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2/5(2x-3)-2/3=1/2(5x+3)
We move all terms to the left:
2/5(2x-3)-2/3-(1/2(5x+3))=0
Domain of the equation: 5(2x-3)!=0
x∈R
Domain of the equation: 2(5x+3))!=0We calculate fractions
x∈R
(-20x^22/(5(2x-3)*2(5x+3))*3)+(12x5/(5(2x-3)*2(5x+3))*3)+(-(1*5(2x-3)*3)/(5(2x-3)*2(5x+3))*3)=0
We calculate terms in parentheses: +(-20x^22/(5(2x-3)*2(5x+3))*3), so:
-20x^22/(5(2x-3)*2(5x+3))*3
We do not support expression: x^22
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