(1/3y)+3=(1/7y)

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Solution for (1/3y)+3=(1/7y) equation:



(1/3y)+3=(1/7y)
We move all terms to the left:
(1/3y)+3-((1/7y))=0
Domain of the equation: 3y)!=0
y!=0/1
y!=0
y∈R
Domain of the equation: 7y))!=0
y!=0/1
y!=0
y∈R
We add all the numbers together, and all the variables
(+1/3y)-((+1/7y))+3=0
We get rid of parentheses
1/3y-((+1/7y))+3=0
We calculate fractions
7y/21y^2+(-((+1*3y)/21y^2+3=0
We calculate fractions
We do not support eypression: y^4

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