Rewrite each equation in the form $y = mx + b$ and identify the slopes. How can you use these slopes to enforce the "no solution" condition?

Set the two slopes equal to make the lines parallel and solve the resulting equation for $k$. This gives a quadratic with two solutions—these are the two $k$-values you need.

Once you have a quadratic equation whose two solutions are the needed $k$-values, think about how to get the product of its roots directly from the coefficients, without solving for each root separately.

In Desmos, use $x$ in place of $k$. Enter the expression for when the lines are parallel: y = (x - 2)(x + 1) - 12. This corresponds to setting the slopes equal and rearranging to get $x^2 - x - 14$.

On the graph of y = (x - 2)(x + 1) - 12, tap the points where the curve crosses the $x$-axis. The $x$-coordinates of these intercepts are the two $k$-values for which the system has no solution.

In a new Desmos line, type the product of the two $x$-intercepts you found (for example, x1 * x2 if you stored them or directly multiply their decimal values). The resulting number shown by Desmos is the product of the two $k$-values.

The two equations represent two lines in the $xy$-plane.

• A system has no solution when the lines are parallel and distinct (same slope, different $y$-intercepts).
• So we need the values of $k$ that make the two lines parallel but not the same line.

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

First equation: $(k-2)x + 4y = 8$

Solve for $y$:

So the slope of the first line is $m_1 = -\dfrac{k-2}{4}$.

Second equation: $3x + (k+1)y = 5k - 1$

Solve for $y$:

So the slope of the second line is $m_2 = -\dfrac{3}{k+1}$.

For the lines to be parallel, their slopes must be equal:

Solve the slope equation:

Cross-multiply:

The two solutions to $k^2 - k - 14 = 0$ are the values of $k$ that make the lines parallel.

Now we must check that for these $k$ the lines are not the same line (otherwise there would be infinitely many solutions instead of none).

• From the first equation, the $y$-intercept is always $2$.
• From the second equation, the $y$-intercept is $\dfrac{5k-1}{k+1}$.

Set these equal to find when the lines would be identical:

But $k = 1$ does not satisfy $k^2 - k - 14 = 0$, so the two solutions of $k^2 - k - 14 = 0$ give parallel distinct lines, meaning the system has no solution for exactly those two $k$-values.

Let the two values of $k$ that satisfy $k^2 - k - 14 = 0$ be $k_1$ and $k_2$.

For a quadratic $ak^2 + bk + c = 0$ with roots $k_1$ and $k_2$:

• The product of the roots is $k_1k_2 = \dfrac{c}{a}$.

Here, $a = 1$, $b = -1$, and $c = -14$, so

Thus, the product of the two values of $k$ for which the system has no solution is $-14$, which corresponds to choice C.