(2t)/(k-3)= 8/(k-2t)

解

\frac{2 t}{k - 3} = \frac{8}{k - 2 t}

t = - \frac{- k + k ^{2} + 32 ( - k + 3 )}{4}, t = \frac{k + k ^{2} + 32 ( - k + 3 )}{4}; k \neq = 3

解答ステップ

\frac{2 t}{k - 3} = \frac{8}{k - 2 t}

たすき掛け

2 t (k - 2 t) = (k - 3) \cdot 8

解く

2 t (k - 2 t) = (k - 3) \cdot 8 : t = - \frac{- k + k ^{2} + 32 ( - k + 3 )}{4}, t = \frac{k + k ^{2} + 32 ( - k + 3 )}{4}

t = - \frac{- k + k ^{2} + 32 ( - k + 3 )}{4}, t = \frac{k + k ^{2} + 32 ( - k + 3 )}{4}; k \neq = 3