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1/2(y-3)=1/4(y+1)
We move all terms to the left:
1/2(y-3)-(1/4(y+1))=0
Domain of the equation: 2(y-3)!=0
y∈R
Domain of the equation: 4(y+1))!=0We calculate fractions
y∈R
(4yy/(2(y-3)*4(y+1)))+(-2yy/(2(y-3)*4(y+1)))=0
We calculate terms in parentheses: +(4yy/(2(y-3)*4(y+1))), so:
4yy/(2(y-3)*4(y+1))
We multiply all the terms by the denominator
4yy
Back to the equation:
+(4yy)
We calculate terms in parentheses: +(-2yy/(2(y-3)*4(y+1))), so:We get rid of parentheses
-2yy/(2(y-3)*4(y+1))
We multiply all the terms by the denominator
-2yy
Back to the equation:
+(-2yy)
4yy-2yy=0
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