1/3k-(I+1/2)=1/6(k+3)

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


x in (-oo:+oo)

(1/3)*k-I-(1/2) = (1/6)*(k+3) // - (1/6)*(k+3)

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

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

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

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

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

2*k-6*I-k-3-3 = 0

k-6*I-3-3 = 0

k-6*I-6 = 0

(k-6*I-6)/(2*3) = 0

(k-6*I-6)/(2*3) = 0 // * 2*3

k-6*I-6 = 0

x belongs to the empty set

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