1/5(n+3)+4=3/10(2n-2)

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Solution for 1/5(n+3)+4=3/10(2n-2) equation: 



1/5(n+3)+4=3/10(2n-2)

We move all terms to the left:

1/5(n+3)+4-(3/10(2n-2))=0



Domain of the equation: 5(n+3)!=0

n∈R





Domain of the equation: 10(2n-2))!=0

n∈R



We calculate fractions


(10n2/(5(n+3)*10(2n-2)))+(-15nn/(5(n+3)*10(2n-2)))+4=0



We calculate terms in parentheses: +(10n2/(5(n+3)*10(2n-2))), so:

10n2/(5(n+3)*10(2n-2))

We multiply all the terms by the denominator


10n2

We add all the numbers together, and all the variables

10n^2

Back to the equation:

+(10n^2)





We calculate terms in parentheses: +(-15nn/(5(n+3)*10(2n-2))), so:

-15nn/(5(n+3)*10(2n-2))

We multiply all the terms by the denominator


-15nn

Back to the equation:

+(-15nn)



We get rid of parentheses

10n^2-15nn+4=0

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

10n^2-15nn=-4

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