Let $t$ be TNYWR.

Each of the three lines having equations $x+t y+8=0,5 x-t y+4=0$, and $3 x-k y+1=0$ passes through the same point. What is the value of $k$ ?

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Adding the equations $x+t y+8=0$ and $5 x-t y+4=0$, we get $x+t y+8+5 x-t y+4=0+0$ or $6 x+12=0$.

Solving for $x$ gives $x=-2$. Note that we have found the value of $x$ without knowing the value of $t$.

Substituting into the first equation gives $-2+t y+8=0$ or $t y=-6$, which means $y=-\frac{6}{t}$. Substituting the values of $x$ and $y$ into the third equation and rearranging, we have

$$
\begin{aligned}
3(-2)-k\left(-\frac{6}{t}\right)+1 & =0 \\
-6+\frac{6 k}{t} & =-1 \\
\frac{6 k}{t} & =5 \\
k & =\frac{5 t}{6} .
\end{aligned}
$$

The answer in part (a) is 6 , so we have that $t=6$. Therefore, $k=\frac{5 \times 6}{6}=5$.