7. Given that $x+y+z=1, x+y-z=2$ and $x-y-z=3$, what is the value of $x y z$ ?
A -2
B $-\frac{1}{2}$
C 0
D $\frac{1}{2}$
E 2
Solution: D
Adding the equations, $x+y+z=1$ and $x-y-z=3$, gives $2 x=4$, from which it follows that $x=2$. Adding the equations $x+y+z=1$ and $x+y-z=2$, gives $2(x+y)=3$. Hence $x+y=\frac{3}{2}$. Therefore, as $x=2$, it follows that $y=-\frac{1}{2}$. Since $x+y+z=1$, it now follows that $z=1-(x+y)=1-\frac{3}{2}=-\frac{1}{2}$. Therefore, $x y z=2 \times-\frac{1}{2} \times-\frac{1}{2}=\frac{1}{2}$.