Answer
The solution of the system is $\left( 2,2,2 \right)$.
Work Step by Step
Now, it is given that
$x+y=4$ ...... (I)
$x+z=4$ ...... (II)
$y+z=4$ ...... (III)
From equation (I) $y=4-x$ ...... (IV)
From equation (II) $z=4-x$ ...... (V)
From equation (III) $y=4-z$ ...... (VI)
From equation (IV) and equation (VI):
$4-x=4-z$
Putting the value of z from equation (V):
$\begin{align}
& 4-x=4-\left( 4-x \right) \\
& 4-x=4-4+x \\
& 4-x=x \\
& 2x=4
\end{align}$
$x=2$
Putting the value of $x=2$ in equation (I):
$\begin{align}
& 2+y=4 \\
& y=2
\end{align}$
Putting the value of $y=2$ in equation (III):
$\begin{align}
& 2+z=4 \\
& z=4
\end{align}$
Thus, the ordered triple $\left( 2,2,2 \right)$ satisfies the three systems.
