Applications of First-Degree Equations

Financial Application

• Scenario: A financial manager has $15,000 to invest, with the goal of earning $1,020 in annual interest.
• Investment Options:
  • Tax-free bonds: 5% annual interest
  • Taxable bonds: 8% annual interest
• Objective: Determine how much to invest in each option to achieve the desired interest.

Solution Steps

1. Define Variables:
   • Let $x$ be the amount invested at 5%.
   • Remaining amount: $15,000 - x$ at 8%.
2. Calculate Interest:
   • Interest from $x$: $0.05x$.
   • Interest from $15,000 - x$: $0.08(15,000 - x)$.
3. Set Up Equation:
   • Total interest: $0.05x + 0.08(15,000 - x) = 1,020$.
4. Solve for $x$:
   • Simplify: $0.05x - 0.08x + 1,200 = 1,020$.
   • Rearrange: $-0.03x = 1,020 - 1,200$.
   • Simplify: $-0.03x = -180$.
   • Solve: $x = 6,000$.
5. Conclusion:
   • Invest $6,000 in tax-free bonds (5%) and $9,000 in taxable bonds (8%).

Car Speed Application

• Scenario: Two cars on a straight road.
  • First car speed: 60 mph
  • Second car speed: 45 mph
  • Distance between them: 90 miles
• Objective: Determine how long it will take the first car to catch up to the second car.

Solution Steps

1. Define the Problem:
   • First car travels faster and will catch up.
   • Determine the time $t$ for both cars to meet.
2. Set Up Equations:
   • Distance first car travels: $90 + D$
   • Distance second car travels: $D$
   • Equations:
     • $45t = D$
     • $60t = 90 + D$
3. Solve the Equations:
   • Substitute $D = 45t$ into $60t = 90 + 45t$.
   • Simplify: $15t = 90$.
   • Solve: $t = 6$ hours.
4. Verify:
   • First car travels $360$ miles (60 mph * 6 hours).
   • Second car travels $270$ miles (45 mph * 6 hours).
   • Consistent: $90 + 270 = 360$.

Conclusion

• Financial Application: Investment strategy to meet interest goals.
• Car Speed Application: Calculate time for one vehicle to catch up to another.
• Both examples demonstrate practical uses of first-degree equations in real-world scenarios.