(7/6)-(4x-4)=(1/3)x-10

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Solution for (7/6)-(4x-4)=(1/3)x-10 equation: 



(7/6)-(4x-4)=(1/3)x-10

We move all terms to the left:

(7/6)-(4x-4)-((1/3)x-10)=0



Domain of the equation: 3)x-10)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

-(4x-4)-((+1/3)x-10)+(+7/6)=0

We get rid of parentheses

-4x-((+1/3)x-10)+4+7/6=0

We calculate fractions


-4x+()/3x-10)*6)+21x/3x-10)*6)+4=0

We calculate fractions


-4x+(()*3x)/9x^2+(-10)*6)+21x*3x)/9x^2=0

We multiply all the terms by the denominator


-4x*9x^2+(()*3x)+(-10)*6)+21x*3x)=0



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

()*3x



Wy multiply elements

-36x^3+(()*3x)+(-10)*6)+21x*3x)=0

We get rid of parentheses

-36x^3+(()*3x)-10)*6)+21x*3x=0



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

()*3x



We do not support expression: x^3

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