(2n+1.2)+(2n+1.2)+(2n+1.2)=(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)

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Solution for (2n+1.2)+(2n+1.2)+(2n+1.2)=(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4) equation: 



(2n+1.2)+(2n+1.2)+(2n+1.2)=(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)

We move all terms to the left:

(2n+1.2)+(2n+1.2)+(2n+1.2)-((3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4))=0

We get rid of parentheses

2n+2n+2n-((3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4))+1.2+1.2+1.2=0



We calculate terms in parentheses: -((3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)), so:

(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)

We get rid of parentheses

3n+3n+3n+3n+3n+3n-4-4-4-4-4-4

We add all the numbers together, and all the variables

18n-24

Back to the equation:

-(18n-24)



We add all the numbers together, and all the variables

6n-(18n-24)+3.6=0

We get rid of parentheses

6n-18n+24+3.6=0

We add all the numbers together, and all the variables

-12n+27.6=0

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

-12n=-27.6

n=-27.6/-12

n=2+1/3.3333333333333

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