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Ch. 1 - Equations and Inequalities
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 2, Problem 37

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.

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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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