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

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



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



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

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

3a!=2

a!=2/3

a!=2/3

a∈R





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

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

5a!=-3

a!=-3/5

a!=-3/5

a∈R



We calculate fractions


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



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

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

We multiply all the terms by the denominator


1*(5a+3))

Back to the equation:

+(1*(5a+3)))





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

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

We multiply all the terms by the denominator


1*(3a-2))

Back to the equation:

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





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

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

We add all the numbers together, and all the variables

1*(5a+3)))+(1*(3a

Back to the equation:

+(1*(5a+3)))+(1*(3a)

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