Problem 270 - Elementary Algebra

Question

A business has two loans totaling \(\$ 85,000\). One loan has a rate of \(6 \%\) and the other has a rate of \(4.5 \%\). This year, the business expects to pay \(\$ 4650\) in interest on the two loans. How much is each loan?

Step-by-Step Solution

Answer

The loan amounts are \( \$55000 \) and \( \$30000 \).

Step 1 - Set up variables

Let the amount of the loan with a 6% interest rate be denoted as \( x \), and the amount of the loan with a 4.5% interest rate be denoted as \( y \).

Step 2 - Establish the total loan equation

The total amount of the two loans is \( x + y = 85000 \).

Step 3 - Establish the total interest equation

The total interest paid this year is expected to be \$4650. Therefore, the interest equation is \( 0.06x + 0.045y = 4650 \).

Step 4 - Solve for one variable

Solve the first equation for \( y \): \( y = 85000 - x \).

Step 5 - Substitute and solve

Substitute \( y = 85000 - x \) into the second equation: \( 0.06x + 0.045(85000 - x) = 4650 \).

Step 6 - Simplify the equation

Distribute and combine like terms: \( 0.06x + 3825 - 0.045x = 4650 \).

Step 7 - Isolate the variable x

Combine the \( x \) terms: \( 0.015x + 3825 = 4650 \), then solve for \( x \): \( 0.015x = 4650 - 3825 \) \( 0.015x = 825 \) \( x = 55000 \).

Step 8 - Find the value of the second loan

Substitute \( x = 55000 \) back into \( y = 85000 - x \): \( y = 85000 - 55000 \) \( y = 30000 \).

Step 9 - Verify the solution

Check that these amounts satisfy the interest equation: \( 0.06 \times 55000 + 0.045 \times 30000 = 3300 + 1350 = 4650 \). Both the total loan amount and the interest match the given conditions.

Key Concepts

linear equations

Linear equations are fundamental in algebra, representing relationships between variables with a straight line when graphed. They typically take the form of \( ax + b = c \), where \( x \) is the variable, and \( a \), \( b \), and \( c \) are constants.

In the given problem, we have two linear equations. The first one is derived from the total amount of the two loans:

\( x + y = 85000 \)

This denotes that the sum of the two loans equals \(85,000. The second equation comes from the total interest payment expectation:

\( 0.06x + 0.045y = 4650 \).

This indicates that the sum of the interests paid on both loans equals \)4650. These equations are crucial for solving the problem and represent the basis of creating and interpreting relationships in word problems. They help us find the unknown values by setting up proper relationships.

interest calculation

Interest calculation is an essential aspect of understanding financial problems. Interest is the cost of borrowing money, often expressed as a percentage of the loan amount. In this problem, each loan has a different interest rate:

• The first loan has a 6% interest rate.
• The second loan has a 4.5% interest rate.

To calculate the interest paid on each loan:

• For the first loan:
  \( 0.06x \)
• For the second loan:
  \( 0.045y \)

Combining these as per the problem statement gives us the total interest equation: \( 0.06x + 0.045y = 4650 \).

Understanding these calculations helps manage finances and solve more complex financial scenarios.

systems of equations

A system of equations is a set of two or more equations with the same set of variables. The goal is to find the values of these variables that satisfy all equations simultaneously. In the given problem, we have the system:

• \( x + y = 85000 \)
• \( 0.06x + 0.045y = 4650 \)

Solving systems of equations typically involves methods like:

• Substitution
• Elimination
• Graphical methods

By applying these methods, we can find the values of \( x \) and \( y \) that satisfy both equations. Systems of equations are powerful tools in various real-world applications, allowing us to find solutions to multiple constraints simultaneously.

variable substitution

Variable substitution is a technique used to solve systems of equations by replacing one variable with an equivalent expression derived from another equation. In our problem, we solve the first equation for \( y \):

\( y = 85000 - x \)

We then substitute this back into the second equation to simplify and solve for \( x \):

\( 0.06x + 0.045(85000 - x) = 4650 \).

This way, we turn the system of equations into a single equation with one variable, which is easier to solve. After finding \( x \), we can find \( y \) by substituting \( x \) back into the original equation. Variable substitution helps break down complex problems and make them more manageable.