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9+2/3(6x-18)=3x-1/5(40-10x)
We move all terms to the left:
9+2/3(6x-18)-(3x-1/5(40-10x))=0
Domain of the equation: 3(6x-18)!=0
x∈R
Domain of the equation: 5(40-10x))!=0We add all the numbers together, and all the variables
x∈R
2/3(6x-18)-(3x-1/5(-10x+40))+9=0
We calculate fractions
(10x(-)/(3(6x-18)*5(-10x+40)))+(-(3x-3x6)/(3(6x-18)*5(-10x+40)))+9=0
We calculate terms in parentheses: +(10x(-)/(3(6x-18)*5(-10x+40))), so:
10x(-)/(3(6x-18)*5(-10x+40))
We add all the numbers together, and all the variables
10x0/(3(6x-18)*5(-10x+40))
We multiply all the terms by the denominator
10x0
We add all the numbers together, and all the variables
10x
Back to the equation:
+(10x)
We calculate terms in parentheses: +(-(3x-3x6)/(3(6x-18)*5(-10x+40))), so:
-(3x-3x6)/(3(6x-18)*5(-10x+40))
We add all the numbers together, and all the variables
-(+3x-3x^6)/(3(6x-18)*5(-10x+40))
We multiply all the terms by the denominator
-(+3x-3x^6)
We do not support expression: x^6
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