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3/(x+1)-(1/5)=1/(2x+2)
We move all terms to the left:
3/(x+1)-(1/5)-(1/(2x+2))=0
Domain of the equation: (x+1)!=0
We move all terms containing x to the left, all other terms to the right
x!=-1
x∈R
Domain of the equation: (2x+2))!=0We add all the numbers together, and all the variables
x∈R
3/(x+1)-(1/(2x+2))-(+1/5)=0
We get rid of parentheses
3/(x+1)-(1/(2x+2))-1/5=0
We calculate fractions
30x/((x+1)*(2x+2))*5)+(-(1*(x+1)*5)/((x+1)*(2x+2))*5)+(-1*(x+1)*(2x+2)))/((x+1)*(2x+2))*5)=0
We add all the numbers together, and all the variables
30x/((x+1)*(2x+2))*5)+(-(1*(x+1)*5)/((x+1)*(2x+2))*5)+(-1*(x+1)*(2x+2)))/((x+1)*(2x=0
We calculate fractions
1)*(2x+(30x-(1*(x+1)*5)*(x/(((x+1)*(2x+2))*5)+(*(x+(-1*(x+1)*(2x+2)))*((x+1)*2x/(((x+1)*(2x+2))*5)+(*(x=0
We can not solve this equation
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