Rewrite each equation in the form $y = mx + b$ so you can clearly see the slope ($m$) and y-intercept ($b$) for each line.

Set the slopes of the two lines equal to each other to find $k$. Then check whether the y-intercepts are the same or different to decide if there is no solution or infinitely many solutions.

In Desmos, type 4x - 2y = 6 and press Enter. This graphs the first line.

For each answer choice, enter the corresponding second equation in Desmos:

• For A) type 0x - 3y = 12
• For B) type 3x - 3y = 12
• For C) type 6x - 3y = 12
• For D) type 9x - 3y = 12 Turn them on one at a time so you can compare with the first line.

For each tested equation, see whether Desmos shows an intersection point between that line and 4x - 2y = 6. The correct value of $k$ is the one for which the two lines are parallel (no intersection point appears), meaning the system has no solution.

For a system of two linear equations, there is no solution when the lines are parallel and distinct.

• Parallel lines have the same slope.
• Distinct lines have different y-intercepts, so they never cross.

So we need to find the value of $k$ that makes the two equations represent parallel lines with different intercepts.

Rewrite the first equation $4x - 2y = 6$ in the form $y = mx + b$:

• Subtract $4x$ from both sides: $-2y = -4x + 6$
• Divide by $-2$: $y = 2x - 3$

So the first line has slope $m_1 = 2$ and y-intercept $-3$.

Now rewrite the second equation $kx - 3y = 12$:

• Subtract $kx$ from both sides: $-3y = -kx + 12$
• Divide by $-3$: $y = \dfrac{k}{3}x - 4$

So the second line has slope $m_2 = \dfrac{k}{3}$ and y-intercept $-4$ (which does not depend on $k$).

For the lines to be parallel, their slopes must be equal:

Solve for $k$ by multiplying both sides by 3:

With $k = 6$, the first line is $y = 2x - 3$ and the second is $y = 2x - 4$. They have the same slope (2) but different y-intercepts ($-3$ and $-4$), so they are parallel and never intersect. Therefore, the system has no solution when $k = 6$.