Ta có: \(x+3z+x+2y=8+9\)
\(\Rightarrow2x+2y+3z=17\)
\(\Rightarrow2x+2y+2z+z=17\)
\(\Rightarrow2\left(x+y+z\right)=17-z\)
Mà \(x+y+z\) có GTLN
\(\Rightarrow17-z\) cũng có GTLN
Mà \(z\ge0\Rightarrow-z\le0\)
\(\Rightarrow17-z\le17\)
\(\Rightarrow17-z\) đạt GTLN là 17 tại \(z=0\)
+) \(x+3z=8\)
Thay \(z=0\)
\(\Rightarrow x+0=8\)
\(\Rightarrow x=8\)
+) \(x+2y=9\)
Thay \(x=8\)
\(\Rightarrow8+2y=9\)
\(\Rightarrow2y=1\)
\(\Rightarrow y=\frac{1}{2}\)
Vậy \(x=8;y=\frac{1}{2};z=0\)