I've been thinking about the following:
For (1) I drew $y= 0$, $y=2x$ and $y = 2x$, $y = 4x$. The expression I wrote down if $y = mx + b$ is one of the lines then $y_2 = (m \pm 2)x + b$ is the other.
For (2) I solved $y= mx + b$ at $y=0$ for $x$. Then we get $x_2 = \frac{-b}{m} \pm 2$ for the new $x$-intercepts and hence two new lines $y_2 = mx + b \pm 2m$. I drew $y = x$ and $y= x \pm 2$.
For (3) we get two new lines $y_2 = mx + b \pm 2$. Again I drew $y = x$ and $y= x \pm 2$.
Now (4) is where I am missing something. I combined (1),(2) and (3) to get lines that are supposed to differ by $2$ in slope, $x$- and $y$-intercept, $y_2 = (m+2)x + b + 2 + 2m$ (if all differ by $+2$), for example:
$y_1 = x + 2, y_2 = 3x + 6$. But this is wrong since the $y$-intercept now differs by $4$.
Question 1: How do I answer (4)?
Question 2: What is special about $2$? I don't understand what the puzzle is getting at. It appears to me that I could replace $2$ by $n$ and do the exact same questions.
Thank you!