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(x+2)/x=x/(2x-5)
We move all terms to the left:
(x+2)/x-(x/(2x-5))=0
Domain of the equation: x!=0
x∈R
Domain of the equation: (2x-5))!=0We calculate fractions
x∈R
-3x^2)+(-(x*x)/(-3x^2)+((x+2)*(2x-5)))/(=0
We add all the numbers together, and all the variables
-3x^2)+(-(+x*x)/(-3x^2)+((x+2)*(2x-5)))/(=0
We calculate fractions
-3x^2)+(+(-(+x*x)*()/((-3x^2)*()+(-3x^2)+((x+2)*(2x-5)))*()/((-3x^2)*()=0
We calculate terms in parentheses: +(-(+x*x)*()/((-3x^2)*()+(-3x^2)+((x+2)*(2x-5)))*()/((-3x^2)*(), so:
-(+x*x)*()/((-3x^2)*()+(-3x^2)+((x+2)*(2x-5)))*()/((-3x^2)*(
We can not solve this equation
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