(2x+9)/(3x^2-7x-20)-(2x+9)/(6x^2+7x-5)=0

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Solution for (2x+9)/(3x^2-7x-20)-(2x+9)/(6x^2+7x-5)=0 equation: 



(2x+9)/(3x^2-7x-20)-(2x+9)/(6x^2+7x-5)=0



Domain of the equation: (3x^2-7x-20)!=0

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

3x^2-7x!=20

x∈R





Domain of the equation: (6x^2+7x-5)!=0

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

6x^2+7x!=5

x∈R



We calculate fractions


((2x+9)*(6x^2+7x-5))/((3x^2-7x-20)*(6x^2+7x-5))+(-(2x+9)*(3x^2-7x-20))/((3x^2-7x-20)*(6x^2+7x-5))=0



We calculate terms in parentheses: +((2x+9)*(6x^2+7x-5))/((3x^2-7x-20)*(6x^2+7x-5)), so:

(2x+9)*(6x^2+7x-5))/((3x^2-7x-20)*(6x^2+7x-5)

We multiply all the terms by the denominator


(2x+9)*(6x^2+7x-5))

Back to the equation:

+((2x+9)*(6x^2+7x-5)))





We calculate terms in parentheses: +(-(2x+9)*(3x^2-7x-20))/((3x^2-7x-20)*(6x^2+7x-5)), so:

-(2x+9)*(3x^2-7x-20))/((3x^2-7x-20)*(6x^2+7x-5)

We multiply all the terms by the denominator


-(2x+9)*(3x^2-7x-20))

Back to the equation:

+(-(2x+9)*(3x^2-7x-20)))

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