Ta có:
\(x-y=8\)
\(y-z=10\)
\(x+z=12\)
\(\Rightarrow\left(x-y\right)+\left(y-z\right)+\left(x+z\right)=8+10+12\)
\(\Rightarrow x-y+y-z+x+z=30\)
\(\Rightarrow x+x-y+y-z+z=30\)
\(\Rightarrow2x=30\)
\(\Rightarrow x=30\div2\)
\(\Rightarrow x=15\)
Thay \(x=15\) vào \(x-y=8\) ta được:
\(15-y=8\)
\(\Rightarrow y=15-8\)
\(\Rightarrow y=7\)
Thay \(y=7\) vào \(y-z=10\) ta được:
\(7-z=10\)
\(\Rightarrow z=7-10\)
\(\Rightarrow z=-3\)
\(\Rightarrow x+y+z=15+7+\left(-3\right)=19\)
Vậy tổng \(x+y+z=19\)
\(\left\{\begin{matrix}x-y=8\left(1\right)\\y-z=10\\x+z=12\left(3\right)\end{matrix}\right.\left(2\right)\)
(2)+(3)=> x+y=22 (4)
(3)-(1)=> y+z=4 (5)
(3)+(4)+(5)=2(x+y+z)=12+22+4
=> (x+y+z)=6+11+2=19
