Rewrite each equation in the form $y = mx + b$ (being careful with the parameter $k$) to identify the slopes of both lines in terms of $k$.

Set the slopes equal to each other and solve for $k$; these $k$-values make the lines parallel. Then, check whether the corresponding lines are actually different (so they don't intersect) rather than being the same line.

After you find the algebraic condition on $k$, see which of the given choices satisfy it and give parallel but distinct lines.

Type the two equations into Desmos exactly as given: (k+1)x - 3y = 4 and 2x - (k-1)y = 5. Desmos will create a slider for k so you can adjust its value.

For each choice ($-2$, $0$, $1$, and $\sqrt{7}$), set the slider k to that value (for $\sqrt{7}$, type sqrt(7) in the slider input). Look at the graphs of the two lines in each case.

For each tested value of k, check whether the two lines intersect. The value of k that makes the lines parallel and not intersect anywhere on the graph corresponds to the correct answer choice.

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

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

So we want values of $k$ that make the two equations represent parallel, distinct lines.

Start with the first equation:

Solve for $y$:

So the slope of the first line is

Now take the second equation:

Solve for $y$ (this works as long as $k\neq 1$ so we are not dividing by 0):

So the slope of the second line is

For the lines to be parallel (and thus possibly have no solution), their slopes must be equal:

Cross-multiply:

Expand the left side:

Add 1 to both sides:

So any value of $k$ with $k^2=7$ makes the lines parallel. These are our candidate values for producing no solution.

Parallel lines give no solution only if they are different lines. If they were the same line, there would be infinitely many solutions.

Using slope-intercept form from Step 2:

• First line: $y = \dfrac{k+1}{3}x - \dfrac{4}{3}$ has $y$-intercept $b_1 = -\dfrac{4}{3}$.
• Second line: $y = \dfrac{2}{k-1}x - \dfrac{5}{k-1}$ has $y$-intercept $b_2 = -\dfrac{5}{k-1}$.

For the two lines to be the same, we would need both the slopes and the intercepts to be equal. We already forced the slopes to match by using $k^2 = 7$.

Now check whether the intercepts could also be equal:

Drop the negative signs:

Cross-multiply:

This value does not satisfy $k^2=7$, so when $k^2=7$ the intercepts are different. That means for $k^2=7$ the lines are parallel and distinct, so the system has no solution.

The equation $k^2 = 7$ gives $k = \pm \sqrt{7}$, and among the answer choices the matching value is $k = \sqrt{7}$, which is choice D.