(X-A)/2=(B+X)/3-C

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Solution for (X-A)/2=(B+X)/3-C equation:


x in (-oo:+oo)

(X-A)/2 = (B+X)/3-C // - (B+X)/3-C

(X-A)/2-((B+X)/3)+C = 0

(X-A)/2+(-1*(B+X))/3+C = 0

(3*(X-A))/(2*3)+(-1*2*(B+X))/(2*3)+(2*3*C)/(2*3) = 0

3*(X-A)-1*2*(B+X)+2*3*C = 0

6*C-3*A-2*B+X = 0

(6*C-3*A-2*B+X)/(2*3) = 0

(6*C-3*A-2*B+X)/(2*3) = 0 // * 2*3

6*C-3*A-2*B+X = 0

x belongs to the empty set

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