1/3(x-6)-1/9(x-9)=1/9(-4x+27)

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Solution for 1/3(x-6)-1/9(x-9)=1/9(-4x+27) equation: 



1/3(x-6)-1/9(x-9)=1/9(-4x+27)

We move all terms to the left:

1/3(x-6)-1/9(x-9)-(1/9(-4x+27))=0



Domain of the equation: 3(x-6)!=0

x∈R





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

x∈R





Domain of the equation: 9(-4x+27))!=0

x∈R



We calculate fractions


(-27x^2x/(3(x-6)*9(x-9)*9(-4x+27)))+(-27x^2x/(3(x-6)*9(x-9)*9(-4x+27)))+(81x^2x/(3(x-6)*9(x-9)*9(-4x+27)))=0



We calculate terms in parentheses: +(-27x^2x/(3(x-6)*9(x-9)*9(-4x+27))), so:

-27x^2x/(3(x-6)*9(x-9)*9(-4x+27))

We multiply all the terms by the denominator


-27x^2x

Back to the equation:

+(-27x^2x)





We calculate terms in parentheses: +(-27x^2x/(3(x-6)*9(x-9)*9(-4x+27))), so:

-27x^2x/(3(x-6)*9(x-9)*9(-4x+27))

We multiply all the terms by the denominator


-27x^2x

Back to the equation:

+(-27x^2x)





We calculate terms in parentheses: +(81x^2x/(3(x-6)*9(x-9)*9(-4x+27))), so:

81x^2x/(3(x-6)*9(x-9)*9(-4x+27))

We multiply all the terms by the denominator


81x^2x

Back to the equation:

+(81x^2x)



We get rid of parentheses

-27x^2x-27x^2x+81x^2x=0

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