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1/2(2x-3)=1/3(x+4)
We move all terms to the left:
1/2(2x-3)-(1/3(x+4))=0
Domain of the equation: 2(2x-3)!=0
x∈R
Domain of the equation: 3(x+4))!=0We calculate fractions
x∈R
(3xx/(2(2x-3)*3(x+4)))+(-2x2/(2(2x-3)*3(x+4)))=0
We calculate terms in parentheses: +(3xx/(2(2x-3)*3(x+4))), so:
3xx/(2(2x-3)*3(x+4))
We multiply all the terms by the denominator
3xx
Back to the equation:
+(3xx)
We calculate terms in parentheses: +(-2x2/(2(2x-3)*3(x+4))), so:We get rid of parentheses
-2x2/(2(2x-3)*3(x+4))
We multiply all the terms by the denominator
-2x2
We add all the numbers together, and all the variables
-2x^2
Back to the equation:
+(-2x^2)
-2x^2+3xx=0
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