(3b+2)/4=(b-3)/5

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



(3b+2)/4=(b-3)/5

We move all terms to the left:

(3b+2)/4-((b-3)/5)=0

We calculate fractions


3b/()+(-((b-3)*4)/()=0



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

-((b-3)*4)/(

We multiply all the terms by the denominator


-((b-3)*4)



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

(b-3)*4

We multiply parentheses

4b-12

Back to the equation:

-(4b-12)



We get rid of parentheses

-4b+12

Back to the equation:

+(-4b+12)



We get rid of parentheses

3b/()-4b+12=0

We multiply all the terms by the denominator


3b-4b*()+12*()=0

We add all the numbers together, and all the variables

3b-4b*()=0

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