(2x+5)/(3x-1)-(3x+2)/(5x+2)=1/15

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Solution for (2x+5)/(3x-1)-(3x+2)/(5x+2)=1/15 equation: 



(2x+5)/(3x-1)-(3x+2)/(5x+2)=1/15

We move all terms to the left:

(2x+5)/(3x-1)-(3x+2)/(5x+2)-(1/15)=0



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





Domain of the equation: (5x+2)!=0

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

5x!=-2

x!=-2/5

x!=-2/5

x∈R



We add all the numbers together, and all the variables

(2x+5)/(3x-1)-(3x+2)/(5x+2)-(+1/15)=0

We get rid of parentheses

(2x+5)/(3x-1)-(3x+2)/(5x+2)-1/15=0

We calculate fractions


((2x+5)*(5x+2)*15)/((3x-1)*(5x+2)*15)+(-(3x+2)*(3x-1)*15)/((3x-1)*(5x+2)*15)+(-1*(3x-1)*(5x+2))/((3x-1)*(5x+2)*15)=0



We calculate terms in parentheses: +((2x+5)*(5x+2)*15)/((3x-1)*(5x+2)*15), so:

(2x+5)*(5x+2)*15)/((3x-1)*(5x+2)*15

We multiply all the terms by the denominator


(2x+5)*(5x+2)*15)

Back to the equation:

+((2x+5)*(5x+2)*15))





We calculate terms in parentheses: +(-(3x+2)*(3x-1)*15)/((3x-1)*(5x+2)*15), so:

-(3x+2)*(3x-1)*15)/((3x-1)*(5x+2)*15

We multiply all the terms by the denominator


-(3x+2)*(3x-1)*15)

Back to the equation:

+(-(3x+2)*(3x-1)*15))





We calculate terms in parentheses: +(-1*(3x-1)*(5x+2))/((3x-1)*(5x+2)*15), so:

-1*(3x-1)*(5x+2))/((3x-1)*(5x+2)*15

We multiply all the terms by the denominator


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

Back to the equation:

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



We add all the numbers together, and all the variables

((2x+5)*(5x+2)*15))+(-(3x+2)*(3x-1)*15))+(-1*(3x-1)*(5x=0

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