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1/2(x-2)=3/1(x+5)
We move all terms to the left:
1/2(x-2)-(3/1(x+5))=0
Domain of the equation: 2(x-2)!=0
x∈R
Domain of the equation: 1(x+5))!=0We calculate fractions
x∈R
(xx/(2(x-2)*1(x+5)))+(-6xx/(2(x-2)*1(x+5)))=0
We calculate terms in parentheses: +(xx/(2(x-2)*1(x+5))), so:
xx/(2(x-2)*1(x+5))
We multiply all the terms by the denominator
xx
Back to the equation:
+(xx)
We calculate terms in parentheses: +(-6xx/(2(x-2)*1(x+5))), so:We get rid of parentheses
-6xx/(2(x-2)*1(x+5))
We multiply all the terms by the denominator
-6xx
Back to the equation:
+(-6xx)
xx-6xx=0
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