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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We calculate fractions


3x/()+(-((x-4)*5)/()=0



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

-((x-4)*5)/(

We multiply all the terms by the denominator


-((x-4)*5)



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

(x-4)*5

We multiply parentheses

5x-20

Back to the equation:

-(5x-20)



We get rid of parentheses

-5x+20

Back to the equation:

+(-5x+20)



We get rid of parentheses

3x/()-5x+20=0

We multiply all the terms by the denominator


3x-5x*()+20*()=0

We add all the numbers together, and all the variables

3x-5x*()=0

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