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(1/6)(n-12)=(1/4)(n+8)
We move all terms to the left:
(1/6)(n-12)-((1/4)(n+8))=0
Domain of the equation: 6)(n-12)!=0
n∈R
Domain of the equation: 4)(n+8))!=0We add all the numbers together, and all the variables
n∈R
(+1/6)(n-12)-((+1/4)(n+8))=0
We multiply parentheses ..
(+n^2+1/6*-12)-((+1/4)(n+8))=0
We calculate fractions
((+n^2+1*4)(n+8)))/(-12)*4)(n+8))+6*)+()/(-12)*4)(n+8))+6*)=0
We calculate fractions
(((+n^2+1*4)(n+8)))*(-12)*4)(n+8))+6*))/((*(-12)*4)n)+(-12)*4)(n+8))+6*)+()*()/((*(-12)*4)n)=0
We can not solve this equation
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