Two lines have no solution when they are parallel but not the same line. How can you express "parallel" using the slopes of the two lines?

Setting the slopes equal will give you an equation in $k$. This equation may have two solutions; think about how you can find the sum of these solutions without necessarily solving for each individually.

After finding the equation that makes the slopes equal, be sure the $y$-intercepts of the two lines are not equal for those $k$-values, so that the lines are distinct and the system really has no solution.

In Desmos, type the expression f(x) = (x-1)/2 - 3/(x+2). This represents the difference between the slopes of the two lines as a function of $x$ (standing in for $k$).

Look at the graph of $f(x)$. The $x$-coordinates where the graph crosses the $x$-axis (where $f(x)=0$) are the values of $k$ that make the two slopes equal, so the lines are parallel.

You can check that these $k$-values do not also make the $y$-intercepts equal (for example, by quickly computing $2$ and $6/(k+2)$ for each value). Then add the $x$-coordinates (the $k$-values) you found from the intercepts of $f(x)$ to get the required sum.

Rewrite each equation to solve for $y$.

First equation:

Subtract $(k-1)x$ from both sides:

Divide by 2:

So the slope of the first line is $m_1 = -\dfrac{k-1}{2}$ and the $y$-intercept is $2$.

Second equation:

Subtract $3x$ from both sides:

Divide by $k+2$ (note $k \neq -2$ so we are not dividing by zero):

So the slope of the second line is $m_2 = -\dfrac{3}{k+2}$ and the $y$-intercept is $\dfrac{6}{k+2}$.

For the system to have no solution, the lines must be parallel and distinct. Parallel means their slopes are equal:

so

Multiply both sides by $-1$:

Now cross-multiply:

Expand the left side:

Subtract 6 from both sides to get a quadratic equation in $k$:

Any $k$ that makes the lines parallel must satisfy this equation.

Parallel lines have no solution only if they are different lines. Two lines with the same slope are the same line if their $y$-intercepts are also equal.

From Step 1:

• First line has $y$-intercept $b_1 = 2$.
• Second line has $y$-intercept $b_2 = \dfrac{6}{k+2}$.

For them to be distinct, we need:

If they were equal,

Multiply both sides by $k+2$:

But check whether $k=1$ satisfies the quadratic $k^2 + k - 8 = 0$:

So $k=1$ is not one of the solutions of the quadratic. Therefore, every solution of $k^2 + k - 8=0$ automatically gives two parallel, distinct lines, so those are exactly the $k$-values that make the system have no solution.

We now need the sum of all real values of $k$ that satisfy

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

Here, $a = 1$, $b = 1$, $c = -8$, so the sum of the roots (the two valid $k$-values) is

Answer: $-1$.