\(\Leftrightarrow x-y+y-z=\left(-9\right)+10=1\)
\(\Leftrightarrow x-z=1\)
Vậy: \(x-z=1\)
\(\Leftrightarrow x+z+x-z=11+1=12\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=\frac{12}{2}=6\)
\(\Leftrightarrow6-y=-9\)
\(\Leftrightarrow-y=-9-6=-15\)
\(\Leftrightarrow y=15\)
\(\Leftrightarrow15-z=10\)
\(\Leftrightarrow-z=10-15=-5\)
\(\Leftrightarrow z=5\)
Vậy: \(x=6;y=15;z=5\)
Ta có: x-y+y-z+x+z=(-9)+10+11
x+x=12
x.2=12
x=12:2
x=6
y=6-(-9)=15
z=15-10=5