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?
 A: 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.
A: Since the goal is to describe the relationship between the two lines quickly, I will give a one-sentence way to do this:
The difference between the graphs of $3x-2y=4$ and $3x-2y=10$ is that the graph of the latter equation is a dilation, centered at the origin, of the graph of the former by a factor of $10/4$.
That means that every point on the original graph "moves" to a point that is $2.5$ times farther away from the origin.
(The detailed explanation for why this is true is contained in Amaan M's answer.)
A: You can consider the $x$-intercepts instead: for the first line, it is the  point $\bigl(\frac23, 0\bigr)$, and for the second line the point$\bigl(\frac53,0\bigr)$. The difference is equal to $1$, so you  obtain the second line through a horizontal translation with vector $(1,0)$ of the first.
A: Consider this.  For every point $x_0 \in \mathbb R$ both lines will have a point in it with the $x$ coordinate equal to $x_0$.
That is to say, if Let $(x_0, y_a)$ be a point in Line 1: $3x -6y = 4$ and lte $(x_0, y_b)$ be a point in Line 2: $3x - 5y = 10$.
The you have two equations
$3x_0 - 6y_a = 4$ and $3x_0 - 6y_b = 10$
but one equation is just $6$ than then the other so so we have
$3x_0 - 6y_b = 10 = 4+ 6 = 3x_0 -6y_a + 6$ and
$-6y_b = -6y_a + 6$
$y_b = y_a - 1$.
So for every point $(x_0, y_a)$ in Line 1: there will be a corresponding point $(x_0, y_a - 1)$ on Line 2.

In other words the graph of Line 2 is simply the graph of Line 1 transposed down by a unit.

.......
By this argument we can know that in general two lines of the form $ax + by = M_1$ and $ax+by = M_2$ will always be parallel.  (We can take this as a known result--- and not a surprising you as $ax + by = M_{1,2}\implies y = -\frac ab+\frac {M_{1,2}}b$ have the same friggin' slope after all )
And we can take it in the future $(ax_0 + by_2) - (ax_0 +by_1) = M_2 - M_1$ so
$y_2 = y_1 + \frac {M_2 - M_1}b$.
Once done forever, informed.
......
From now on, I'd accept the following as an acceptable answer:

$3x - 6y = 4$ and $3x -6y = 10$ are parallel lines as the have the same $x,y$ coefficients.  The shift is the difference of the constants, $10-4 = 6$ divided by the $y$ coefficient $-6$ so the second line is shifted by $\frac {6}{-6} = -1$ or down by $1$.

=====
Note:  You can do the same thing to by picking an $y_0$ and comparing $x_a$ to $x_b$ and get $x_b = x_a + 2$ or a shift to the right by $2$.
The "distance" between the two lines is $10 -4=6$ so that means if you pick a point $(x_0, y_0)$ on line one the nearest point on line 2 will be $6$ away.  The nearest point is in a perpendiculare path so the whole line is shifted in a perpendicular direction by $6$ units.
That is as the slope is $y = \frac 12 x + C$ the slope of the perpendicular is $-2$ so a point $(x_0, y_0)$ shifts to $(x_0 + k, y_0 -2k)$ where $k$ makes the distance between the two lines $6$.
That is $\sqrt{k^2 + 4k^2} = 6$ or $k = 6\sqrt {\frac 15}$.....
.... but, I can't say that in the end that was a quick or clear answer.  It started out nice "shift to line $6$ units perpendicularly" but the exact expression $(x,y) \mapsto (x + 6\sqrt {\frac 15}, y - 12\sqrt{\frac 15})$ is ... not so nice.
Oh, well.  That's mathematics.
