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1/3(2x+1)-1/2(x-1)=5/6
We move all terms to the left:
1/3(2x+1)-1/2(x-1)-(5/6)=0
Domain of the equation: 3(2x+1)!=0
x∈R
Domain of the equation: 2(x-1)!=0We add all the numbers together, and all the variables
x∈R
1/3(2x+1)-1/2(x-1)-(+5/6)=0
We get rid of parentheses
1/3(2x+1)-1/2(x-1)-5/6=0
We calculate fractions
(12xx/(3(2x+1)*2(x-1)*6)+(-18x2/(3(2x+1)*2(x-1)*6)+(-30x^22/(3(2x+1)*2(x-1)*6)=0
We calculate terms in parentheses: +(12xx/(3(2x+1)*2(x-1)*6)+(-18x2/(3(2x+1)*2(x-1)*6)+(-30x^22/(3(2x+1)*2(x-1)*6), so:
12xx/(3(2x+1)*2(x-1)*6)+(-18x2/(3(2x+1)*2(x-1)*6)+(-30x^22/(3(2x+1)*2(x-1)*6
We do not support expression: x^22
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