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6+1/3(x-9)=1/2x(2-x)
We move all terms to the left:
6+1/3(x-9)-(1/2x(2-x))=0
Domain of the equation: 3(x-9)!=0
x∈R
Domain of the equation: 2x(2-x))!=0We add all the numbers together, and all the variables
x∈R
1/3(x-9)-(1/2x(-1x+2))+6=0
We calculate fractions
(2x(-)/(3(x-9)*2x(-1x+2)))+(-3xx/(3(x-9)*2x(-1x+2)))+6=0
We calculate terms in parentheses: +(2x(-)/(3(x-9)*2x(-1x+2))), so:
2x(-)/(3(x-9)*2x(-1x+2))
We add all the numbers together, and all the variables
2x0/(3(x-9)*2x(-1x+2))
We multiply all the terms by the denominator
2x0
We add all the numbers together, and all the variables
2x
Back to the equation:
+(2x)
We calculate terms in parentheses: +(-3xx/(3(x-9)*2x(-1x+2))), so:We get rid of parentheses
-3xx/(3(x-9)*2x(-1x+2))
We multiply all the terms by the denominator
-3xx
Back to the equation:
+(-3xx)
2x-3xx+6=0
We move all terms containing x to the left, all other terms to the right
2x-3xx=-6
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