Rewrite each equation in the form $y = mx + b$ so you can see the slope of each line as a function of $k$. Then, set those slopes equal to find an equation in $k$.

After you set the slopes equal, you will get a quadratic equation in $k$. Instead of fully solving for each root, think about how to find the sum of the roots directly from the quadratic’s coefficients.

Even after finding the $k$-values that make the slopes equal, you must still check whether the two equations become the same line (infinitely many solutions) or two different parallel lines (no solution). Compare the ratios of all coefficients, including the constants.

From your algebra work (after setting the slopes equal), you should have the quadratic equation $3k^2 - 6k - 36 = 0$. In Desmos, enter the function

treating $x$ as the variable standing in for $k$.

Use Desmos to find the $x$-intercepts (where the graph crosses the $x$-axis). These $x$-coordinates are the solutions to $3k^2 - 6k - 36 = 0$ and therefore are the $k$-values that make the two lines parallel.

If you want to see this geometrically, create a slider by defining k = (any starting value), then enter the two equations

$(3k - 6)x + 4y = 8$ and $9x + ky = 6$.

Move the slider to each of the $k$-values you found from the quadratic graph. You should see the two lines become parallel and not coincide. Then, outside Desmos, add those $k$-values to get the required sum.

For a system of two linear equations in $x$ and $y$:

• There is one solution if the lines intersect once (different slopes).
• There are infinitely many solutions if the two equations describe the same line (all coefficients are in the same ratio).
• There is no solution if the lines are parallel but different (same slope but not the same line).

So our first goal is to find all $k$ for which the two lines have the same slope.

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

First equation:

Solve for $y$:

So the slope is

Second equation:

Solve for $y$ (assuming $k \neq 0$; we will check $k=0$ separately):

So the slope is

For the lines to be parallel (same slope), set $m_1 = m_2$:

Cross-multiply:

So any $k$ that makes the slopes equal must satisfy

When $3k^2 - 6k - 36 = 0$, the lines are parallel. We still must rule out the possibility that they are actually the same line (which would give infinitely many solutions instead of none).

If two equations describe the same line, every coefficient (including the constant term) must be in the same ratio. Here, the constant terms have ratio

From the second equation, the ratio of the $y$-coefficients is $\dfrac{4}{k}$ (since the first equation’s $y$-coefficient is $4$ and the second’s is $k$). For the lines to be identical, we would need

which gives $k = 3$.

But $k = 3$ does not satisfy $3k^2 - 6k - 36 = 0$ (plugging in gives $27 - 18 - 36 \neq 0$), so there is no value of $k$ that makes the equations identical.

Therefore, every solution of $3k^2 - 6k - 36 = 0$ corresponds to a system with no solution (parallel, distinct lines).

We now need the sum of all real solutions of

For a quadratic $ak^2 + bk + c = 0$, the sum of its roots is $-\dfrac{b}{a}$.

Here, $a = 3$ and $b = -6$, so the sum of the roots is

The discriminant $b^2 - 4ac = (-6)^2 - 4(3)(-36) = 468 > 0$, so there are two real roots, and both correspond to no-solution cases. Thus, the sum of all real values of $k$ for which the system has no solution is 2.