(4y+1)/5=(y-4)/20

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Solution for (4y+1)/5=(y-4)/20 equation: 



(4y+1)/5=(y-4)/20

We move all terms to the left:

(4y+1)/5-((y-4)/20)=0

We calculate fractions


4y/()+(-((y-4)*5)/()=0



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

-((y-4)*5)/(

We multiply all the terms by the denominator


-((y-4)*5)



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

(y-4)*5

We multiply parentheses

5y-20

Back to the equation:

-(5y-20)



We get rid of parentheses

-5y+20

Back to the equation:

+(-5y+20)



We get rid of parentheses

4y/()-5y+20=0

We multiply all the terms by the denominator


4y-5y*()+20*()=0

We add all the numbers together, and all the variables

4y-5y*()=0

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