n,41+n=4(5+n)

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Solution for n,41+n=4(5+n) equation: 



n.41+n=4(5+n)

We move all terms to the left:

n.41+n-(4(5+n))=0

We add all the numbers together, and all the variables

n.41+n-(4(n+5))=0

We add all the numbers together, and all the variables

n+n.41-(4(n+5))=0



We calculate terms in parentheses: -(4(n+5)), so:

4(n+5)

We multiply parentheses

4n+20

Back to the equation:

-(4n+20)



We get rid of parentheses

n+n.41-4n-20=0

We add all the numbers together, and all the variables

-3n+n.41-20=0

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

-3n+n.41=20

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