(n+2/n-1)=5.C(n,1)

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Solution for (n+2/n-1)=5.C(n,1) equation: 



(n+2/n-1)=5.(n.1)

We move all terms to the left:

(n+2/n-1)-(5.(n.1))=0



Domain of the equation: n-1)!=0

n∈R



We add all the numbers together, and all the variables

(n+2/n-1)-(5.(+n.1))=0

We get rid of parentheses

n+2/n-(5.(+n.1))-1=0

We multiply all the terms by the denominator


n*n-((5.(+n.1)))*n-1*n+2=0

We add all the numbers together, and all the variables

-1n+n*n-((5.(+n.1)))*n+2=0

Wy multiply elements

n^2-1n-((5.(+n.1)))*n+2=0

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

n^2-1n-((5.(+n.1)))*n=-2

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