Помогите очень надо завтра контрольная (((

$\frac{x-1}{5-x} = \frac{2}{9} \\\frac{x-1}{5-x} - \frac{2}{9} = 0 \\\frac{9(x-1) }{9(5-x)} - \frac{2(5-x)}{9(5-x)} = 0 \\\frac{9x-9-10+2x}{9(5-x)} = 0 \\\frac{11x-19}{9(5-x)} = 0$
$\left \{ {{11x-19 =0} \atop {9(5-x) \neq 0)}} \right. \left \{ {{11x = 19} \atop {x \neq 5}} \right. \left \{ {{x = \frac{19}{11} } \atop {x \neq 5}} \right.$

Ответ: $\frac{19}{11}$