Example 12. Solve the inequality

$$
\sqrt{25-x^{2}}+\sqrt{x^{2}+7 x}>3
$$
Solution. The domain of definition of the given inequality is determined by the system

$$
\left\{\begin{array}{l}
25-x^{2} \geqslant 0 \\
x^{2}+7 x \geqslant 0
\end{array}\right.
$$

from which $0 \leqslant x \leqslant 5$. Both sides of the inequality (17) are non-negative; therefore, it is equivalent to the system

$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant 5 \\
25-x^{2}+2 \sqrt{25-x^{2}} \sqrt{x^{2}+7 x}+x^{2}+7 x>9
\end{array}\right.
$$

i.e., the system

$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant 5 \\
2 \sqrt{25-x^{2}} \cdot \sqrt{x^{2}+7 x}>-16-7 x
\end{array}\right.
$$

For $0 \leqslant x \leqslant 5$, we have $-16-7 x<0$ and $2 \sqrt{25-x^{2}} \sqrt{x^{2}+7 x} \geqslant 0$; therefore, the solution to the last system, and hence to the inequality (17), are all numbers from the interval $0 \leqslant x \leqslant 5$.