-(1/9)(x-27)+1/3(x+3)=x-17

Solution for -(1/9)(x-27)+1/3(x+3)=x-17 equation:

-(1/9)(x-27)+1/3(x+3)=x-17

We move all terms to the left:

-(1/9)(x-27)+1/3(x+3)-(x-17)=0


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

x∈R



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

x∈R


We add all the numbers together, and all the variables

-(+1/9)(x-27)+1/3(x+3)-(x-17)=0

We get rid of parentheses

-(+1/9)(x-27)+1/3(x+3)-x+17=0

We multiply parentheses ..

-(+x^2+1/9*-27)+1/3(x+3)-x+17=0

We calculate fractions


-x+(-(x^2+3xx)/x-27)*3()+()/x+17-27)*3()=0


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

-(x^2+3xx)/x-27)*3(

We add all the numbers together, and all the variables

-(x^2+3xx)/x

We multiply all the terms by the denominator


-(x^2+3xx)

We get rid of parentheses

-x^2-3xx

We add all the numbers together, and all the variables

-1x^2-3xx

Back to the equation:

+(-1x^2-3xx)


We add all the numbers together, and all the variables

(-1x^2-3xx)-1x+()/x=0

We get rid of parentheses

-1x^2-3xx-1x+()/x=0

We multiply all the terms by the denominator


-1x^2*x-3xx*x-1x*x+()=0

We add all the numbers together, and all the variables

-1x^2*x-3xx*x-1x*x=0

Wy multiply elements

-1x^3-3x^2-1x^2=0

We do not support expression: x^3