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1/3(9x+3)=2/5(10x+5)
We move all terms to the left:
1/3(9x+3)-(2/5(10x+5))=0
Domain of the equation: 3(9x+3)!=0
x∈R
Domain of the equation: 5(10x+5))!=0We calculate fractions
x∈R
(5x1/(3(9x+3)*5(10x+5)))+(-6x9/(3(9x+3)*5(10x+5)))=0
We calculate terms in parentheses: +(5x1/(3(9x+3)*5(10x+5))), so:
5x1/(3(9x+3)*5(10x+5))
We multiply all the terms by the denominator
5x1
We add all the numbers together, and all the variables
5x
Back to the equation:
+(5x)
We calculate terms in parentheses: +(-6x9/(3(9x+3)*5(10x+5))), so:
-6x9/(3(9x+3)*5(10x+5))
We multiply all the terms by the denominator
-6x9
We add all the numbers together, and all the variables
-6x^9
We do not support expression: x^9
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