Problem 73 - Intermediate Algebra

Question

Managing an Apartment. The manager of an apartment complex is also a tenant. He pays only three-fourths of the monthly rent that each of the remaining 5 tenants pays. Fach month, a total of \(\$ 5,520\) in rent is paid by the 6 occupants. How much rent does the manager pay?

Step-by-Step Solution

Answer

The manager pays $720 in rent each month.

Step 1: Define Variables

Let the rent paid by each tenant be \( x \). Therefore, the manager pays \( \frac{3}{4} x \) as he pays only three-fourths of what others pay.

Step 2: Write an Equation for Total Rent

The total rent paid by 5 regular tenants is \( 5x \) and the rent paid by the manager is \( \frac{3}{4}x \). Thus, the total rent for all 6 tenants can be represented as: \[ 5x + \frac{3}{4}x = 5520 \]

Step 3: Find a Common Denominator

To combine the terms on the left side of the equation, convert \( 5x \) to quarters, which is \( \frac{20}{4}x \). Then the equation becomes:\[ \frac{20}{4}x + \frac{3}{4}x = 5520 \]

Step 4: Simplify the Equation

Simplify the equation by combining like terms:\[ \frac{23}{4}x = 5520 \]

Step 5: Solve for x

Multiply both sides of the equation by 4 to eliminate the fraction:\[ 23x = 22080 \]Now divide by 23:\[ x = 960 \]

Step 6: Calculate the Manager's Rent

The manager pays three-fourths of the rent of a regular tenant. Thus:\[ \text{Manager's Rent} = \frac{3}{4} \times 960 = 720 \]

Key Concepts

Linear Equations

Linear equations are crucial in solving problems like the one about the apartment complex manager's rent. At the heart of these problems is finding the value of an unknown variable that makes an equation true. In this scenario, the equation involved represents the total rent paid by all tenants.

In general, a linear equation takes the form:

• \( ax + b = c \)

where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable that we need to solve for. Linear equations can often describe real-world situations like budgets, expenses, and, as shown here, rent payments.

Understanding linear equations allows us to break down complex problems and express relationships between numbers in a straightforward way. They are often the stepping stone to more advanced algebra topics, and mastering them is foundational for continued math success.

Fraction Operations

Dealing with fractions is an essential skill in algebra, especially when equations involve parts of a whole. In this rent problem, fractions help express exactly how much rent the manager pays compared to regular tenants. The manager's rent is a fraction (three-fourths) of the other tenants' rent.

Working with fractions involves these key operations:

• Addition and subtraction, requiring a common denominator.
• Multiplication and division, which are more straightforward.

In the exercise, adding fractions requires converting each term in the equation to a common denominator. For example, turning \( 5x \) into \( \frac{20}{4}x \) allows combining it directly with \( \frac{3}{4}x \).

This step highlights the importance of understanding fraction operations. It ensures that expressions are comparable, aiding in simplifying and eventually solving equations.

Variable Substitution

Variable substitution in algebra involves replacing variables with known values to verify or simplify equations. This technique is particularly useful when initially defining variables or checking solutions.

In this problem, defining \( x \) as the rent for one regular tenant sets the stage for calculating both the total rent and the manager's portion. Once you determine that \( x = 960 \), you can substitute back to see how it fits the equation and solve specifically for the manager's rent.

Variable substitution is valuable because it allows re-evaluating equations step-by-step. By substituting \( x \) back into relevant expressions, you can ensure each part of the solution is accurate. In our case, substituting \( x = 960 \) back into the manager's rent formula gives:

• \[ \text{Manager's Rent} = \frac{3}{4} imes 960 = 720 \]

This straightforward approach verifies that the solution is consistent with the original problem conditions.