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1/2(z+3)=1/3(z-7)
We move all terms to the left:
1/2(z+3)-(1/3(z-7))=0
Domain of the equation: 2(z+3)!=0
z∈R
Domain of the equation: 3(z-7))!=0We calculate fractions
z∈R
(3zz/(2(z+3)*3(z-7)))+(-2zz/(2(z+3)*3(z-7)))=0
We calculate terms in parentheses: +(3zz/(2(z+3)*3(z-7))), so:
3zz/(2(z+3)*3(z-7))
We multiply all the terms by the denominator
3zz
Back to the equation:
+(3zz)
We calculate terms in parentheses: +(-2zz/(2(z+3)*3(z-7))), so:We get rid of parentheses
-2zz/(2(z+3)*3(z-7))
We multiply all the terms by the denominator
-2zz
Back to the equation:
+(-2zz)
3zz-2zz=0
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