After finding each slope, write equations for $f(x)$ and $g(x)$ in the form $y = mx + b$ by plugging in one of the given points to solve for $b$.

Set your expressions for $f(x)$ and $g(x)$ equal to each other and solve the resulting linear equation for $x$.

When solving $3x - 16 = -\dfrac{5}{2}x + 25$, combine the x-terms correctly (convert $3x$ to a fraction with denominator 2) and then isolate $x$.

In Desmos, type (17 - (-7))/(11 - 3) on one line and (5 - 20)/(8 - 2) on another to verify the slopes $3$ and $-5/2$. Then, using these slopes and one point from each table, write and enter y = 3x - 16 and y = (-5/2)x + 25 as two separate functions.

With both lines graphed, tap or click on the point where the two lines cross. Desmos will display the coordinates of this intersection; the x-coordinate of that point is the value of $x$ where $f(x) = g(x)$.

Alternatively, enter a new function h(x) = (3x - 16) - ((-5/2)x + 25). Then find the x-intercept of this graph (where it crosses the x-axis); that x-value is the solution of $3x - 16 = -\dfrac{5}{2}x + 25$ and thus the x where $f(x) = g(x)$.

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

For $f$ (points $(3,-7)$ and $(11,17)$):

For $g$ (points $(2,20)$ and $(8,5)$):

So $f$ has slope $3$ and $g$ has slope $-\dfrac{5}{2}$.

Use the slope-intercept form $y = mx + b$.

For $f$:

• Assume $f(x) = 3x + b$.
• Plug in point $(3,-7)$: $-7 = 3(3) + b$, so $-7 = 9 + b$, giving $b = -16$.
• Thus $f(x) = 3x - 16$.

For $g$:

• Assume $g(x) = -\dfrac{5}{2}x + b$.
• Plug in point $(2,20)$: $20 = -\dfrac{5}{2}(2) + b = -5 + b$, so $b = 25$.
• Thus $g(x) = -\dfrac{5}{2}x + 25$.

We want the x-value where $f(x) = g(x)$, so set the equations equal:

Move all x-terms to one side and constants to the other:

• Add $\dfrac{5}{2}x$ to both sides.

Now $3x + \dfrac{5}{2}x = 25 + 16$.

Rewrite $3x$ with denominator 2: $3x = \dfrac{6}{2}x$, so

which gives $\dfrac{11}{2}x = 41$.

We have

Multiply both sides by $\dfrac{2}{11}$:

So the value of $x$ where $f(x) = g(x)$ is $\boxed{\dfrac{82}{11}}$, which corresponds to choice B.