Question 30·Hard·Linear Equations in One Variable

After distributing, group the $x$-terms together on each side and the constant (number-only) terms together on each side.

Move all $x$-terms to one side and constants to the other so the equation looks like something times $x$ equals a constant. Then think: for what value of $k$ does this kind of equation fail to have any solution?

For an equation like $ax = b$, consider what happens when $a$ is zero and when $a$ is not zero. Apply that idea to the coefficient of $x$ you found, which depends on $k$.

Type the left side as y = ((k-4)/3)*(6x+2) - 5 and the right side as y = (2/3)*(3x + k) + (k-4)*x. Desmos will create a slider for k.

For each option ($k = 0, 3, 6, 9$), set the slider to that value and look at the two graphs. If the graphs intersect at some point, that value of $k$ gives at least one solution for $x$.

The correct choice is the value of $k$ for which the two lines are distinct and never cross (no intersection point on the graph). That value of $k$ makes the original equation have no solutions.

Start by expanding the parentheses on both sides.

Left side:

This simplifies to

Right side:

Combine like terms on the right to get

Now set the simplified left and right sides equal:

Subtract $(k-2)x$ from both sides:

The coefficient becomes $(2k-8)-(k-2) = k-6$, so

Now subtract $\tfrac{2k-23}{3}$ from both sides:

So the equation is equivalent to

Think about the simple equation $ax = b$.

• If $a \ne 0$, then $x = \dfrac{b}{a}$ and there is exactly one solution.
• If $a = 0$ and $b = 0$, then every $x$ works (infinitely many solutions).
• If $a = 0$ and $b \ne 0$, then no value of $x$ can satisfy the equation, so there are no solutions.

In our equation $(k-6)x = \tfrac{23}{3}$, the constant $\tfrac{23}{3}$ is not zero. For the equation to have no solutions, the coefficient of $x$ must be zero:

So the value of $k$ that makes the original equation have no solutions is $\boxed{6}$.