Find $t$ such that $(t,5)$ lies on the line through $(0,3)$ and $(-8,0)$.
The slope of the line through $(0,3)$ and $(-8,0)$ is $(0-3)/(-8-0) = 3/8$. If $(t,5)$ is also on this line, then the slope of  the line through $(t,5)$ and $(0,3)$ must also be $3/8$.  Therefore, we must have \[\frac{3-5}{0-t} = \frac{3}{8} \implies \frac{2}{t} = \frac{3}{8} \implies (2)(8) = 3(t) \implies t = \boxed{\frac{16}{3}}.\]
\frac{16}{3}
Level 4
Algebra
Problem: Find $t$ such that $(t,5)$ lies on the line through $(0,3)$ and $(-8,0)$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $t$, we first need to determine the equation of the line that passes through the points $(0,3)$ and $(-8,0)$. 

1. Calculate the slope of the line using the formula: 
   Slope $m = \frac{y_2 - y_1}{x_2 - x_1}$. 
   Here, $(x_1, y_1) = (0, 3)$ and $(x_2, y_2) = (-8, 0)$.
   Thus, $m = \frac{0 - 3}{-8 - 0} = \frac{-3}{-8} = \frac{3}{8}$.

2. Use the point-slope form of the line equation, which is $y - y_1 = m(x - x_1)$. 
   Substituting $(x_1, y_1) = (0, 3)$ and $m = \frac{3}{8}$ gives:
   $y - 3 = \frac{3}{8}(x - 0)$.
   Simplifying this, we get:
   $y - 3 = \frac{3}{8}x$,
   or $y = \frac{3}{8}x + 3$.

3. To find $t$ such that the point $(t, 5)$ lies on this line, substitute $y = 5$ into the line equation:
   $5 = \frac{3}{8}t + 3$.

4. Rearranging this equation to solve for $t$:
   $5 - 3 = \frac{3}{8}t$,
   $2 = \frac{3}{8}t$.

5. Multiply both sides by $\frac{8}{3}$ to isolate $t$:
   $t = 2 \cdot \frac{8}{3} = \frac{16}{3}$.

Thus, the value of $t$ that satisfies the condition is $\frac{16}{3}$.<|END_OF_LATENT|> The final answer is $\frac{16}{3}$.