Let's determine the equation of the line whose segment between the lines

$$
3 x-5 y=6 \quad \text { and } \quad 4 x+y+6=0
$$

is bisected by the origin.
The sought line passes through the origin, so its equation is $y=a x$, where we need to determine the slope $a$ so that the line meets the requirement. This happens when the

$3 x-5 y=6 \quad$ and $y=a x$ lines intersect at point $M_{1}$

furthermore, the $4 x+y+6=0$ and $y=a x \quad, \quad M_{2}$

are symmetric with respect to the origin, that is, if the abscissa of $M_{1}$ is $x_{1}$ and that of $M_{2}$ is $x_{2}$, then $x_{1}+x_{2}=0$. [In this case, $y_{1}+y_{2}=$ $\left.a\left(x_{1}+x_{2}\right)=0\right]$.

$$
\begin{aligned}
& M_{1} \text { for } 3 x_{1}-5 a x_{1}=6, \quad \text { that is } \quad x_{1}=\frac{6}{3-5 a} \text {, } \\
& M_{2} \text { for } \quad " \quad 4 x_{2}+a x_{2}+6=0, \quad, \quad x_{2}=-\frac{6}{4+a},
\end{aligned}
$$

thus

$$
\frac{6}{3-5 a}-\frac{6}{4+a}=0 \text { and from here } a=-\frac{1}{6}
$$

The equation of the sought line is $y=-\frac{1}{6} x$.

$$
\left[x_{1}=\frac{36}{23}, \quad x_{2}=-\frac{36}{23}, \quad x_{1}+x_{2}=0\right]
$$