\[y = \sqrt{16 - x²} + \sqrt{7 - 5x}\]
\[1)\ 16 - x^{2} \geq 0\]
\[x^{2} - 16 \leq 0\]
\[(x + 4)(x - 4) \leq 0.\]
\[2)\ 7 - 5x \geq 0\]
\[- 5x \geq - 7\]
\[x \leq 1,4.\]
\[Ответ:x \in \lbrack - 4;1,4\rbrack.\]