## Elementary Linear Algebra 7th Edition

There is one solution to the system. $x = 3/2$ $y = 3/2$ $z = 0$
From Equation 3, we can see that $z = 0$. Substituting $z = 0$ into Equation 2 gives us $2y = 3$. We can then conclude that $y = 3/2$. By substituting $z = 0$ and $y = 3/2$ into Equation 1, we can see that $-x + 3/2 = 0$. We can solve this equation to then find that $x = 3/2$.