Rewrite each equation in the form $y = mx + b$ so you can clearly see the slope of each line. What condition must their slopes satisfy for the lines to be parallel?

Parallel lines have the same slope, but for no solution they must not be the same line. After matching the slopes, what else must be different between the two equations?

For any chosen value of $k$ from the options, substitute that number into both equations and type them into Desmos. For example, for a candidate value $k = k_0$, enter: (2*k0 - 3)x + 4y = 5 and 3x + 6y = k0, replacing k0 with the specific number.

Look at the graphs of the two lines for that value of $k$. If the lines cross at exactly one point, that $k$ gives a unique solution and is not correct. If the lines lie on top of each other (infinitely many solutions), that $k$ is also not correct. You are looking for the value of $k$ where the two lines are parallel and never meet, meaning there is no solution. Test each answer choice until you find the one with no intersection point.

For two linear equations in $x$ and $y$, having no solution means their graphs are two parallel lines that never intersect.

• Parallel lines have the same slope.
• To have no solution (not infinitely many), they must also have different y-intercepts.

Start with the system:

First equation:

• $(2k-3)x + 4y = 5$
• $4y = -(2k-3)x + 5$
• $y = -\dfrac{2k-3}{4}x + \dfrac54$

So the slope of the first line is

Second equation:

• $3x + 6y = k$
• Divide everything by $3$: $x + 2y = \dfrac{k}{3}$
• $2y = -x + \dfrac{k}{3}$
• $y = -\dfrac12 x + \dfrac{k}{6}$

So the slope of the second line is

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

Multiply both sides by $4$ to clear the denominator:

Now solve this linear equation for $k$:

Keep this result and solve for $k$ in the next step, then check the intercepts.

From the previous step:

Now check the y-intercepts when $k = \frac52$:

• First line’s intercept: $b_1 = \dfrac54$ (from $y = -\dfrac{2k-3}{4}x + \dfrac54$, which does not depend on $k$).
• Second line’s intercept: $b_2 = \dfrac{k}{6} = \dfrac{\frac52}{6} = \dfrac{5}{12}$.

Since $\dfrac54 \ne \dfrac{5}{12}$, the lines are parallel with different y-intercepts, so the system has no solution when