For two linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, parallel lines occur when $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$. Identify $a_1, b_1, a_2, b_2$ from the given system and write this ratio equation.

First, simplify $\dfrac{b_1}{b_2} = \dfrac{-3/5}{-24/5}$. Then set $\dfrac{k - 1/2}{4}$ equal to that value and solve that simple equation for $k$.

After finding $k$, compare $\dfrac{c_1}{c_2}$ with the common ratio of $\dfrac{a_1}{a_2}$ and $\dfrac{b_1}{b_2}$. If the constant ratio is different, the system has no solution.

Type the first equation as (k - 1/2)x - 3/5 y = 7/2 and the second as 4x - 24/5 y = 14. Desmos will prompt you to add a slider for k; create it.

Adjust the k slider and watch the two lines. For most values of k, you should see the lines intersect at a point (the system has a solution).

Move the k slider until the two lines have exactly the same slope (they look parallel) but cross the y-axis at different points. At that slider value, the lines never meet, so the system has no solution. Read that value of k from the slider.

After solving for k by hand, plug that value into the first equation in Desmos (replace k with your number), keep the second equation the same, and confirm that the lines are parallel and do not intersect. If an intersection point still appears, recheck your algebra.

For a system of two linear equations

there is no solution when the lines are parallel but different. This means:

• The coefficient ratios for $x$ and $y$ are equal: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ (same slope), but
• The constant ratio is not equal to that same value: $\dfrac{c_1}{c_2} \ne \dfrac{a_1}{a_2}$.

So we will use $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ to find $k$.

From the given system

identify:

• First equation: $a_1 = k - \tfrac12$, $b_1 = -\tfrac35$, $c_1 = \tfrac72$.
• Second equation: $a_2 = 4$, $b_2 = -\tfrac{24}{5}$, $c_2 = 14$.

Compute the $y$-coefficient ratio:

For parallel lines, we need

so

This equation will let us solve for $k$.

Solve

Multiply both sides by 4:

Add $\tfrac12$ to both sides to isolate $k$:

Now check the constants ratio:

which is not equal to $\tfrac18$. So with $k = 1$, the two equations have matching $x$- and $y$-coefficient ratios but a different constant ratio, meaning the lines are parallel and distinct. Therefore, the value of $k$ that makes the system have no solution is $1$.