Elementary Linear Algebra 7th Edition

There is one solution to the system. $x=6$ $y=2$ $z=2$
By dividing both sides of Equation 3 by 3, we can conclude that $z = 2$. Substituting this value into Equation 2 gives us the equation $2y + 2 = 6$. We can then solve this to conclude that $y = 2$. Substituting this value into Equation 1 gives us the equation $x - 2 = 4$. We can then solve this to conclude that $x=6$.