For every number k, the line (1 + 2k)x + (2 - k)y - (8 + k) = 0 passes through the same fixed point…

Step 1. Separate. $(x + 2y - 8) + k(2x - y - 1) = 0$.

Step 2. Solve both brackets. $x + 2y = 8$ and $2x - y = 1$. From the second, $y = 2x - 1$. Then $x + 4x - 2 = 8$, so $x = 2$ and $y = 3$.

Step 3. Parallel to the $x$-axis. The coefficient of $x$ must be $0$: $1 + 2k = 0$, so $k = -0.5$. The line is then $2.5y - 7.5 = 0$, which is $y = 3$.