Four. (25 points) Let the real number $x$ satisfy
$$
\frac{3 x-1}{2}-\frac{4 x-2}{3} \geqslant \frac{6 x-3}{5}-\frac{13}{10} \text {. }
$$

Find the minimum value of $2|x-1|+|x+4|$.
Four, multiplying both sides of the original inequality by 30 gives
$$
15(3 x-1)-10(4 x-2) \geqslant 6(6 x-3)-39 \text {. }
$$

Solving this yields $x \leqslant 2$.
Let $y=2|x-1|+|x+4|$.
(1) When $x \leqslant-4$,
$$
y=-2(x-1)-(x+4)=-3 x-2 \text {. }
$$

Therefore, the minimum value of $y$ is $(-3) \times(-4)-2=$ 10, at this point $x=-4$.
(2) When $-4 \leqslant x \leqslant 1$,
$$
y=-2(x-1)+(x+4)=-x+6 \text {. }
$$

Therefore, the minimum value of $y$ is 5, at this point $x=1$.
(3) When $1 \leqslant x \leqslant 2$,
$$
y=2(x-1)+(x+4)=3 x+2 \text {. }
$$

Therefore, the minimum value of $y$ is 5, at this point $x=1$.
In summary, the minimum value of $2|x-1|+|x+4|$ is 5, which is achieved when $x=1$.