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1/6(x+3)=1/5(x-4)=1
We move all terms to the left:
1/6(x+3)-(1/5(x-4))=0
Domain of the equation: 6(x+3)!=0
x∈R
Domain of the equation: 5(x-4))!=0We calculate fractions
x∈R
(5xx/(6(x+3)*5(x-4)))+(-6xx/(6(x+3)*5(x-4)))=0
We calculate terms in parentheses: +(5xx/(6(x+3)*5(x-4))), so:
5xx/(6(x+3)*5(x-4))
We multiply all the terms by the denominator
5xx
Back to the equation:
+(5xx)
We calculate terms in parentheses: +(-6xx/(6(x+3)*5(x-4))), so:We get rid of parentheses
-6xx/(6(x+3)*5(x-4))
We multiply all the terms by the denominator
-6xx
Back to the equation:
+(-6xx)
5xx-6xx=0
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