4x/(3x-12)=1/x-4

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Solution for 4x/(3x-12)=1/x-4 equation: 



4x/(3x-12)=1/x-4

We move all terms to the left:

4x/(3x-12)-(1/x-4)=0



Domain of the equation: (3x-12)!=0

We move all terms containing x to the left, all other terms to the right

3x!=12

x!=12/3

x!=4

x∈R





Domain of the equation: x-4)!=0

x∈R



We get rid of parentheses

4x/(3x-12)-1/x+4=0

We calculate fractions


4x^2/(3x^2-12x)+(-1*(3x-12))/(3x^2-12x)+4=0



We calculate terms in parentheses: +(-1*(3x-12))/(3x^2-12x), so:

-1*(3x-12))/(3x^2-12x

We add all the numbers together, and all the variables

-12x-1*(3x-12))/(3x^2

We multiply all the terms by the denominator


-12x*(3x^2-1*(3x-12))

Back to the equation:

+(-12x*(3x^2-1*(3x-12)))



We multiply all the terms by the denominator


4x^2+((-12x*(3x^2-1*(3x-12))))*(3x^2-12x)+4*(3x^2-12x)=0



We calculate terms in parentheses: +((-12x*(3x^2-1*(3x-12))))*(3x^2-12x), so:

(-12x*(3x^2-1*(3x-12))))*(3x^2-12x

We add all the numbers together, and all the variables

-12x+(-12x*(3x^2-1*(3x-12))))*(3x^2

Back to the equation:

+(-12x+(-12x*(3x^2-1*(3x-12))))*(3x^2)



We multiply parentheses

4x^2+12x^2+(-12x+(-12x*(3x^2-1*(3x-12))))*3x^2-48x=0

We add all the numbers together, and all the variables

16x^2-48x+(-12x+(-12x*(3x^2-1*(3x-12))))*3x^2=0

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