3(3x-1)/(2x+3)-(2x+3)/(3x-1)=5

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Solution for 3(3x-1)/(2x+3)-(2x+3)/(3x-1)=5 equation: 



3(3x-1)/(2x+3)-(2x+3)/(3x-1)=5

We move all terms to the left:

3(3x-1)/(2x+3)-(2x+3)/(3x-1)-(5)=0



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

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

2x!=-3

x!=-3/2

x!=-1+1/2

x∈R





Domain of the equation: (3x-1)!=0

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

3x!=1

x!=1/3

x!=1/3

x∈R



We calculate fractions


(3(3x-1)*(3x-1))/((2x+3)*(3x-1))+(-(2x+3)*(2x+3))/((2x+3)*(3x-1))-5=0



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

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

We multiply all the terms by the denominator


3(3x-1)*(3x-1))

Back to the equation:

+(3(3x-1)*(3x-1)))





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

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

We multiply all the terms by the denominator


-(2x+3)*(2x+3))

Back to the equation:

+(-(2x+3)*(2x+3)))

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