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2/3(5x+2)=1/2(x-3)
We move all terms to the left:
2/3(5x+2)-(1/2(x-3))=0
Domain of the equation: 3(5x+2)!=0
x∈R
Domain of the equation: 2(x-3))!=0We calculate fractions
x∈R
(4xx/(3(5x+2)*2(x-3)))+(-3x5/(3(5x+2)*2(x-3)))=0
We calculate terms in parentheses: +(4xx/(3(5x+2)*2(x-3))), so:
4xx/(3(5x+2)*2(x-3))
We multiply all the terms by the denominator
4xx
Back to the equation:
+(4xx)
We calculate terms in parentheses: +(-3x5/(3(5x+2)*2(x-3))), so:
-3x5/(3(5x+2)*2(x-3))
We multiply all the terms by the denominator
-3x5
We add all the numbers together, and all the variables
-3x^5
We do not support expression: x^5
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