Rewrite each equation in the form $y = mx + b$ so you can see the slopes clearly. Then set the slopes equal and solve for $k$ to find when the lines are parallel.

When you set the slopes equal, you should get an equation in $k$ that simplifies to a quadratic. Once you have that quadratic, think about how to find the sum of its solutions without fully solving for each root.

For a quadratic $ak^2 + bk + c = 0$, the sum of the two roots is $-\dfrac{b}{a}$. Apply this to the quadratic you found in $k$.

In Desmos, type (k+2)x + 3y = 7 and 2x + (k-1)y = 4. Desmos will prompt you to add a slider for k; create that slider so you can vary $k$.

Move the $k$ slider and watch the two lines. You are looking for values of $k$ where the lines have the same slope (they never meet) but are not on top of each other (different intercepts). Record all such $k$-values you observe.

Once you have the two $k$-values that make the lines parallel and non-intersecting, add them together (you can type k1 + k2 into Desmos or a calculator) to verify the sum.

Think of each equation as a line in the $xy$-plane.

• A unique solution means the lines intersect once.
• No solution means the lines are parallel and distinct (same slope, different intercepts).
• Infinitely many solutions means the two equations describe the same line.

So we must find all $k$ for which the two equations represent parallel but different lines.

Start with the first equation:

Solve for $y$:

So its slope is $m_1 = -\dfrac{k+2}{3}$.

Now the second equation:

Solve for $y$ (noting $k \neq 1$ to avoid division by zero):

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

For no solution, we need the lines to be parallel, so their slopes must be equal:

which simplifies to

Solve

Cross-multiply:

Expand the left-hand side:

Set equal to 6 and rearrange:

Any $k$ that makes the system have no solution must satisfy this quadratic.

(If $k = 1$, the second equation becomes $2x = 4$ or $x = 2$, which intersects the first line at exactly one point, so it does not give no solution.)

Also, if the intercepts were equal (same $b$ in $y = mx + b$), the lines would be identical and give infinitely many solutions. Solving $\dfrac{7}{3} = \dfrac{4}{k-1}$ gives $k = \dfrac{19}{7}$, which does not satisfy $k^2 + k - 8 = 0$, so our quadratic’s solutions really correspond to parallel, distinct lines (no solution).

We now know the relevant $k$-values are the solutions of

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

Here, $a = 1$ and $b = 1$, so the sum of all real $k$ satisfying this equation is

Therefore, the sum of all real values of $k$ for which the system has no solution is $-1$.