Solve each problem. See Example 1. Michael must build a rectangular storage shed. He wants the length to be 6 ft greater than the width, and the perimeter will be 44 ft. Find the length and the width of the shed.
Table of contents
- 0. Review of Algebra4h 18m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations1h 43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 5m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 22m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Linear Equations
Problem 37
Textbook Question
Solve each problem. See Example 4. In planning her retirement, Kaya deposits some money at 2.5% interest, and twice as much money at 3%. Find the amount deposited at each rate if the total annual interest income is \$850.
Verified step by step guidance1
Define variables for the amounts deposited: let \(x\) be the amount deposited at 2.5%, and since twice as much is deposited at 3%, the amount at 3% is \$2x$.
Write expressions for the annual interest earned from each deposit: the interest from the 2.5% deposit is \$0.025 \times x\(, and the interest from the 3% deposit is \)0.03 \times 2x$.
Set up an equation representing the total annual interest income: \$0.025x + 0.03 \times 2x = 850$.
Simplify the equation by combining like terms: \$0.025x + 0.06x = 850\(, which simplifies to \)0.085x = 850$.
Solve for \(x\) by dividing both sides of the equation by 0.085, then find the amount deposited at 3% by calculating \$2x$.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Simple Interest Calculation
Simple interest is calculated using the formula I = P × r, where I is the interest earned, P is the principal amount, and r is the interest rate expressed as a decimal. Understanding how to compute interest for different principal amounts and rates is essential to solve problems involving total interest income.
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Setting Up Algebraic Equations
Translating word problems into algebraic equations involves defining variables for unknown quantities and expressing relationships between them. In this problem, assigning variables to the amounts deposited and using given ratios helps form equations that represent the total interest earned.
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Solving Systems of Linear Equations
When multiple unknowns are involved, systems of linear equations can be solved using substitution or elimination methods. Here, two equations derived from the interest amounts and total interest allow finding the exact deposits at each interest rate.
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