n/2(2(40000)+(n-1)(3500))=n(2(10000)+(n-1)(2000))

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Solution for n/2(2(40000)+(n-1)(3500))=n(2(10000)+(n-1)(2000)) equation: 



n/2(2(40000)+(n-1)(3500))=n(2(10000)+(n-1)(2000))

We move all terms to the left:

n/2(2(40000)+(n-1)(3500))-(n(2(10000)+(n-1)(2000)))=0



Domain of the equation: 2(240000+(n-1)3500)!=0

n∈R



We multiply all the terms by the denominator


n-((n(210000+(n-1)2000)))*2(240000+(n-1)3500)=0



We calculate terms in parentheses: -((n(210000+(n-1)2000)))*2(240000+(n-1)3500), so:

(n(210000+(n-1)2000)))*2(240000+(n-1)3500

We multiply parentheses

(n(210000+(n-1)2000)))*2(240000+3500n-3500

We add all the numbers together, and all the variables

3500n+(n(210000+(n-1)2000)))*2(240000-3500

Back to the equation:

-(3500n+(n(210000+(n-1)2000)))*2(240000-3500)



We add all the numbers together, and all the variables

n-(3500n+(n(210000+(n-1)2000)))*2236500=0

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