3\end{cases}\Rightarrow x\in(3;\;5]" alt="\begin{cases}x^2-8x+15\leq0\\\frac{3-x}3<0\end{cases}\\x^2-8x+15=0\\D=64-4\cdot15=4\\x_1=5,\;x_2=3\\\begin{cases}(x-3)(x-5)\leq0\\3-x<0\end{cases}\Rightarrow\begin{cases}(x-3)(x-5)\leq0\\x>3\end{cases}\Rightarrow x\in(3;\;5]" align="absmiddle" class="latex-formula">