Rearrange each equation into the form $y = mx + b$ so you can express each slope $m$ in terms of $k$. How can you use those expressions to find when the lines are parallel?

After you set the slopes equal and solve the resulting equation for $k$, you should get two values. Make sure those values really give lines with no solution (not the same line), then add the two $k$-values and match that sum to one of the answer choices.

In Desmos, type k = 0 and click the slider icon so that k becomes a slider. Then enter the two equations exactly as given, using k as the parameter:

• (k + 2)x - 3y = 6
• 2x - (k - 4)y = k

Move the k slider and watch how the two lines change. Look for values of k where the lines have the same steepness (same slope) and never cross each other on the graph—this indicates no solution. You should see two distinct values of k where this happens.

Read off the two approximate k values from the slider at the points where the lines are parallel and separate. Use Desmos (for example, by typing their decimal approximations into a new expression line and adding them) or mental math to find their sum, then choose the answer choice whose value matches that sum.

Each equation represents a line in the $xy$-plane.

• A single solution means the lines intersect at one point.
• Infinitely many solutions means the two equations are the same line.
• No solution means the lines are parallel but distinct, so they never intersect.

So we need values of $k$ that make the two lines parallel (same slope) but not the same line.

Start with the first equation:

Solve for $y$:

• Subtract $(k+2)x$ from both sides: $-3y = -(k+2)x + 6$.
• Divide by $-3$: $y = \dfrac{k+2}{3}x - 2$.

So the first line has slope

Now the second equation:

Rewrite and solve for $y$:

• $-(k-4)y = -2x + k$
• Divide by $-(k-4)$ (for now assume $k \neq 4$ so we can divide):

So the second line has slope

For the lines to be parallel (and not vertical), their slopes must be equal:

Cross-multiply:

Expand the left side:

Simplify:

Move all terms to one side:

Now solve this quadratic using the quadratic formula:

So there are two values of $k$ that make the slopes equal: $k = 1 + \sqrt{15}$ and $k = 1 - \sqrt{15}$ (both different from 4).

We must ensure these parallel lines are not the same line.

• The first line has $y$-intercept $-2$.
• The second line has $y$-intercept $-\dfrac{k}{k-4}$.

The lines would be the same only if both slope and $y$-intercept matched. Setting the intercepts equal,

leads to $k = 8$, which is different from $1 \pm \sqrt{15}$. So for $k = 1 + \sqrt{15}$ and $k = 1 - \sqrt{15}$, the lines are parallel and distinct, giving no solution.

We also check the special case $k = 4$: then the second equation becomes $2x = 4$, or $x = 2$, a vertical line that clearly intersects the first line once, so $k = 4$ does not give no solution.

Thus the only $k$-values that make the system have no solution are $1 + \sqrt{15}$ and $1 - \sqrt{15}$. Their sum is

so the correct answer is 2, which corresponds to choice B.