6. If real numbers $x, y, z \geqslant 0$, and
$$
x+y+z=30,3 x+y-z=50 \text {, }
$$

then the range of $T=5 x+4 y+2 z$ is $\qquad$
6. $[120,130]$.

Notice,
$$
\begin{array}{l}
T=5 x+4 y+2 z \\
=(x+y+z)+(4 x+3 y+z) \\
=30+(4 x+3 y+z) .
\end{array}
$$

From $4 x+2 y=(x+y+z)+(3 x+y-z)=80$
$\Rightarrow T=110+(y+z)$.
Also, $20=(3 x+y-z)-(x+y+z)=2(x-z)$,
then $x-z=10$.
Since $x, z \geqslant 0$, therefore, $x \geqslant 10$.
Thus, from $x+y+z=30$, we know $y+z \leqslant 20$.
Therefore, $T=110+(y+z) \leqslant 130$ (equality holds when $z=0$, $x=10$, $y=20$).
From $4 x+2 y=80, y \geqslant 0$, we know $x \leqslant 20$.
Thus, $y+z=30-x \geqslant 10$.
Therefore, $T=110+(y+z) \geqslant 120$ (equality holds when $x=20$, $y=0$, $z=10$).
Hence, $120 \leqslant T \leqslant 130$.