Problem 32 - Introductory Algebra for College Students

Question

Involve dual investments. A bank loaned out 250,000 dollar, part of it at the rate of \(8 \%\) annual mortgage interest and the rest at the rate of \(18 \%\) annual credit card interest. The interest received on both loans totaled 23,000 dollar How much was loaned at each rate? Organize your work in the following table. (TABLE CAN NOT COPY)

Step-by-Step Solution

Answer

The bank loaned $100,000 at an annual mortgage of \(8 \%\) and $150,000 at annual credit card interest of \(18 \%\).

Step one: Identify variables and constants

First, consider the total amount loaned as the sum of the two different loans. Let's represent the amount loaned at an annual mortgage of \(8 \%\) by 'x' and the amount loaned at annual credit card interest of \(18 \%\) by 'y'. From the problem, we know that x + y = 250,000. This is our first equation.

Step two: Formulate the second equation

Now, consider the total interest. From the problem, we know that the 8% interest from the first loan and the 18% interest from the second loan equals $23,000. We can write this as \(0.08x + 0.18y = 23,000\). This is the second equation.

Step three: Solve the system of equations

Substitute the first equation into the second to find the value of x. The solution is 100000 for x. Substitute x's value into the first equation to find y. The solution is 150000 for y.

Key Concepts

Algebraic Word Problems

Solving algebraic word problems involves translating real-world scenarios into mathematical expressions and equations. Here's a simplified breakdown of the process:

• Understand the problem: Carefully read the problem to comprehend what is asked and identify the relevant information.
• Define the variables: Assign variables to the unknown quantities. In our exercise, 'x' represents the amount loaned at 8% interest, and 'y' represents the amount loaned at 18% interest.
• Translate into equations: Convert the words into algebraic equations using the information given. In the example, we have two key pieces of information resulting in the equations:
  
  • Total loaned amount:
    \(x + y = 250,000\)
  • Total interest from both loans:
    \(0.08x + 0.18y = 23,000\)
• Solve the system: Use various mathematical techniques like substitution, elimination, or graphical methods to solve for the variables.
• Interpret the solution: Once you have the values of your variables, relate them back to the real-world context to ensure they make sense.

Remember to check your solution by plugging the variables back into original equations to ensure that they satisfy all conditions given in the problem statement.

Simple Interest Calculations

Simple interest is a way to calculate the interest charge on a loan, often depicted by the formula:
\[ Interest = Principal \times Rate \times Time \]
In this formula, the Principal is the initial amount of money loaned or invested, the Rate is the percentage of the principal charged as interest per period, and Time is the duration for which the money is loaned or invested.

Applying Simple Interest to Word Problems

In our exercise with dual investments, we didn't just calculate simple interest directly; instead, we used the concept of simple interest to set up our second equation. Given an 8% interest rate and an 18% interest rate, we had to find the respective amounts loaned that would result in a total interest of $23,000. Here simple interest calculations are integrated into the algebraic framework to find the solution.

Solving Linear Equations

Linear equations express a straight-line relationship between two variables, typically 'x' and 'y'. The general form is \(ax + by = c\), where 'a', 'b', and 'c' are constants. Solving linear equations usually involves finding the values for the variables that make the equation true.

Methods of Solving Linear Equations

Some common methods include:

• Graphing: Plotting the equations on a coordinate plane and finding the point where the lines intersect.
• Substitution: Solve one equation for one variable and substitute this into the other equation, as we did in our exercise solution where we found 'x' and substituted it to find 'y'.
• Elimination: Add or subtract equations to eliminate one variable and solve for the other.

In the context of our exercise, the substitution method led us to find the investment amounts with ease. It's a powerful technique that can simplify the process of dealing with multiple linear equations.