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4/(2x-3)=2/(3x+5)
We move all terms to the left:
4/(2x-3)-(2/(3x+5))=0
Domain of the equation: (2x-3)!=0
We move all terms containing x to the left, all other terms to the right
2x!=3
x!=3/2
x!=1+1/2
x∈R
Domain of the equation: (3x+5))!=0We calculate fractions
x∈R
12x/((2x-3)*(3x+5)))+(-(2*(2x-3))/((2x-3)*(3x+5)))=0
We calculate terms in parentheses: -(2*(2x-3))/((2x-3)*(3x+5))), so:We add all the numbers together, and all the variables
2*(2x-3))/((2x-3)*(3x+5))
We multiply all the terms by the denominator
2*(2x-3))
We multiply parentheses
4x+
We add all the numbers together, and all the variables
4x
Back to the equation:
-(4x)
-4x+12x/((2x-3)*(3x+5)))+(=0
We multiply all the terms by the denominator
-4x*((2x-3)*(3x+5)))+(+12x=0
We add all the numbers together, and all the variables
12x-4x*((2x-3)*(3x+5)))+(=0
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