Chapter 3 - Section 3.3 - Systems of Linear Equations in Three Variables - Exercise Set - Page 216: 16

The solution is $x=2 $, $y=2 $, and $z=2 $

The given equations are $x+y=4 $ ... (1) $x+z=4 $ ... (2) $y+z=4 $ ... (3) In equation (1) isolate $x$. $x=4-y $ Substitute into equation (2). $4-y+z=4 $ $ -y+z=0 $ $z=y $ Substitute into equation (3). $ y+y=4 $ $ 2y=4 $ $y=2 $ Therefore, $z=2 $ Substitute above value into equation (2). $ x+2=4 $ $ x=2 $.