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(2)/(3)(5n-1)=-(3)/(5)(n+2)
We move all terms to the left:
(2)/(3)(5n-1)-(-(3)/(5)(n+2))=0
Domain of the equation: 3(5n-1)!=0
n∈R
Domain of the equation: 5(n+2))!=0We calculate fractions
n∈R
(10nn/(3(5n-1)*5(n+2)))+(-(-9n5)/(3(5n-1)*5(n+2)))=0
We calculate terms in parentheses: +(10nn/(3(5n-1)*5(n+2))), so:
10nn/(3(5n-1)*5(n+2))
We multiply all the terms by the denominator
10nn
Back to the equation:
+(10nn)
We calculate terms in parentheses: +(-(-9n5)/(3(5n-1)*5(n+2))), so:
-(-9n5)/(3(5n-1)*5(n+2))
We add all the numbers together, and all the variables
-(-9n^5)/(3(5n-1)*5(n+2))
We multiply all the terms by the denominator
-(-9n^5)
We do not support enpression: n^5
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