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1/x+1/(x+4)=1/3
We move all terms to the left:
1/x+1/(x+4)-(1/3)=0
Domain of the equation: x!=0
x∈R
Domain of the equation: (x+4)!=0We add all the numbers together, and all the variables
We move all terms containing x to the left, all other terms to the right
x!=-4
x∈R
1/x+1/(x+4)-(+1/3)=0
We get rid of parentheses
1/x+1/(x+4)-1/3=0
We calculate fractions
(-1x^2*()/(3x^2+12x)+(3x+12)/(3x^2+12x)+9x/(3x^2+12x)=0
We calculate terms in parentheses: +(-1x^2*()/(3x^2+12x)+(3x+12)/(3x^2+12x)+9x/(3x^2+12x), so:
-1x^2*()/(3x^2+12x)+(3x+12)/(3x^2+12x)+9x/(3x^2+12x
We calculate fractions
We do not support expression: x^3
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