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