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1/5(2x+1)-1/3(x-2)=x-73/3
We move all terms to the left:
1/5(2x+1)-1/3(x-2)-(x-73/3)=0
Domain of the equation: 5(2x+1)!=0
x∈R
Domain of the equation: 3(x-2)!=0We add all the numbers together, and all the variables
x∈R
1/5(2x+1)-1/3(x-2)-(+x-73/3)=0
We get rid of parentheses
1/5(2x+1)-1/3(x-2)-x+73/3=0
We calculate fractions
-x+(9xx/(5(2x+1)*3(x-2)*3)+(-5x2/(5(2x+1)*3(x-2)*3)+(365x2/(5(2x+1)*3(x-2)*3)=0
We calculate terms in parentheses: +(9xx/(5(2x+1)*3(x-2)*3)+(-5x2/(5(2x+1)*3(x-2)*3)+(365x2/(5(2x+1)*3(x-2)*3), so:
9xx/(5(2x+1)*3(x-2)*3)+(-5x2/(5(2x+1)*3(x-2)*3)+(365x2/(5(2x+1)*3(x-2)*3
We can not solve this equation
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