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