For a system of two linear equations in $x$ and $y$ to have no solution, what must be true about the slopes of the lines? How does that translate into a relationship between the coefficients in standard form $Ax + By = C$?

Once both equations are in the form $Ax + By = C$, set $\dfrac{A_1}{A_2} = \dfrac{B_1}{B_2}$ and use the values $A_1 = 6$, $A_2 = 4$, $B_1 = -21$, and $B_2 = k$ to write an equation you can solve for $k$.

After you form the proportion with $k$, simplify the fraction $\dfrac{6}{4}$ first, then cross-multiply. Pay close attention to the negative sign when you solve for $k$.

Algebraically solve each equation for $y$ first:

• From $6x - 21y = 18$, get $y = \frac{2}{7}x - \frac{6}{7}$.
• From $4x + ky = 1$, get $y = \frac{-4}{k}x + \frac{1}{k}$.

In Desmos, enter the first equation as:

y = (2/7)x - 6/7

This draws the line corresponding to the first equation.

In a new Desmos line, type:

y = (-4/k)x + 1/k

Desmos will create a slider for k. This represents the second equation for different values of $k$.

Use the k slider to change the second line. Look for the value of $k$ where the second line has exactly the same slope as the first line (the lines are parallel) but different y-intercepts (they never overlap). The corresponding value of k on the slider is the solution.

Start with the first equation:

Move all the $y$-terms to the left side by subtracting $12y$ from both sides:

Now the system is:

Both are in the standard form $Ax + By = C$.

Two linear equations in $x$ and $y$ have no solution when they represent parallel but distinct lines.

For two lines in standard form $A_1x + B_1y = C_1$ and $A_2x + B_2y = C_2$:

• They are parallel if $\dfrac{A_1}{A_2} = \dfrac{B_1}{B_2}$.
• They are not the same line if this common ratio is not equal to $\dfrac{C_1}{C_2}$.

From our system:

• First equation: $A_1 = 6$, $B_1 = -21$, $C_1 = 18$.
• Second equation: $A_2 = 4$, $B_2 = k$, $C_2 = 1$.

Set the ratios of the $x$- and $y$-coefficients equal:

This equation will allow us to solve for $k$.

Start with the proportion:

Simplify $\dfrac{6}{4}$ to $\dfrac{3}{2}$:

Now cross-multiply:

This is a simple linear equation in $k$.

Solve $3k = -42$ by dividing both sides by $3$:

Now check that the ratio of the constants is not equal to the ratio of the coefficients of $x$:

• Coefficient ratio: $\dfrac{6}{4} = \dfrac{3}{2}$.
• Constant ratio: $\dfrac{18}{1} = 18$.

Since $\dfrac{3}{2} \ne 18$, the lines are parallel and distinct, so the system has no solution when