Move all $x$ terms and constant terms to one side of the equation so it looks like $ax + b = 0$.

For a linear equation of the form $ax + b = 0$, what must be true about $a$ and $b$ so that there is no possible value of $x$ that makes the equation true?

Identify the expressions that play the roles of $a$ and $b$ in your simplified equation, then set up equations/inequalities that match the 'no solution' condition.

In Desmos, enter the two expressions as functions of $x$:

• y = k(x - 4) + 8
• y = 3x + 2 When you type k, Desmos will prompt you to add a slider for $k$; add that slider.

Adjust the $k$ slider and watch how the graph of $y = k(x - 4) + 8$ changes. You want the two lines to have the same steepness (same slope) but different vertical positions so they never intersect. When the lines are clearly parallel and do not cross, note the value of $k$ on the slider.

Click near where the lines might meet; Desmos normally labels intersection points. For the $k$ value you found, verify that Desmos does not show any intersection point between the two lines. The corresponding $k$ value is the one that makes the original equation have no solution.

Start with the given equation:

Distribute $k$ on the left:

Subtract $3x$ and $2$ from both sides to set the equation equal to $0$:

Combine like terms:

• Combine $kx$ and $-3x$ to get $(k - 3)x$.
• Combine $8 - 2$ to get $6$.

So the equation becomes:

Think of the equation in the form

• If $a \neq 0$, there is one solution: $x = -\dfrac{b}{a}$.
• If $a = 0$ and $b = 0$, then $0 = 0$ and there are infinitely many solutions.
• If $a = 0$ and $b \neq 0$, we get something impossible like $0 = 5$, so there are no solutions.

In our equation

we have:

• Coefficient of $x$: $a = k - 3$
• Constant term: $b = -4k + 6$

For no solution, we need:

• $k - 3 = 0$ (so the $x$ term disappears), and
• $-4k + 6 \neq 0$ (so we are left with a false statement like $0 = \text{nonzero}$).

First, set the coefficient of $x$ equal to $0$:

Now check the constant term when $k = 3$:

So the equation becomes something like $0x - 6 = 0$, that is $-6 = 0$, which is impossible. Therefore, the original equation has no solution only when $k = 3$. The correct answer is $k = 3$ (choice D).