Example 4 Given $y=\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5}$, and $y \leqslant|x|$. Find the range of values for $x$.
(1995, National Junior High School Competition (Ethnic Group))
Analysis: From $y \leqslant|x|$, we know that when $x \geqslant 0$, $y \leqslant x$; when $x<0$, $y \leqslant -x$. Therefore, these two cases should be discussed separately.
Solution: When $x \geqslant 0$, we have $y \leqslant x$, i.e., $\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5} \leqslant x$, which also means $(x-2)(x-9) \leqslant 0$.
Solving this, we get $2 \leqslant x \leqslant 9$.
When $x<0$, we have $y \leqslant -x$, i.e., $\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5} \leqslant -x$, which means $(x+3)(x+6) \leqslant 0$.
Solving this, we get $-6 \leqslant x \leqslant -3$.
In summary, the range of $x$ that satisfies the condition is
$$
2 \leqslant x \leqslant 9 \text { or } -6 \leqslant x \leqslant -3 .
$$