Suppose you have entered a 109​-mile biathlon that consists of a run and a bicycle race. During your​ run, your average velocity is 5 miles per​ hour, and during your bicycle​ race, your average velocity is 28 miles per hour. You finish the race in 8 hours. What is the distance of the​ run? What is the distance of the bicycle​ race?

To solve this problem, let's define the variables:

- Let \( r \) be the distance of the run (in miles).
- Let \( b \) be the distance of the bicycle race (in miles).

We know the following:

1. The total distance of the biathlon is 109 miles:
   \[
   r + b = 109
   \]
   
2. The time taken to complete the run is given by the distance divided by the velocity. Since the average velocity during the run is 5 miles per hour, the time spent running is:
   \[
   \text{Time for run} = \frac{r}{5} \text{ hours}
   \]
   
3. Similarly, the time taken to complete the bicycle race, with an average velocity of 28 miles per hour, is:
   \[
   \text{Time for bicycle race} = \frac{b}{28} \text{ hours}
   \]
   
4. The total time for the entire race is 8 hours:
   \[
   \frac{r}{5} + \frac{b}{28} = 8
   \]

Now, we have two equations:

\[
r + b = 109 \quad \text{(1)}
\]
\[
\frac{r}{5} + \frac{b}{28} = 8 \quad \text{(2)}
\]

### Step 1: Solve Equation (1) for one variable

Let's solve for \( b \) in terms of \( r \):

\[
b = 109 - r
\]

### Step 2: Substitute into Equation (2)

Substitute \( b = 109 - r \) into the second equation:

\[
\frac{r}{5} + \frac{109 - r}{28} = 8
\]

### Step 3: Clear the fractions

To eliminate the fractions, multiply every term by the least common multiple (LCM) of 5 and 28, which is 140:

\[
140 \left(\frac{r}{5}\right) + 140 \left(\frac{109 - r}{28}\right) = 140 \times 8
\]

Simplifying this:

\[
28r + 5(109 - r) = 1120
\]

Distribute and combine like terms:

\[
28r + 545 - 5r = 1120
\]
\[
23r + 545 = 1120
\]

### Step 4: Solve for \( r \)

Subtract 545 from both sides:

\[
23r = 575
\]

Finally, divide by 23:

\[
r = 25 \text{ miles}
\]

### Step 5: Find \( b \)

Substitute \( r = 25 \) into the equation \( b = 109 - r \):

\[
b = 109 - 25 = 84 \text{ miles}
\]

### Conclusion

- The distance of the run is **25 miles**.
- The distance of the bicycle race is **84 miles**.

Would you like more details on any part of this solution or have any other questions? Here are some related questions:

1. How would the solution change if the total time were different?
2. What if the running speed was increased to 6 miles per hour?
3. How would the problem change if the race was 120 miles long instead of 109 miles?
4. What if the average biking speed decreased to 20 miles per hour?
5. How would you solve the problem using a graphical method?

**Tip:** When solving problems involving time, distance, and speed, it's helpful to set up clear equations that relate these variables, allowing you to solve for the unknowns step by step.

Learn how to solve a biathlon problem involving distances of a run and a bicycle race. Understand the application of time-speed-distance formulas with specific average velocities for each segment. Find out the distances of each segment with a step-by-step solution. Suitable for high school level mathematics.

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