[Algebra]
Problem: $\quad$ Let $T=40$. If $x+9 y=17$ and $T x+(T+1) y=T+2$, compute $20 x+14 y$.

Solution: Multiply each side of the first equation by $T$ to obtain $T x+9 T y=17 T$. Subtract the second equation to yield $9 T y-T y-y=16 T-2 \Rightarrow y(8 T-1)=2(8 T-1)$. Hence either $T=\frac{1}{8}$ (in which case, the value of $y$ is not uniquely determined) or $y=2$. Plug $y=2$ into the first equation to obtain $x=-1$. Hence $20 x+14 y=-20+28=\mathbf{8}$.

Final answer: 8