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

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Solution for 1/(3x-2)+1/(5x+3)=0 equation: 



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



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

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

3x!=2

x!=2/3

x!=2/3

x∈R





Domain of the equation: (5x+3)!=0

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

5x!=-3

x!=-3/5

x!=-3/5

x∈R



We calculate fractions


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



We calculate terms in parentheses: +(1*(5x+3))/((3x-2)*(5x+3)), so:

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

We multiply all the terms by the denominator


1*(5x+3))

Back to the equation:

+(1*(5x+3)))





We calculate terms in parentheses: +(1*(3x-2))/((3x-2)*(5x+3)), so:

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

We multiply all the terms by the denominator


1*(3x-2))

Back to the equation:

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





We calculate terms in parentheses: +(1*(5x+3)))+(1*(3x-2))), so:

1*(5x+3)))+(1*(3x-2))

We add all the numbers together, and all the variables

1*(5x+3)))+(1*(3x

Back to the equation:

+(1*(5x+3)))+(1*(3x)

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