<s>[INST] How can you quickly use the graph of $3x-6y=4$ to produce the graph of $3x-6y=10$? To start with, I am asked to graph the lines $3x-2y=k$ for $k=1,2,3$ and $4$.
I find that the equations produce different parallel lines.
For each line where $k=2,3$ and $4$, the coordinates points are $0.5k$ below the coordinates points of the line $3x-2y=1$.
If we let $x=0$, we can find the $y$-intercept of the lines.
\begin{align*}
-2y&=k\\
y &=\frac{k}{-2}
\end{align*}
and if we increase $k$ by $1$ we decrease $y$ by $0.5$.
\begin{align*}
-2y&=k+1\\
y &=\frac{k}{-2}-\frac{1}{2}
\end{align*}
I'm not certain the above logic suffices in quickly using the graph of $3x -6y = 4$ to produce the graph of $3x-6y=10$.
\begin{align*}
-6y&=4+6\\
y&=-\frac{2}{3}-1 \\
y&=-\frac{5}{3}
\end{align*}
Can my answer be improved?
 [/INST] Great question. Let's start with a couple of general lines here:
$$ ax + by = k_1 , $$
$$ ax + by = k_2 . $$
As you note, these lines are parallel because their slopes are the same:
$$ ax + by = k_1 \Rightarrow y = -\frac{a}{b}x + \frac{k_1}{b} , $$
$$ ax + by = k_1 \Rightarrow y = -\frac{a}{b}x + \frac{k_2}{b} . $$
And again, you're right to focus on the $x$ and $y$-intercepts.
Compare the $y$-intercepts from the slope-intercept form above. Your approach was to look at the difference, which is perfectly valid:
$$ \frac{k_2}{b} - \frac{k_1}{b} = \frac{k_2 - k_1}{b} . $$
Note, the $y$-intercept increases by $\frac{1}{b}$ times the increase in the constant. A similar pattern holds for the $x$-intercept. If you plug in $y=0$, you'll find the $x$-intercepts are, respectively, $\frac{k_1}{a}$ and $\frac{k_2}{a}$. Their difference is then:
$$ \frac{k_2}{a} - \frac{k_1}{a} = \frac{k_2 - k_1}{a} .$$
So, generally, as the constant increases by $\Delta k = k_2 - k_1$, the intercepts increase by the same amount, scaled down by their respective coefficients. The $x$-intercept increases by $\frac{1}{a}*\Delta k$, and the $y$-intercept increases by $\frac{1}{b}*\Delta k$.
In this case, every time you increase $k$ by 1, you increase the $x$-intercept by $\frac{1}{a} = \frac{1}{3}$ and you increase the $y$-intercept by $\frac{1}{b} = -\frac{1}{6}$, i.e., you decrease the $y$-intercept by $\frac{1}{6}$.
You can also look at what happens to the intercepts multiplicatively. Let's look at the ratio of the $y$-intercepts:
$$ \frac{k_2}{b} \bigg/ \frac{k_1}{b} = \frac{k_2}{b} * \frac{b}{k_1} = \frac{k_2}{k_1} . $$
Similarly, the ratio of the $x$-intercepts is
$$ \frac{k_2}{a} \bigg/ \frac{k_1}{a} = \frac{k_2}{a} * \frac{a}{k_1} = \frac{k_2}{k_1} . $$
So, as the constant scales up by some factor $\kappa = \frac{k_2}{k_1} $, both the $x$ and $y$ intercepts also get scaled up by that same factor.
You can use either approach, looking at the difference $k_2 - k_1$ or the ratio $k_2 / k_1$, depending on which is easier with the intercepts. You just have to calculate the intercepts once.
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