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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We calculate fractions


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



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

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

We multiply all the terms by the denominator


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



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

(-3r+4)*5

We multiply parentheses

-15r+20

Back to the equation:

-(-15r+20)



We get rid of parentheses

15r-20

Back to the equation:

+(15r-20)



We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

2r+15r*()=0

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