Problem 66 - College Algebra

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, and 46 students failed.

Step 1: Define the Variables

Let \( x \) represent the number of students who passed the class, and \( y \) represent the number of students who failed the class.

Step 2: Set Up the Equations Based on the Problem

We know from the problem statement two things: the total number of students enrolled, and the relation between those who passed and those who failed. This gives us the two equations we need:1. \( x + y = 276 \) (Total number of students)2. \( x = 5y \) (Five times as many students passed as failed)

Step 3: Substitute the Expression for One Variable

Using the second equation \( x = 5y \), substitute \( x \) in the first equation:\( 5y + y = 276 \)

Step 4: Solve for One Variable

Simplify and solve the equation:\( 6y = 276 \)Divide both sides by 6:\( y = 46 \)This means 46 students failed the class.

Step 5: Solve for the Other Variable

Substitute \( y = 46 \) back into the expression for \( x \):\( x = 5y \)\( x = 5 \times 46 \)\( x = 230 \)So 230 students passed the class.

Key Concepts

Linear Equations

Linear equations are mathematical equations that graph as straight lines. These equations have the standard form, which is often written as \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables. Linear equations are crucial in solving real-world problems because they help us understand relationships between different quantities.
In the given problem, the system of linear equations involves two equations that describe the relationship between the number of students who passed and failed a class. The simplicity of linear equations allows us to efficiently find solutions and understand the dynamics between variables. By combining both equations, students can visualize how two separate conditions can intersect and help identify respective values.

Variables

Variables are placeholders for numbers and are essential in forming equations. In this exercise, we assign variables to represent unknown quantities we want to solve for:

• \( x \) represents the number of students who passed the chemistry class
• \( y \) represents the number of students who failed the class

Using these two variables, the problem is framed as a system of equations. This allows us to express the given relationships and conditions algebraically. It is important to define variables clearly, ensuring every step is based on correct assumptions and calculations.

Substitution Method

The substitution method is one of the most common techniques for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one unknown, making it simpler to solve.
In our problem, the substitution method is used as follows:

• From the second equation \( x = 5y \), solve for \( x \).
• Substitute \( 5y \) into the first equation \( x + y = 276 \).
• This results in \( 5y + y = 276 \), which simplifies to \( 6y = 276 \).

This technique is powerful, especially when one equation can be easily manipulated to express a variable in terms of others. It ensures accuracy and efficiency in finding the solutions.

Solving Equations

Solving equations forms the core of this exercise. Once the substitution has been made, and a single equation is obtained (\( 6y = 276 \)), the next step involves isolating the variable:

• Divide both sides by 6 to solve for \( y \), giving \( y = 46 \).
• Use the solution for \( y \) to find \( x \) by substituting back into the equation \( x = 5y \).
• Thus, \( x = 5 \times 46 = 230 \).

This process of solving equations shows how each step builds on the one before it. It highlights the logical progression needed to find the solution. Consistently checking each step against given conditions ensures that mistakes are caught and the solution accurately reflects the problem's requirements.