Both equations are already in the form $ax + by = c$. For two lines to be parallel, how must the ratios of their x-coefficients and y-coefficients be related?

Write an equation that sets the ratio of the x-coefficients equal to the ratio of the y-coefficients, then simplify that equation to get a quadratic in $k$.

Once you have a quadratic equation whose solutions are the possible values of $k$, recall how to find the sum of its roots directly from its coefficients without solving for each root individually.

Treat $k$ as the x-variable in Desmos. Enter two functions: y = (x+2)/4 and y = 3/(x-1). These represent the two ratios $\frac{k+2}{4}$ and $\frac{3}{k-1}$ as functions of $k$.

Look for the intersection points of the two graphs. Click on each intersection; the x-coordinates are the values of $k$ where the lines in the original system are parallel (the condition for having either no solution or infinitely many solutions).

To see that these k-values do not make the equations identical, add a third function y = 12/9 (which equals $\frac{4}{3}$, the constant ratio). Verify that at your intersection x-values, the y-values of (x+2)/4 and 3/(x-1) are not both equal to 12/9; this confirms the system has no solution for those k-values, not infinitely many solutions.

Once you have the two x-coordinates from the intersections, type a new expression adding them (for example, k1 + k2 using the numbers you found). The result shown by Desmos is the sum of all possible k-values that make the system have no solution.

A system of two linear equations in x and y has no solution when the two lines are parallel but distinct.

For lines written as

"parallel" means the ratios of the x- and y-coefficients are equal:

but to avoid being the same line, that common ratio must not equal $c_1/c_2$.

Here,

• $a_1 = k+2$, $b_1 = 3$, $c_1 = 12$,
• $a_2 = 4$, $b_2 = k-1$, $c_2 = 9$.

So the parallel condition is

Now solve the equation

Cross-multiply:

Expand the left side:

So the equation becomes

Move 12 to the left to get a standard quadratic:

The values of $k$ that make the lines parallel are exactly the roots of this quadratic.

"Infinitely many solutions" would require all three ratios to be equal:

The condition we used, $k^2 + k - 14 = 0$, came only from

and if we instead set

we get a different equation and a different value of $k$ that does not satisfy $k^2 + k - 14 = 0$.

So the two roots of $k^2 + k - 14 = 0$ correspond to lines that are parallel but not the same line, which means the system has no solution for both of those k-values.

If a quadratic is written as

then the sum of its roots is $-\dfrac{b}{a}$.

Here the quadratic is

so $a = 1$ and $b = 1$, and the sum of the two possible values of $k$ is

Therefore, the sum of all possible values of $k$ is $-1$.