Зависимость уровня потребления у некоторого вида товаров от уровня дохода семьи х...

Решите задачу:

$\begin{cases}a-\frac b{50+c}=0\\a-\frac b{74+c}=0,8\\a-\frac b{326+c}=2,3\end{cases}\Rightarrow\begin{cases}a=\frac b{50+c}\\\frac b{50+c}-\frac b{74+c}=0,8\\\frac b{50+c}-\frac b{326+c}=2,3\end{cases}\Rightarrow\\\Rightarrow\begin{cases}a=\frac b{50+c}\\b=\frac{0,8}{\frac1{50+c}-\frac1{74+c}}\\\frac b{50+c}-\frac b{326+c}=2,3\end{cases}\Rightarrow\begin{cases}a=\frac b{50+c}\\b=\frac{(50+c)(74+c)}{30}\\\frac {(50+c)(74+c)}{30(50+c)}-\frac{(50+c)(74+c)}{30(326+c)}=2,3\end{cases}$
$\frac {(50+c)(74+c)}{30(50+c)}-\frac{(50+c)(74+c)}{30(326+c)}=2,3\\\frac{74+c}{30}-\frac{(50+c)(74+c)}{30(326+c)}=2,3\\\frac{(74+c)(326+c)-(50+c)(74+c)}{30(326+c)}=2,3\\(74+c)(326+c-50-c)=69(326+c)\\(74+c)276=69(326+c)\\276c+20424=69c+22494\\207c=2070\\c=10\\\begin{cases}a=2,8\\b=168\\c=10\end{cases}$

$y=2,8-\frac{168}{x+10}\\\\y(158)=2,8-\frac{168}{158+10}=2,8-\frac{168}{168}=2,8-1=1,8$