(2r+3)/5=(4r-3)/7

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Solution for (2r+3)/5=(4r-3)/7 equation: 



(2r+3)/5=(4r-3)/7

We move all terms to the left:

(2r+3)/5-((4r-3)/7)=0

We calculate fractions


2r/()+(-((4r-3)*5)/()=0



We calculate terms in parentheses: +(-((4r-3)*5)/(), so:

-((4r-3)*5)/(

We multiply all the terms by the denominator


-((4r-3)*5)



We calculate terms in parentheses: -((4r-3)*5), so:

(4r-3)*5

We multiply parentheses

20r-15

Back to the equation:

-(20r-15)



We get rid of parentheses

-20r+15

Back to the equation:

+(-20r+15)



We get rid of parentheses

2r/()-20r+15=0

We multiply all the terms by the denominator


2r-20r*()+15*()=0

We add all the numbers together, and all the variables

2r-20r*()=0

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