1. Suppose natural numbers $a$ and $b$ are such that the numbers $a x + 2$ and $b x + 3$ are not coprime for any $x \in \mathbb{N}$. What can $a / b$ be equal to?
1. If $\frac{b}{a}=\frac{3}{2}$, the condition of the problem is satisfied.

Assume that $\frac{b}{a} \neq \frac{3}{2}$ and take $x=|3 a-2 b| \neq 0$. If $\gcd(a x+2, b x+3)=d$, then

$$
\begin{array}{ll}
d \mid a(b x+3)-b(a x+2) & =3 a-2 b= \pm x \quad \text { and } \\
d \mid(b x+3)-(a x+2) & =(b-a) x+1
\end{array}
$$

However, then $d \mid(b-a) x+1-(b-a) x=1$, which is impossible.