4(3n+1)=11(n+2)-19

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Solution for 4(3n+1)=11(n+2)-19 equation: 



4(3n+1)=11(n+2)-19

We move all terms to the left:

4(3n+1)-(11(n+2)-19)=0

We multiply parentheses

12n-(11(n+2)-19)+4=0



We calculate terms in parentheses: -(11(n+2)-19), so:

11(n+2)-19

We multiply parentheses

11n+22-19

We add all the numbers together, and all the variables

11n+3

Back to the equation:

-(11n+3)



We get rid of parentheses

12n-11n-3+4=0

We add all the numbers together, and all the variables

n+1=0

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

n=-1

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