Solution
= Solution
Here $\alpha\ne3$, so the same elimination forces $z=0$. The first two <linear equations> would then require $x-y=1$ and $x-y=-1$ simultaneously. Thus
\b[There is no solution for any value of $\beta$.]
= Solution
Here $\alpha\ne3$, so the same elimination forces $z=0$. The first two <linear equations> would then require $x-y=1$ and $x-y=-1$ simultaneously. Thus
\b[There is no solution for any value of $\beta$.]