At the y-intercept, what is the x-coordinate? Use this fact with the point $(k-5,b)$ to determine the value of $k$.

Once you know $k$, you know a specific point on the line, like $(5,13)$. Write a slope equation between this point and the intercept $(0,b)$, then solve that equation for $b$.

Use the fact that the y-intercept has x-coordinate $0$. Since the intercept is $(k-5,b)$, set $k-5=0$ to find $k=5$. The two given points become $(5,13)$ and $(12,-15)$.

Create a table in Desmos. In the first column (x1), enter $5$ and $12$. In the second column (y1), enter $13$ and $-15$. You should see the two points plotted.

In a new expression line, type y1 ~ m x1 + c. Desmos will perform a linear regression through the two points and display values for $m$ and $c$; $c$ is the y-intercept of the line.

The value shown for $c$ in the regression equation is the y-value where the line crosses the y-axis. That y-value is the same as $b$ in the problem.

The two given points are $(k,13)$ and $(k+7,-15)$.

Use the slope formula $m = \dfrac{y_2 - y_1}{x_2 - x_1}$:

So, the slope of the line is $-4$.

The y-intercept is where the line crosses the y-axis, so its x-coordinate must be $0$.

The y-intercept is given as $(k-5, b)$, so we set

and solve for $k$:

Now we know one concrete point on the line is $(5,13)$ (substituting $k=5$ into $(k,13)$).

The y-intercept has coordinates $(0,b)$. We already know another point on the line is $(5,13)$ and the slope is $-4$.

Use the slope formula between $(0,b)$ and $(5,13)$:

This equation connects $b$ with the known slope.

Start from the equation

Multiply both sides by $5$:

Subtract $13$ from both sides:

Multiply both sides by $-1$:

So, the value of $b$ is $33$.