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6-2/3(3x-12)=2/5(10x+50)
We move all terms to the left:
6-2/3(3x-12)-(2/5(10x+50))=0
Domain of the equation: 3(3x-12)!=0
x∈R
Domain of the equation: 5(10x+50))!=0We calculate fractions
x∈R
(-10x1/(3(3x-12)*5(10x+50)))+(-6x3/(3(3x-12)*5(10x+50)))+6=0
We calculate terms in parentheses: +(-10x1/(3(3x-12)*5(10x+50))), so:
-10x1/(3(3x-12)*5(10x+50))
We multiply all the terms by the denominator
-10x1
We add all the numbers together, and all the variables
-10x
Back to the equation:
+(-10x)
We calculate terms in parentheses: +(-6x3/(3(3x-12)*5(10x+50))), so:
-6x3/(3(3x-12)*5(10x+50))
We multiply all the terms by the denominator
-6x3
We add all the numbers together, and all the variables
-6x^3
We do not support expression: x^3
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