Rewrite $U=d+1$ to get $d$ in terms of $U$, and rewrite $V=3(t-5)$ to get $t$ in terms of $V$.

Substitute your expressions for $d$ and $t$ into the best-fit line equation, then solve for $V$.

The $V$-intercept happens when $U=0$. Plug $U=0$ into your equation for $V$.

In Desmos, compute the slope from the labeled points:

m=(9-15)/(8-2)

Then find the intercept $b$ using $(2,15)$:

b=15-m*2

Enter the line:

t = m d + b

Use $d=U-1$ and $t=V/3+5$. In Desmos you can type the substituted equation:

V/3 + 5 = m*(U-1) + b

Then solve (algebraically) for $V$ in terms of $U$.

After you have $V$ as a function of $U$, evaluate it at $U=0$ to get the $V$-intercept.

From the graph, the best-fit line passes through the labeled points $(2,15)$ and $(8,9)$.

The slope is

Using $(2,15)$ in $t=-d+b$:

So an equation for the line of best fit is

From $U=d+1$, we have

From $V=3(t-5)$, we have

Substitute $d=U-1$ and $t=\frac{V}{3}+5$ into $t=-d+17$:

Simplify the right side:

So

Multiply by $3$:

At the $V$-intercept, $U=0$, so

Therefore, the $V$-intercept is $39$.