Look at the coefficients of $x$ and $y$ in both equations. Is there a single number you can multiply the first equation’s coefficients by to get the second equation’s coefficients?

Once you find the multiplier that turns $6x - 8y$ into $9x - 12y$, apply that same multiplier to $24$ to find $k$.

In Desmos, enter the first equation as 6x - 8y = 24. This will draw the line representing the first equation.

Notice that $9$ is obtained from $6$ by multiplying by $9/6$. In Desmos, type 9/6 to see this multiplier as a simplified fraction or decimal.

To find $k$, the constant must be multiplied by the same factor. In Desmos, enter 24 * (9/6) and look at the output; that numerical result is the value of $k$ that makes the two equations represent the same line.

For a system of two linear equations to have infinitely many solutions, the two equations must represent the same line.

That happens when the ratios of the x-coefficients, the y-coefficients, and the constant terms are all equal:

Here, the first equation is $6x - 8y = 24$ and the second is $9x - 12y = k$.

Compare the $x$-coefficients and $y$-coefficients.

• For $x$: $6$ in the first equation and $9$ in the second.
  • The ratio is $\dfrac{9}{6} = \dfrac{3}{2}$.
• For $y$: $-8$ in the first equation and $-12$ in the second.
  • The ratio is $\dfrac{-12}{-8} = \dfrac{3}{2}$.

So the second equation is obtained by multiplying every term in the first equation by $\dfrac{3}{2}$ (which is $1.5$). To have infinitely many solutions, the constant term must also be multiplied by this same factor.

The constant in the first equation is $24$. Since the whole equation is scaled by $\dfrac{3}{2}$, the constant in the second equation must satisfy

Multiply $24$ by $\dfrac{3}{2}$ to get $k$:

So the value of $k$ that makes the system have infinitely many solutions is 36.