Think about what must be true about the two lines in the coordinate plane if the system has no solution. How are their slopes and intercepts related?

Rewrite both equations in the form $y = mx + b$. Set the slopes equal to find values of $k$ that make the lines parallel, then check whether the resulting equations are actually the same line or different lines.

Pick one answer choice for $k$ (for example, $k = -2$) and type the two equations into Desmos with that value substituted. For instance, if $k=-2$, enter -x + 2y = 8 and 3x + 12y = 24 as two separate lines.

Look at the graph: if the two lines cross at one point, the system has one solution; if they lie exactly on top of each other, there are infinitely many solutions; if they are parallel and never meet, that $k$ value gives no solution.

Repeat the substitution-and-graph process for $k = 0$, $k = 1$, and $k = 2$, each time checking whether the two lines intersect, coincide, or are parallel and distinct. The value of $k$ for which the lines are parallel but do not coincide is the correct answer.

Start by simplifying the second equation so it looks more like the first.

From

divide every term by $3$:

Now the system is

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

• If the lines intersect once, there is one solution.
• If they are the same line, there are infinitely many solutions.
• If they are parallel but different, they never meet, so there is no solution.

So we want $k$ such that the two lines are parallel but not the same line. That means:

• their slopes are equal, but
• their y-intercepts are different.

Rewrite each equation in slope–intercept form $y=mx+b$ to see the slopes.

From $(k+1)x + 2y = 8$:

• Move $(k+1)x$ to the right: $2y = -(k+1)x + 8$.
• Divide by $2$:

So slope $m_1 = -\dfrac{k+1}{2}$.

From $x + (2-k)y = 8$:

• Move $x$ to the right: $(2-k)y = -x + 8$.
• Divide by $(2-k)$ (assuming $k \ne 2$ for now):

So slope $m_2 = -\dfrac{1}{2-k}$.

Set slopes equal for parallel lines:

Multiply both sides by $-1$:

Cross-multiply:

Expand the left side:

Set equal to $2$:

Subtract $2$ from both sides:

Multiply by $-1$:

Factor:

so the possible values are $k = 0$ or $k = 1$.

Now test $k=0$ and $k=1$ in the simplified system.

For $k = 0$:

• First equation: $(0+1)x + 2y = 8 \Rightarrow x + 2y = 8$.
• Second equation (already simplified): $x + (2-0)y = 8 \Rightarrow x + 2y = 8$.

Both equations are the same line, so there are infinitely many solutions, not no solution.

For $k = 1$:

• First equation: $(1+1)x + 2y = 8 \Rightarrow 2x + 2y = 8 \Rightarrow x + y = 4$.
• Second equation: $x + (2-1)y = 8 \Rightarrow x + y = 8$.

These are two parallel lines ($x+y$ on the left in both), but with different constants ($4$ vs. $8$), so they never intersect: the system has no solution.

Therefore, the value of $k$ that makes the system have no solution is $k = 1$, which corresponds to choice C.