Problem 32 - Elementary Algebra

Question

Suppose that Lou invested a certain amount of money at \(3 \%\) interest, and he invested \(\$ 750\) more than that amount at \(5 \%\). His total yearly interest was \(\$ 157.50\). How much did he invest at each rate?

Step-by-Step Solution

Answer

Lou invested $1500 at 3% and $2250 at 5%.

Step 1: Understand what we know

We have two investments. The first is at a 3% interest rate, and the second, which is $750 more than the first, is at a 5% interest rate. Their combined interest is $157.50.

Step 2: Define the variable

Let \( x \) be the amount invested at 3%. This means the amount invested at 5% is \( x + 750 \).

Step 3: Write the interest equations

The interest from the first investment is \( 0.03x \) and from the second investment is \( 0.05(x + 750) \). These together total $157.50.

Step 4: Set up the equation

Set up the equation for total interest: \[ 0.03x + 0.05(x + 750) = 157.50 \]

Step 5: Expand and simplify

Expand the equation:\[ 0.03x + 0.05x + 37.5 = 157.50 \]Simplify to:\[ 0.08x + 37.5 = 157.50 \]

Step 6: Solve for x

Subtract 37.5 from both sides:\[ 0.08x = 120 \]Divide by 0.08:\[ x = 1500 \]

Step 7: Find the amounts invested

The amount invested at 3% is $1500. The amount invested at 5% is \( 1500 + 750 = 2250 \).

Key Concepts

Simple Interest

Simple interest is a basic form of interest calculation that is applied to invest money or take a loan. It's an easy and widely used method to determine how much interest will earn or owe over a specific period. When using simple interest, the interest amount is calculated based only on the principal amount, or the initial amount of money.

• The formula for simple interest is: \[ I = P \times r \times t \] where \( I \) is the interest, \( P \) is the principal amount, \( r \) is the rate of interest per period, and \( t \) is the time in periods.
• In Lou's problem, the principal amounts were invested at two different interest rates, 3% and 5% respectively.
• This method can quickly illustrate potential gains from investments or costs of loans.

By understanding the concept of simple interest, one can easily figure out the interest generated or paid, simply by multiplying the principal by the interest rate and the time. This makes it a vital concept for personal finance management and basic algebra word problems.

Linear Equations

Linear equations form the backbone of many types of algebraic problems. They are equations of the first order and involve constants and a single variable. The solutions to such equations provide the values of the unknown we're trying to find.

• The standard form of a linear equation in one variable is \[ ax + b = c \], where \(a\) and \(b\) are constants, and \( x \) is the variable.
• In Lou's example, our equation representing the total interest is a linear equation: \[ 0.03x + 0.05(x + 750) = 157.50 \].
• Such equations are solved by isolating the variable on one side through operations like addition, subtraction, multiplication, and division.

Linear equations can be interpreted and used to solve real-life problems easily. They transform complex word problems into manageable mathematical statements allowing one to deduce viable solutions.By systematically solving the equation, we found out that Lou invested \( \\( 1500 \) at 3% and \( \\) 2250 \) at 5%.

Mathematical Modeling

Mathematical modeling involves using mathematical structures and equations to represent real-world scenarios succinctly. It involves translating word problems into mathematical languages that can be analyzed and solved.

• This begins with defining variables representing unknown quantities. In our case, we defined \( x \) as the amount invested at 3% interest.
• The relationships and constraints between different components then guide the formulation of equations. For instance, the total interest Lou earned set up the foundational equation for our model.
• Upon setting the equation, we utilize algebraic principles to solve the model consistently and accurately.

By employing mathematical modeling, complex problems are decomposed into structured forms, so intricate relations could be evaluated straightforwardly. It enhances decision-making processes by providing a clear mathematical foundation from which conclusions can be drawn efficiently.