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(2x-1)/x-5=(3x-2)/5-x
We move all terms to the left:
(2x-1)/x-5-((3x-2)/5-x)=0
Domain of the equation: x!=0
x∈R
Domain of the equation: 5-x)!=0We calculate fractions
We move all terms containing x to the left, all other terms to the right
-x)!=-5
x!=-5/1
x!=-5
x∈R
(10x-x)-5)/(5x-x))+(-((3x-2)*x)/(5x-x))-5=0
We add all the numbers together, and all the variables
(+9x)+4x))+(-((3x-2)*x)/(+4x))-5-5)/(=0
We get rid of parentheses
9x+4x))+(-((3x-2)*x)/(+4x))-5-5)/(=0
We multiply all the terms by the denominator
9x*(+4x))+(-((3x-2)*x)+4x))-5-5)=0
We calculate terms in parentheses: -((3x-2)*x), so:We get rid of parentheses
(3x-2)*x
We multiply parentheses
3x^2-2x
Back to the equation:
-(3x^2-2x)
-3x^2+9x*(+4x))+(+2x+4x))-5-5)=0
We add all the numbers together, and all the variables
-3x^2+9x*(+4x))+(+6x))-5-5)=0
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