1/2(2k-4)=3(k+2)-2k

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Solution for 1/2(2k-4)=3(k+2)-2k equation:


k in (-oo:+oo)

(1/2)*(2*k-4) = 3*(k+2)-(2*k) // - 3*(k+2)-(2*k)

(1/2)*(2*k-4)-(3*(k+2))+2*k = 0

(1/2)*(2*k-4)-3*(k+2)+2*k = 0

1/2*(2*k-4)-3*(k+2)+2*k = 0

1/2*(2*k-4)-3*(k+2)+2*k = 0

2*k-2*k-8 = 0

-8 = 0

-8 = 0

k belongs to the empty set

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