(2x+1)/5x=(6x-2)/(9x+2)

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Solution for (2x+1)/5x=(6x-2)/(9x+2) equation: 



(2x+1)/5x=(6x-2)/(9x+2)

We move all terms to the left:

(2x+1)/5x-((6x-2)/(9x+2))=0



Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R





Domain of the equation: (9x+2))!=0

x∈R



We calculate fractions


((2x+1)*(9x+2)))/55x^2+(-((6x-2)*5x)/55x^2=0

We calculate fractions


(((2x+1)*(9x+2)))*55x^2)/(55x^2+(*55x^2)+(-((6x-2)*5x)*55x^2)/(55x^2+(*55x^2)=0



We calculate terms in parentheses: +(-((6x-2)*5x)*55x^2)/(55x^2+(*55x^2), so:

-((6x-2)*5x)*55x^2)/(55x^2+(*55x^2

We multiply all the terms by the denominator


-((6x-2)*5x)*55x^2)+((*55x^2)*(55x^2

Back to the equation:

+(-((6x-2)*5x)*55x^2)+((*55x^2)*(55x^2)



We get rid of parentheses

(((2x+1)*(9x+2)))*55x^2)/(55x^2+*55x^2+(-((6x-2)*5x)*55x^2)+((*55x^2)*55x^2=0

We multiply all the terms by the denominator


(((2x+1)*(9x+2)))*55x^2)+(*55x^2)*(55x^2+((-((6x-2)*5x)*55x^2))*(55x^2+(((*55x^2)*55x^2)*(55x^2=0

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