1+(x-3)/3=(2x+1)/6-2

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Solution for 1+(x-3)/3=(2x+1)/6-2 equation: 


x in (-oo:+oo)

(x-3)/3+1 = (2*x+1)/6-2 // - (2*x+1)/6-2

(x-3)/3-((2*x+1)/6)+1+2 = 0

(x-3)/3+(-1*(2*x+1))/6+1+2 = 0

(6*(x-3))/(3*6)+(-1*3*(2*x+1))/(3*6)+(1*3*6)/(3*6)+(2*3*6)/(3*6) = 0

6*(x-3)-1*3*(2*x+1)+1*3*6+2*3*6 = 0

18-21+36 = 0

36-3 = 0

33 = 0

33/(3*6) = 0

33/(3*6) = 0 // * 3*6

33 = 0

x belongs to the empty set

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