Question

Решив систему неравенств: {-10(×-4)<3(4-2×)-3×, {1,6(0,1×-3)>3(-6+0,1×)+0,3×

Answer 1

$\begin{cases}-10(x-4)<3(4-2x)-3x\\1.6(0.1x-3)>3(-6+0.1x)+0.3x\end{cases}\Rightarrow\begin{cases}-10x+40<12-6x-3x\\0.16x-4.8>-18+0.3x+0.3x\end{cases}\Rightarrow\\\Rightarrow\begin{cases}-10x+40<12-9x\\0.16x-4.8>-18+0.6x\end{cases}\Rightarrow\begin{cases}10x-9x>40-12\\0.6x-0.16x<18-4.8\end{cases}\Rightarrow\\\Rightarrow\begin{cases}x>28\\0.44x<13.2\end{cases}\Rightarrow\begin{cases}x>7\\x<30\end{cases}\Rightarrow x\in(28;\;30)$