Suppose you have entered a 131-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 29 miles per hour. You finish the race in 7 hours. What is try distance of the run? What is the distance of the bicycle race?

We can solve this problem using the concept of time and velocity. Let’s define the variables and then use the relationship between distance, velocity, and time:

- Let the distance of the run be \( d_r \) miles.
- Let the distance of the bicycle race be \( d_b \) miles.

### Step 1: Establish Equations

We know the following:
- Total distance of the biathlon is 131 miles: 
  \[
  d_r + d_b = 131
  \]
- Average velocity during the run is 5 miles per hour, so the time spent running is \( \frac{d_r}{5} \) hours.
- Average velocity during the bicycle race is 29 miles per hour, so the time spent cycling is \( \frac{d_b}{29} \) hours.
- Total time for both events is 7 hours:
  \[
  \frac{d_r}{5} + \frac{d_b}{29} = 7
  \]

### Step 2: Solve the System of Equations

We have the system of two equations:
1. \( d_r + d_b = 131 \)
2. \( \frac{d_r}{5} + \frac{d_b}{29} = 7 \)

#### From equation (1), solve for \( d_b \):
\[
d_b = 131 - d_r
\]

#### Substitute this into equation (2):
\[
\frac{d_r}{5} + \frac{131 - d_r}{29} = 7
\]

Now, solve for \( d_r \).

### Step 3: Solve for \( d_r \)
To eliminate the fractions, multiply through by the least common denominator of 5 and 29, which is 145:
\[
145 \left( \frac{d_r}{5} \right) + 145 \left( \frac{131 - d_r}{29} \right) = 145 \times 7
\]

This simplifies to:
\[
29d_r + 5(131 - d_r) = 1015
\]

Distribute the 5 on the left:
\[
29d_r + 655 - 5d_r = 1015
\]

Combine like terms:
\[
24d_r + 655 = 1015
\]

Subtract 655 from both sides:
\[
24d_r = 360
\]

Solve for \( d_r \):
\[
d_r = \frac{360}{24} = 15
\]

### Step 4: Solve for \( d_b \)
Now substitute \( d_r = 15 \) into equation (1):
\[
d_b = 131 - 15 = 116
\]

### Final Answer:
- The distance of the run is \( \mathbf{15} \) miles.
- The distance of the bicycle race is \( \mathbf{116} \) miles.

Would you like further details or clarification?

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Here are 5 related questions to consider:
1. How would the solution change if the total time was 8 hours instead of 7?
2. What would happen if the running speed increased to 6 miles per hour?
3. How would you solve the problem if the total distance of the biathlon changed to 150 miles?
4. Can you derive the formula for distance using average velocity and time?
5. How long would the run take if the distance was increased by 5 miles?

**Tip**: When solving systems of equations, always try to simplify one variable before substitution. This minimizes errors.

Suppose you have entered a 131-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 29 miles per hour. You finish the race in 7 hours. What is the distance of the run? What is the distance of the bicycle race?

In this detailed example, learn how to solve a 131-mile biathlon problem involving running and cycling with different average velocities. Using systems of linear equations, we find the distances for each portion of the race, making it suitable for students in Grades 8-10.

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