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Magazine Chapter 5: Problem 28
Solve the systems of equations. In Exercises \(25-32\) it is necessary to set up the appropriate equations. All numbers are accurate to at least three significant digits. A person invests \(\$ 22,500,\) partly at \(5.00 \%,\) partly at \(6.00 \%,\) the remainder at \(6.50 \%,\) with a total annual interest of \(\$ 1308 .\) If the interest received at \(5.00 \%\) equals the interest received at \(6.00 \%\) how much is invested at each rate?
Short Answer
Expert verified
Invest \(\$ 8060.87\) at 5%, \(\$ 6717.39\) at 6%, and \(\$ 7704.8\) at 6.5%.
Step by step solution
01
Define the Variables
Let's denote the amount invested at 5% as \(x\), at 6% as \(y\), and at 6.5% as \(z\). Our task is to determine the values of \(x\), \(y\), and \(z\).
02
Set Up the Equations
Using the information provided, we can set up the following equations:1. Total investment: \(x + y + z = 22500\)2. Total annual interest: \(0.05x + 0.06y + 0.065z = 1308\)3. Equal interest from 5% and 6%: \(0.05x = 0.06y\)
03
Simplify Equation 3
From Equation 3, we know that \(0.05x = 0.06y\). Solving for one of the variables gives us \(x = \frac{0.06}{0.05}y = 1.2y\).
04
Substitute into Equation 1
Substitute \(x = 1.2y\) into the first equation: \(1.2y + y + z = 22500\).This simplifies to \(2.2y + z = 22500\).
05
Substitute into Equation 2 and Solve for y and z
Substitute \(x = 1.2y\) into the interest equation: \(0.05(1.2y) + 0.06y + 0.065z = 1308\).Simplify to get \(0.06y + 0.06y + 0.065z = 1308\) or \(0.12y + 0.065z = 1308\).
06
Solve the Two-Variable System
Use the equations \(2.2y + z = 22500\) and \(0.12y + 0.065z = 1308\).From \(2.2y + z = 22500\), express \(z\) in terms of \(y\): \(z = 22500 - 2.2y\).Substitute \(z\) in the second equation: \(0.12y + 0.065(22500 - 2.2y) = 1308\).
07
Solve for y
Expand and simplify \(0.12y + 1462.5 - 0.143z = 1308\).After expanding and rearranging, \(-0.023y = -154.5\).Solve for \(y\): \(y = 6717.39\).
08
Solve for z
Substitute \(y = 6717.39\) back into the equation \(z = 22500 - 2.2y\):\(z = 22500 - 2.2(6717.39) = 7704.8\).
09
Solve for x
Recall that \(x = 1.2y\). Substitute \(y = 6717.39\) to find \(x\):\(x = 1.2 \times 6717.39 = 8060.87\).
10
Verify the Solution
Verify the amounts by checking the total interest and total investment:1. Total investment: \(8060.87 + 6717.39 + 7704.8 = 22500\).2. Total interest: \(0.05 \times 8060.87 + 0.06 \times 6717.39 + 0.065 \times 7704.8 = 1308\).Both conditions are satisfied.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Investment
Investments are an exciting topic in financial mathematics, focusing on how to allocate money in different financial ventures in hopes of gaining positive returns. In our exercise, the person has decided to invest a total of $22,500 across three different interest rates. Each investment behaves differently, with specific rates or returns usually expressed as percentages.
- An investment involves setting aside an amount of money, often called the principal, in various financial plans or instruments like stocks, bonds, or savings accounts.
- Investment risk and return usually go hand in hand; generally, higher returns come with higher risks.
- In the context of equations, the distribution of this initial investment is crucial to understanding the overall outcome and how interests are calculated.
Interest Rates
Interest rates are a fundamental concept in both personal and corporate finance. They represent the cost of borrowing money or the return on money lent. Expressed as percentages, these rates directly affect the growth of an investment over time. In this scenario:
- The interest rates are 5%, 6%, and 6.5%, applied to different portions of the total investment.
- Calculating interest involves multiplying the principal amount by the interest rate.
- In our exercise, the equality condition between interest from 5% and 6% headings dictates the relationship between the investments under these rates.
Algebraic Solutions
The use of algebra is highlighted in this example to solve systems of equations involving investments and interest calculations. Algebra helps break down the investment scenario into manageable, solvable parts:
- We start by defining variables for each amount invested: let鈥檚 say \(x\) for 5%, \(y\) for 6%, and \(z\) for 6.5%.
- The equations set up illustrate different conditions: the overall sum of investments, total interest expectation, and an equal interest condition for 5% and 6% investments.
- Algebra enables solving these equations for \(x\), \(y\), and \(z\) by manipulations and substitutions, ensuring all given conditions are fulfilled.
Financial Mathematics
Financial mathematics is essential for understanding financial market complexities and personal finance management. It merges various mathematical disciplines, including algebra, calculus, and statistics, into practical applications:
- This discipline teaches us how to model financial markets and instruments, forecast economic trends, and manage financial portfolios effectively.
- It also involves understanding fundamental principles, such as the time value of money, risk assessment, and capital allocation through diverse financial techniques and tools.
- In our scenario, financial mathematics allows for precise modeling of investment scenarios to assess outcomes based on different interest rates and conditions.
