Answer :

Explanation:

Sure, here are five daily life situations represented as linear equations, along with their solutions:

**1. Situation:**

You go for a run and track your distance covered each day. On Day 1, you run 5 kilometers. Every day thereafter, you increase your distance by 0.5 kilometers.

**Linear Equation:**

Let \( D \) represent the distance covered in kilometers after \( n \) days.

\[ D = 5 + 0.5n \]

**Solution:**

- For Day 1 (\( n = 1 \)): \( D = 5 + 0.5 \times 1 = 5.5 \) kilometers

- For Day 2 (\( n = 2 \)): \( D = 5 + 0.5 \times 2 = 6 \) kilometers

- For Day 3 (\( n = 3 \)): \( D = 5 + 0.5 \times 3 = 6.5 \) kilometers

- and so on.

**2. Situation:**

You are saving money each month for a vacation. In the first month, you save $200, and each subsequent month you increase your savings by $50.

**Linear Equation:**

Let \( S \) represent the amount of money saved after \( n \) months.

\[ S = 200 + 50n \]

**Solution:**

- After 1 month (\( n = 1 \)): \( S = 200 + 50 \times 1 = 250 \) dollars

- After 2 months (\( n = 2 \)): \( S = 200 + 50 \times 2 = 300 \) dollars

- After 3 months (\( n = 3 \)): \( S = 200 + 50 \times 3 = 350 \) dollars

- and so on.

**3. Situation:**

You are driving and notice your fuel gauge decreases linearly as you drive. When the tank is full, the gauge shows 100%. After driving 300 kilometers, the gauge shows 75%.

**Linear Equation:**

Let \( F \) represent the fuel gauge percentage after driving \( d \) kilometers.

\[ F = 100 - \frac{25}{300} d \]

**Solution:**

- After driving 0 kilometers (\( d = 0 \)): \( F = 100 \)% (tank is full)

- After driving 300 kilometers (\( d = 300 \)): \( F = 100 - \frac{25}{300} \times 300 = 75 \)%

**4

Other Questions