(2b+1)/5=(3-b)/4

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Solution for (2b+1)/5=(3-b)/4 equation: 



(2b+1)/5=(3-b)/4

We move all terms to the left:

(2b+1)/5-((3-b)/4)=0

We add all the numbers together, and all the variables

(2b+1)/5-((-1b+3)/4)=0

We calculate fractions


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



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

-((-1b+3)*5)/(

We multiply all the terms by the denominator


-((-1b+3)*5)



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

(-1b+3)*5

We multiply parentheses

-5b+15

Back to the equation:

-(-5b+15)



We get rid of parentheses

5b-15

Back to the equation:

+(5b-15)



We get rid of parentheses

2b/()+5b-15=0

We multiply all the terms by the denominator


2b+5b*()-15*()=0

We add all the numbers together, and all the variables

2b+5b*()=0

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