## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( 2,2,2 \right)$.
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.