*3

We multiply parentheses

15x-27

Back to the equation:

+(15x-27)



We get rid of parentheses

15x-9)/27x^2+(-2)*9)+9x/27x^2-27=0

We calculate fractions


15x+(-9)*27x^2-27)/(27x^2+(*27x^2-27)+(-2)*9)+9x*27x^2+()/(27x^2+(*27x^2-27)=0

We get rid of parentheses

15x+(-9)*27x^2-27)/(27x^2+*27x^2+(-2)*9)+9x*27x^2+()/(27x^2+*27x^2-27-27=0

We multiply all the terms by the denominator


15x*(27x^2+((-9)*27x^2)*(27x^2-27)+(*27x^2)*(27x^2+(-2)*9)+9x*27x^2+()+(*27x^2)*(27x^2-27*(27x^2-27*(27x^2=0

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