Problem 43 - Algebra for College Students

Question

Solve each equation. Eva invested a certain amount of money at \(10 \%\) interest and \(\$ 1500\) more than that amount at \(11 \%\). Her total yearly interest was \(\$ 795\). How much did she invest at each rate?

Step-by-Step Solution

Answer

Eva invested $3000 at 10% and $4500 at 11%.

Step 1: Define variables

Let's define two variables. Let \( x \) be the amount Eva invested at \( 10\% \) interest, and \( x + 1500 \) be the amount Eva invested at \( 11\% \) interest.

Step 2: Set up the equation based on yearly interest

According to the problem, the total yearly interest is \( 795 \) dollars. We can express the interest from \( x \) dollars as \( 0.10x \) and the interest from \( x + 1500 \) dollars as \( 0.11(x + 1500) \). Therefore, the equation is:\[ 0.10x + 0.11(x + 1500) = 795 \]

Step 3: Expand and simplify the equation

Expand the expression \( 0.11(x + 1500) \) and simplify:\[ 0.10x + 0.11x + 165 = 795 \]Combine like terms:\[ 0.21x + 165 = 795 \]

Step 4: Solve for x

Subtract 165 from both sides to isolate the term with \( x \):\[ 0.21x = 630 \]Now, divide both sides by \( 0.21 \):\[ x = \frac{630}{0.21} = 3000 \]So, \( x = 3000 \).

Step 5: Calculate the second investment amount

Since \( x = 3000 \), Eva invested \( x + 1500 \) at \( 11\% \), which amounts to:\[ 3000 + 1500 = 4500 \]

Key Concepts

Linear Equations

Linear equations are fundamental tools in mathematics used to find unknown values. They are equations of the first degree, which means they have variables raised to the power of one. In the context of investments, linear equations help determine values like the amount of money invested to achieve a desired interest.

Here's how a linear equation works in investment interest problems:

• We define variables to represent unknown values, like the initial sum Eva invested.
• These variables are used to form an equation based on known relationships, such as the rate of interest and total interest.
• The goal is to solve this equation to find the value of the unknown.

In our problem, the equation formulated was: \[0.10x + 0.11(x + 1500) = 795\] This equation helps us figure out how much Eva invested at each interest rate. Understanding how to construct and solve these equations is crucial for solving similar financial problems.

Percentage Calculations

Percentage calculations are essential in understanding how much interest an investment earns over time. Interest rates are typically given as percentages, and calculating the actual dollar amount earned from these percentages is a key skill.

Here's a breakdown of percentage calculations:

• The percentage describes how much of something (like money) is considered relative to the whole (100%).
• To find the interest earned, we multiply the invested amount by the interest rate (expressed as a decimal).
• For example, an interest rate of 10% becomes 0.10 in decimal form.

In the exercise, to find the interest for each investment: - For the amount invested at 10%, calculate 0.10 times the amount; - For the amount invested at 11%, calculate 0.11 times the amount. These calculations provide the total interest which we use in our linear equation.

Algebraic Equations

Algebraic equations are used to express relationships and solve for unknowns using operations like addition, subtraction, multiplication, and division. They are versatile and appear when dealing with financial problems, such as our investment interest problem.

Key concepts of algebraic equations include:

• Using defined variables to represent unknown quantities (e.g., amount of money invested).
• Setting up an equation that reflects the problem scenario, such as total interest calculation.
• Solving the equation by manipulating both sides to isolate the variable.

In solving for Eva's investments, the algebraic equation \[0.21x + 165 = 795\] was achieved after expanding and combining like terms. Solving this equation involved simple algebraic steps—subtracting a constant and dividing by the coefficient of the variable. Mastery of such concepts ensures reliable solutions to complex real-world problems like investment distribution.