91Ó°ÊÓ

Linear equations are foundational in algebra and describe relationships between two or more variables through equations that represent straight lines on a graph. These equations include only terms that are either constants or products of a constant and a single variable. For example, in our exercise, we deal with the equations

• \( x + y + z = 800,000 \)
• \( 0.08x + 0.09y + 0.10z = 67,000 \)
• \( x = 5z \)

Each of these equations is a linear equation because they satisfy the condition of being a first-degree polynomial equation. Understanding how to set up and manipulate these equations is essential in solving them. In this exercise, our goal is to find values for \( x \), \( y \), and \( z \) that make all three equations true simultaneously.

Interest rate problems involve calculating the amount of interest earned or owed on borrowed or invested money. In these problems, interest is typically calculated using the formula \( I = Prt \), where \( I \) is the interest, \( P \) is the principal (amount of money borrowed or invested), \( r \) is the rate of interest per period, and \( t \) is the time the money is invested or borrowed.
In our exercise, the corporation borrowed different amounts at different rates of interest:

• 8% interest on amount \( x \)
• 9% interest on amount \( y \)
• 10% interest on amount \( z \)

The equation \( 0.08x + 0.09y + 0.10z = 67,000 \) emphasizes that the total interest from these three loans should equal the annual interest owed, which is $67,000. By setting up such an equation, students learn to calculate how different interest rates affect the total interest.

The substitution method is a technique to solve systems of equations where one equation is solved for one variable, and this expression is substituted into the other equations. This method simplifies solving the systems of equations.
For example, in the exercise given, the third equation \( x = 5z \) is used in the substitution process. By replacing \( x \) with \( 5z \) in the first and second equations:

• Substituting in \( x + y + z = 800,000 \), transforms it into \( 5z + y + z = 800,000 \).
• Substituting in \( 0.08x + 0.09y + 0.10z = 67,000 \), becomes \( 0.08 \times 5z + 0.09y + 0.10z = 67,000 \).

These transformations help in reducing the complexity by eliminating one variable, making it easier to solve for the other variables \( y \) and \( z \). Understanding and implementing the substitution method aids in simplifying complex problems into more manageable steps.