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(3(x+2)/5=2x-3
We move all terms to the left:
(3(x+2)/5-(2x-3)=0
Domain of the equation: 5-(2x-3)!=0We multiply all the terms by the denominator
We move all terms containing x to the left, all other terms to the right
-(2x-3)!=-5
x∈R
(3(x+2)=0
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