1/3(n+2)=1/2(6-n)

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Solution for 1/3(n+2)=1/2(6-n) equation:



1/3(n+2)=1/2(6-n)
We move all terms to the left:
1/3(n+2)-(1/2(6-n))=0
Domain of the equation: 3(n+2)!=0
n∈R
Domain of the equation: 2(6-n))!=0
n∈R
We add all the numbers together, and all the variables
1/3(n+2)-(1/2(-1n+6))=0
We calculate fractions
(2n(-)/(3(n+2)*2(-1n+6)))+(-3nn/(3(n+2)*2(-1n+6)))=0
We calculate terms in parentheses: +(2n(-)/(3(n+2)*2(-1n+6))), so:
2n(-)/(3(n+2)*2(-1n+6))
We add all the numbers together, and all the variables
2n0/(3(n+2)*2(-1n+6))
We multiply all the terms by the denominator
2n0
We add all the numbers together, and all the variables
2n
Back to the equation:
+(2n)
We calculate terms in parentheses: +(-3nn/(3(n+2)*2(-1n+6))), so:
-3nn/(3(n+2)*2(-1n+6))
We multiply all the terms by the denominator
-3nn
Back to the equation:
+(-3nn)
We get rid of parentheses
2n-3nn=0

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