(1/2x-3)=3/(3x-2)

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Solution for (1/2x-3)=3/(3x-2) equation: 



(1/2x-3)=3/(3x-2)

We move all terms to the left:

(1/2x-3)-(3/(3x-2))=0



Domain of the equation: 2x-3)!=0

x∈R





Domain of the equation: (3x-2))!=0

x∈R



We get rid of parentheses

1/2x-(3/(3x-2))-3=0

We calculate fractions


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

We add all the numbers together, and all the variables

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

We calculate fractions


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

We get rid of parentheses

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

We calculate fractions


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



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

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

We can not solve this equation

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