Write down the coefficients of $x$, $y$, and the constant term from both equations. How are $3$ and $6$ related? Whatever that relationship is, it must also hold between $2$ and $k$, and between $14$ and $28$.

You know the ratio between $6$ and $3$. Set up an equation so that the ratio between $k$ and $2$ is the same as the ratio between $6$ and $3$, then solve for $k$.

After you find a candidate value for $k$, multiply the entire first equation by the ratio and see if you get the second equation exactly. If you do, that value of $k$ gives infinitely many solutions.

In Desmos, type the first equation solved for $y$: y = (14 - 3x)/2. This is the line representing $3x+2y=14$.

For each option for $k$ (for example, $k=-6$), substitute it into $6x+ky=28$, solve for $y$, and graph that line in Desmos. For instance, with a generic $k$, you would rearrange to y = (28 - 6x)/k when $k \ne 0$, or to x = 28/6 when $k=0$.

Compare each graphed line with the first line. The value of $k$ that makes the second graph lie exactly on top of the first line (they coincide everywhere, not just intersect once) is the value that gives infinitely many solutions in the system.

A system of two linear equations has infinitely many solutions only when both equations describe the same line.

That means:

• The coefficient of $x$ in one equation must be a constant multiple of the coefficient of $x$ in the other.
• The same constant multiple must apply to the coefficient of $y$.
• And the same constant multiple must apply to the constant term (the number on the right side).

Compare the $x$-coefficients in the two equations:

• First equation: $3x+2y=14$ has $3$ as the coefficient of $x$.
• Second equation: $6x+ky=28$ has $6$ as the coefficient of $x$.

The ratio from the first to the second is

So, to represent the same line, every part of the first equation must be multiplied by $2$ to get the second equation.

Using the ratio $2$:

• The $y$-coefficient in the first equation is $2$.
• The constant term in the first equation is $14$.

For the equations to be the same line, we need:

• The $y$-coefficient in the second equation ($k$) to match $2$ times the first $y$-coefficient.
• The constant term $28$ to match $2$ times $14$ (which it already does).

So we must have the condition

Solve the equation from the previous step:

Multiply both sides by $2$:

Check: If we multiply the entire first equation $3x+2y=14$ by $2$, we get

which matches the form $6x + ky = 28$ when $k=4$. This means the two equations are the same line, so the system has infinitely many solutions when $k=4$.