Example 7 Given that $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=a^{2}, \\
x+y+z=b,
\end{array}\right.
$$

where $a, b$ are real constants, and $|b| \leqslant \sqrt{3}|a|$. Try to find the range of values for $x$.
Explanation: According to the problem, the original system of equations is equivalent to
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=a^{2}-x^{2}, \\
y+z=b-x .
\end{array} \text { (considering } x\right. \text { as a parameter) }
$$

Since $y z=\frac{(y+z)^{2}-\left(y^{2}+z^{2}\right)}{2}$
$$
=\frac{1}{2}\left[(b-x)^{2}-\left(a^{2}-x^{2}\right)\right] \text {, }
$$

Therefore, the system of equations (1) is equivalent to
$$
\left\{\begin{array}{l}
y+z=b-x, \\
y z=\frac{1}{2}\left[(b-x)^{2}-\left(a^{2}-x^{2}\right)\right] .
\end{array}\right.
$$

Thus, we can construct a quadratic equation with $y, z$ as its roots:
$$
t^{2}-(b-x) t+\frac{1}{2}\left[(b-x)^{2}-\left(a^{2}-x^{2}\right)\right]=0,
$$

The necessary and sufficient condition for this equation to hold is
$$
\Delta=(b-x)^{2}-4 \times \frac{1}{2}\left[(b-x)^{2}-\left(a^{2}-x^{2}\right)\right] \geqslant 0 \text {, }
$$

which simplifies to $3 x^{2}-2 b x+b^{2}-2 a^{2} \leqslant 0$.
Solving this, we get
$$
\frac{b-\sqrt{2\left(3 a^{2}-b^{2}\right)}}{3} \leqslant x \leqslant \frac{b+\sqrt{2\left(3 a^{2}-b^{2}\right)}}{3} \text {. }
$$

Therefore, the range of $x$ in the original system of equations is
$$
\left[\frac{b-\sqrt{2\left(3 a^{2}-b^{2}\right)}}{3}, \frac{b+\sqrt{2\left(3 a^{2}-b^{2}\right)}}{3}\right] .
$$