3.17. Solve the system

$$
\left\{\begin{array}{l}
x_{1}+2 x_{2}+2 x_{3}+2 x_{4}+2 x_{5}=1 \\
x_{1}+3 x_{2}+4 x_{3}+4 x_{4}+4 x_{5}=2 \\
x_{1}+3 x_{2}+5 x_{3}+6 x_{4}+6 x_{5}=3 \\
x_{1}+3 x_{2}+5 x_{3}+7 x_{4}+8 x_{5}=4 \\
x_{1}+3 x_{2}+5 x_{3}+7 x_{4}+9 x_{5}=5
\end{array}\right.
$$
3.17. Answer: $x_{1}=x_{3}=x_{5}=1, x_{2}=x_{4}=-1$.

Let's first write down the first equation, then the second equation from which the first is subtracted, then the third equation from which the second is subtracted, and so on:

$$
\left\{\begin{aligned}
x_{1}+2 x_{2}+2 x_{3}+2 x_{4}+2 x_{5} & =1 \\
x_{2}+2 x_{3}+2 x_{4}+2 x_{5} & =1 \\
x_{3}+2 x_{4}+2 x_{5} & =1 \\
x_{4}+2 x_{5} & =1 \\
x_{5} & =1
\end{aligned}\right.
$$

Now we can sequentially find $x_{5}, x_{4}, x_{3}, x_{2}, x_{1}$.