2(x+1)/5+1/3=7(x+4)/5

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Solution for 2(x+1)/5+1/3=7(x+4)/5 equation: 



2(x+1)/5+1/3=7(x+4)/5

We move all terms to the left:

2(x+1)/5+1/3-(7(x+4)/5)=0

We calculate fractions


(6x+6)/()+(-(7(x+4)*3)/()+()/()=0



We calculate terms in parentheses: +(-(7(x+4)*3)/()+()/(), so:

-(7(x+4)*3)/()+()/(

We add all the numbers together, and all the variables

-(7(x+4)*3)/()+1

We multiply all the terms by the denominator


-(7(x+4)*3)+1*()



We calculate terms in parentheses: -(7(x+4)*3), so:

7(x+4)*3

We multiply parentheses

21x+84

Back to the equation:

-(21x+84)



We add all the numbers together, and all the variables

-(21x+84)

We get rid of parentheses

-21x-84

Back to the equation:

+(-21x-84)



We get rid of parentheses

(6x+6)/()-21x-84=0

We multiply all the terms by the denominator


(6x+6)-21x*()-84*()=0

We add all the numbers together, and all the variables

(6x+6)-21x*()=0

We get rid of parentheses

6x-21x*()+6=0

We move all terms containing x to the left, all other terms to the right

6x-21x*()=-6

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