(120-3x)/3x=(120-4x)/4x

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Solution for (120-3x)/3x=(120-4x)/4x equation: 



(120-3x)/3x=(120-4x)/4x

We move all terms to the left:

(120-3x)/3x-((120-4x)/4x)=0



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R





Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

(-3x+120)/3x-((-4x+120)/4x)=0

We calculate fractions


(-3x+120)*4x)/12x^2+(-((-4x+120)*3x)/12x^2=0

We calculate fractions


((-3x+120)*4x)*12x^2)/(12x^2+(*12x^2)+(-((-4x+120)*3x)*12x^2)/(12x^2+(*12x^2)=0



We calculate terms in parentheses: +(-((-4x+120)*3x)*12x^2)/(12x^2+(*12x^2), so:

-((-4x+120)*3x)*12x^2)/(12x^2+(*12x^2

We multiply all the terms by the denominator


-((-4x+120)*3x)*12x^2)+((*12x^2)*(12x^2

Back to the equation:

+(-((-4x+120)*3x)*12x^2)+((*12x^2)*(12x^2)



We get rid of parentheses

((-3x+120)*4x)*12x^2)/(12x^2+*12x^2+(-((-4x+120)*3x)*12x^2)+((*12x^2)*12x^2=0

We multiply all the terms by the denominator


((-3x+120)*4x)*12x^2)+(*12x^2)*(12x^2+((-((-4x+120)*3x)*12x^2))*(12x^2+(((*12x^2)*12x^2)*(12x^2=0

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