Problem 57 - College Algebra

Question

Th ee coworkers work for the same employer. Their jobs are warehouse manager, office anager, and truck driver. The sum of the annual salaries of the warehouse manager and office anager is $$\$ 82,000$$. The office anager makes $$\$ 4,000$$ more than the truck driver annually. The annual salaries of the warehouse manager and the truck driver total $$\$ 78,000$$. What is the annual salary of each of the co-workers?

Step-by-Step Solution

Answer

Warehouse manager: $44,000, Office manager: $38,000, Truck driver: $34,000.

Step 1: Define Variables

Let's introduce variables to represent the annual salaries: \( W = \) warehouse manager's salary, \( O = \) office manager's salary, \( T = \) truck driver's salary. Our goal is to find the values of \( W \), \( O \), and \( T \).

Step 2: Set Up Equations

Using the information given, we set up the following equations:1. \( W + O = 82000 \)2. \( O = T + 4000 \)3. \( W + T = 78000 \)

Step 3: Solve for One Variable

Start with equation (2), \( O = T + 4000 \). Now, express \( O \) in terms of \( T \) as \( O = T + 4000 \).

Step 4: Substitute and Simplify

Substitute \( O = T + 4000 \) from Step 3 into Equation (1):\[ W + (T + 4000) = 82000 \]Simplify to get:\[ W + T = 78000 \]This equation is identical to Equation (3), so all conditions are consistent.

Step 5: Solve for Warehouse Manager's Salary

From equation (3):\[ W + T = 78000 \]Using Equation (1) \( W + O = 82000 \) and given \( O = T + 4000 \), substitute to get:\[ W + (T + 4000) = 82000 \]Simplifying:\[ W + T + 4000 = 82000 \]\[ W + T = 78000 \]Since it's the same equation again, solve for \( T \) in terms of another known variable:

Step 6: Solve Equations Simultaneously

With the equations \( W + T = 78000 \) and \( W + O = 82000 \), and knowing \( O = T + 4000 \), we subtract:\[ W + T - (W + T + 4000) = 82000 - 4000 - W - T = 78000 \]Solve to verify consistency and calculate \( T \):\[ T = \frac{(82000 - 4000 - 78000)}{2} = 34000 \]

Step 7: Find Other Salaries

Using \( T = 34000 \): - From \( O = T + 4000 \), substitute \( T = 34000 \):\[ O = 34000 + 4000 = 38000 \] - From equation \( W + T = 78000 \):\[ W + 34000 = 78000 \]\[ W = 78000 - 34000 = 44000 \]

Step 8: Verification Check

Verify all equations with these calculated salaries:- For \( W + O = 82000 \):\( 44000 + 38000 = 82000 \) (True)- For \( O = T + 4000 \):\( 38000 = 34000 + 4000 \) (True)- For \( W + T = 78000 \):\( 44000 + 34000 = 78000 \) (True)

Key Concepts

Variable Substitution

Variable substitution is a fundamental method used in solving systems of equations. This technique helps in replacing a variable with an equivalent expression from another equation. By doing so, we essentially reduce the number of variables in an equation, making it simpler to solve.

Let's consider a real-world problem. For instance, we are given that the office manager's salary is $4000 more than the truck driver's. We can set this relationship using variables. Suppose:

• \( T \) represents the truck driver's salary
• \( O \) represents the office manager's salary

The relationship becomes \( O = T + 4000 \). By substituting \( O \) with \( T + 4000 \) in either related equation, we're primarily working with just one variable, \( T \), which simplifies our calculations significantly.

In this problem, we substitute \( O = T + 4000 \) into the equation involving the warehouse manager and office manager's salaries, \( W + O = 82000 \). Thus it becomes \( W + T + 4000 = 82000 \). The substitution reduces complexity, allowing for clearer mathematical manipulation.

Linear Equations

Linear equations are an essential element in systems of equations. These are mathematical statements where each term is either a constant or the product of a constant and a single variable. Each equation forms a straight line when graphed, and their intersection points are potential solutions to the system.

We express the salaries of coworkers using a set of these linear equations:

• \( W + O = 82000 \)
• \( O = T + 4000 \)
• \( W + T = 78000 \)

In these equations, \( W \), \( O \), and \( T \) each represent one straight line when plotted. The task at hand is finding the intersection of these equations which would represent the salary of each of the coworkers.

By solving these linear equations, especially through substitution and simplification, we identify a singular solution where all these equations meet, which provides the unique salary corresponding to each position.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate or eliminate variables, allowing us to solve for unknowns. This technique is critical when dealing with systems of equations, where multiple variables are interdependent.

In our problem, after setting up the initial equations:

• \( W + O = 82000 \)
• \( O = T + 4000 \)
• \( W + T = 78000 \)

Through algebraic manipulation, we substitute \( O \) from the second equation into the first to eliminate one variable step by step. This simplifies the expression to a point where we have a single equation in terms of\( W + T \).

More manipulation comes when we simplify \( W + (T + 4000) = 82000 \) down to \( W + T = 78000 \). Continuing these manipulation steps with careful rearrangement provides us eventual clarity on values for all involved variables \( W = 44000 \), \( O = 38000 \), and \( T = 34000 \).

With proper handling of the equations, algebraic manipulation makes an otherwise complex problem more manageable.