Question

Помогите пожалуйста решить 1067 #1 спасибо

Answer 1

$\begin{cases}\frac x3+\frac14(x-2)\ \textgreater \ x-\frac{x-1}2\\\frac{0,7x-8,7}{3}-\frac56\ \textgreater \ 0\end{cases}\Rightarrow\begin{cases}\frac x3+\frac x4-\frac12\ \textgreater \ \frac{2x-x+1}2\\\frac{0,7x-8,7}{3}\ \textgreater \ \frac56\end{cases}\Rightarrow\\\\\Rightarrow\begin{cases}\frac{7x}{12}-\frac12\ \textgreater \ \frac{x+1}2\\0,7x-8,7\ \textgreater \ \frac52\end{cases}\Rightarrow\begin{cases}\frac{7x-6}{12}\ \textgreater \ \frac{x+1}2\\0,7x\ \textgreater \ 11,2\end{cases}\Rightarrow\begin{cases}7x-6\ \textgreater \ 6x+6\\x\ \textgreater \ 16\end{cases}\Rightarrow\\\Rightarrow\begin{cases}x\ \textgreater \ 12\\x\ \textgreater \ 16\end{cases}\Rightarrow x\ \textgreater \ 16,\\x\in(16;\;+\infty)$