Problem 65 - College Algebra

Question

Use a system of linear equations with two variables and two equations to solve. A total of 1,595 fi st- and second-year college students gathered at a pep rally. The number of freshmen exceeded the number of sophomores by 15\. How many freshmen and sophomores were in attendance?

Step-by-Step Solution

Answer

There were 805 freshmen and 790 sophomores in attendance.

Step 1: Define Variables

Let \( f \) represent the number of freshmen and \( s \) represent the number of sophomores in attendance. We are tasked with finding the values of \( f \) and \( s \).

Step 2: Set Up Equations

We know from the problem statement that the total number of students is 1,595, which gives us the first equation: \( f + s = 1595 \). Additionally, the number of freshmen exceeds the number of sophomores by 15, which gives us a second equation: \( f = s + 15 \).

Step 3: Substitute and Solve for One Variable

Substitute the expression from the second equation \( f = s + 15 \) into the first equation \( f + s = 1595 \). This gives us: \((s + 15) + s = 1595\). Simplify to solve for \( s \): \( 2s + 15 = 1595 \).

Step 4: Solve for Sophomores

Continue solving the equation from Step 3: \( 2s + 15 = 1595 \). Subtract 15 from both sides to get \( 2s = 1580 \). Divide both sides by 2 to find \( s = 790 \).

Step 5: Solve for Freshmen

Use the expression \( f = s + 15 \) to find \( f \). Substitute \( s = 790 \) into the equation: \( f = 790 + 15 \). Calculate this to find \( f = 805 \).

Step 6: Verify the Solution

Verify the solution by checking both conditions: The total number of students should be \( 1595 \), and \( f = 805 \) and \( s = 790 \) satisfy \( f = s + 15 \). Indeed, \( 805 + 790 = 1595 \) and \( 805 = 790 + 15 \), confirming the solution is correct.

Key Concepts

Variables in Algebra

In algebra, a variable is a symbol, typically a letter, used to represent a number in mathematical expressions or equations. Variables allow us to tackle complex problems by keeping them in a more general form until we know more information about the numbers they represent. In this exercise, we use the variables \( f \) and \( s \) for freshmen and sophomores, respectively.

• \( f \) represents the unknown number of freshmen students.
• \( s \) represents the unknown number of sophomore students.

Using variables helps us to set up equations based on the given situation. Here, understanding what each variable symbolizes is essential to solving the problem correctly. Once we relate these variables through equations, we can proceed to find their numeric values by substitution or elimination methods.

Solving Equations

To solve equations means to find the values of the variables that make the equations true. In our exercise, we are given a system of linear equations:

• \( f + s = 1595 \)
• \( f = s + 15 \)

These represent the relationships outlined in the problem. Solving such equations typically involves expressing one variable in terms of the other and substituting back to find both values.

The process begins by substituting \( f = s + 15 \) into \( f + s = 1595 \), resulting in a single equation in terms of \( s \). Simplifying gives us \( 2s + 15 = 1595 \).

• Step 1: Isolate \( s \) by subtracting 15 from both sides, resulting in \( 2s = 1580 \).
• Step 2: Divide by 2 to obtain \( s = 790 \).

Finally, substitute \( s \) back into \( f = s + 15 \) to find \( f = 805 \). Yay, you've solved for both variables!

College Algebra

College algebra teaches students how to handle more advanced topics in algebra, such as solving systems of linear equations, which is crucial for real-world problem-solving. This topic often involves breaking down problems using variables and setting up equations that describe the relationships between those variables.
In cases like our exercise, college algebra gives us tools like substitution and elimination to find specific solutions. The method involved:

• Setting up two equations based on the problem's descriptions and constraints.
• Simplifying and manipulating these equations to isolate variables.
• Applying techniques quickly to effectively solve for unknowns.

This approach makes it possible to solve problems involving multiple unknowns logically and systematically — an essential skill for students venturing into fields like engineering, economics, or the sciences where algebraic modeling is commonplace.