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