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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We calculate fractions


4b/()+(-((-1b+2)*2)/()=0



We calculate terms in parentheses: +(-((-1b+2)*2)/(), so:

-((-1b+2)*2)/(

We multiply all the terms by the denominator


-((-1b+2)*2)



We calculate terms in parentheses: -((-1b+2)*2), so:

(-1b+2)*2

We multiply parentheses

-2b+4

Back to the equation:

-(-2b+4)



We get rid of parentheses

2b-4

Back to the equation:

+(2b-4)



We get rid of parentheses

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

We multiply all the terms by the denominator


4b+2b*()-4*()=0

We add all the numbers together, and all the variables

4b+2b*()=0

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