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

Lösung

\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

Schritte zur Lösung

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

Kreuzmultiplizieren

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

Löse

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

- \frac{1}{x - 1} = \frac{1}{1 - x}

\frac{2}{3 ( 12 x - 18 )} = 4

\frac{b - 10}{b - 9} = \frac{9}{7}

0 = 2 + \frac{1}{x + 4}

- \frac{15}{24} = \frac{12}{x}