Problem 66 - College Algebra with Corequisite Support

Question

For the following exercises, use a system of linear equations with two variables and two equations to solve. 276 students enrolled in a freshman-level chemistry class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed.

Step-by-Step Solution

Answer

230 students passed, 46 students failed.

Step 1: Define the Variables

Let's define the variables needed to solve this problem. Let \( p \) be the number of students who passed the class, and \( f \) be the number of students who failed the class. Therefore, we have two unknowns: \( p \) and \( f \).

Step 2: Set Up the First Equation

We know the total number of students, both those who passed and those who failed, sums up to 276. This gives us our first equation: \( p + f = 276 \).

Step 3: Set Up the Second Equation

According to the problem, 5 times the number of students who passed is equal to the number of students who failed. This relation can be written as the second equation: \( p = 5f \).

Step 4: Solve the Equations

We now have two equations: \( p + f = 276 \) and \( p = 5f \). Substitute the second equation into the first equation: \( 5f + f = 276 \). Simplify to find \( 6f = 276 \).

Step 5: Solve for Failed Students

Divide both sides of the equation \( 6f = 276 \) by 6 to find \( f \). \( f = \frac{276}{6} = 46 \). Therefore, 46 students failed the class.

Step 6: Solve for Passed Students

Use the relationship \( p = 5f \) to find \( p \). Since \( f = 46 \), calculate \( p = 5 \times 46 = 230 \). Hence, 230 students passed the class.

Key Concepts

Understanding Two Variables

When tackling problems involving systems of linear equations, it's important to understand the concept of **two variables**. In most algebra problems, variables represent quantities that can change.
In our exercise, we have two unknown quantities: the number of students who passed the class, and the number who failed. We define these as:

• \( p \) - representing the number of students who passed
• \( f \) - representing the number of students who failed

By defining these variables clearly, you simplify the process of setting up equations and solving the problem without confusion.

Linear Equations in Algebra

A **linear equation** is an equation that makes a straight line when it is graphed. It has no variables with exponents higher than one. In algebra, linear equations often look like \( ax + by = c \), where \( x \) and \( y \) are variables.
Our problem involves two linear equations:

• The first equation, \( p + f = 276 \), signifies the total number of students.
• The second equation, \( p = 5f \), shows the relationship between those who passed and those who failed.

These equations are linear because the variables \( p \) and \( f \) are not raised to a power other than one. This simplicity helps in problem-solving, allowing us to use straightforward algebraic techniques to find solutions.

Problem Solving Using Algebra

**Problem solving in algebra** often involves setting up equations to represent a given problem and then solving those equations to find the unknown values. In this exercise, once we have our two linear equations:

• \( p + f = 276 \)
• \( p = 5f \)

We use substitution or elimination methods to find the values of \( p \) and \( f \).
We substituted \( p = 5f \) into the first equation to simplify it to \( 6f = 276 \), making it easier to solve for \( f \). Once \( f \) is found, it becomes simple to calculate \( p \) using the given relationship \( p = 5f \). This step-by-step logical process is a core principle in algebraic problem-solving.

Exploring Algebraic Expressions

An **algebraic expression** is a mathematical phrase that includes numbers, variables, and operations. It's important to distinguish between an expression and an equation; the latter includes an equality sign.
In this exercise, we rely on algebraic expressions like \( 5f \) to describe relationships between variables. These expressions are manipulated and combined to set up our equations.
Algebraic expressions allow us to model mathematical problems and apply operations to derive solutions. By expressing the problem in algebraic terms, we can leverage tools like substitution to systematically solve for unknown values, making complex problems more manageable.