After expanding, move all the terms to one side so the equation looks like $\text{(something with }m)x + \text{(something with }m) = 0$.

For an equation like $ax + b = 0$, what happens if $a = 0$? Under what condition on $b$ would that make the equation impossible to satisfy for any $x$?

Identify the expression that plays the role of $a$ (the coefficient of $x$) and set it equal to zero, then ensure the constant part is not zero for that value of $m$.

In Desmos, type L(x) = (3m - 9)(x + 2) and R(x) = 6x + 12 - 4m. Desmos will automatically create a slider for m so you can change its value.

Move the m slider and watch where the graphs of $y = L(x)$ and $y = R(x)$ intersect. Each intersection point corresponds to a solution of the equation $(3m - 9)(x + 2) = 6x + 12 - 4m$ for that value of $m$.

Adjust the slider until the two lines have the same slope (they look parallel) but never cross. At that slider value, the equation has no intersection point, meaning there is no solution for $x$. Read that value of $m$ from the slider and match it to the correct answer choice.

Start with the equation

Expand the left-hand side:

So the equation becomes

Subtract the entire right-hand side from both sides:

Group like terms:

• $x$-terms: $[(3m - 9) - 6]x = (3m - 15)x$.
• Constant terms: $6m + 4m - 18 - 12 = 10m - 30$.

So the equation can be rewritten as

A linear equation of the form $ax + b = 0$ has:

• one solution if $a \ne 0$,
• no solution if $a = 0$ but $b \ne 0$.

Here, $a = 3m - 15$ and $b = 10m - 30$. For there to be no solution for $x$, the $x$-term must disappear, so we need

and we must also have $10m - 30 \ne 0$ so that a nonzero constant remains.

Solve $3m - 15 = 0$:

Check the constant term at this value:

So when $m = 5$, the equation becomes $0x + 20 = 0$, which is impossible (it says $20 = 0$), meaning no value of $x$ works. Thus, the equation has no solution when $m = 5$, which corresponds to choice C.