Example 3 Given three different real numbers $a, b, c$, satisfying $a-b+c=3$, the equations
$$
x^{2}+a x+1=0 \text { and } x^{2}+b x+c=0
$$

have one common real root, and the equations
$$
x^{2}+x+a=0 \text { and } x^{2}+c x+b=0
$$

also have one common real root. Find the values of $a, b, c$.
Let $x_{1}$ be a common real root of the equations
$x^{2}+a x+1=0$ and $x^{2}+b x+c=0$.
Then
$$
\left\{\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}\right.
$$

Subtracting the two equations, we get
$$
x_{1}=\frac{c-1}{a-b} \text {. }
$$

Let $x_{2}$ be a common real root of the equations
$$
x^{2}+x+a=0 \text { and } x^{2}+c x+b=0
$$

Then
$$
\left\{\begin{array}{l}
x_{2}^{2}+x_{2}+a=0, \\
x_{2}^{2}+c x_{2}+b=0 .
\end{array}\right.
$$

Subtracting the two equations, we get
$$
x_{2}=\frac{a-b}{c-1} \text {. }
$$

Therefore, $x_{1} x_{2}=1$.
Since the product of the roots of the equation $x^{2}+a x+1=0$ is 1, it follows that $x_{2}$ is also a root of the equation $x^{2}+a x+1=0$.
Thus, $x_{2}^{2}+a x_{2}+1=0$.
Subtracting (1) from (2), we get
$(a-1) x_{2}=a-1$.
If $a=1$, then the equation $x^{2}+a x+1=0$ has no real roots.
Therefore, $a \neq 1$.
Hence, $x_{2}=1$.
Thus, $a=-2, b+c=-1$.
Also, $a-b+c=3$, solving these, we get $b=-3, c=2$.
Therefore, $a=-2, b=-3, c=2$.