Problem 236 - Elementary Algebra

Question

The first graders at one school went on a field trip to the zoo. The total number of children and adults who went on the field trip was 115 . The number of adults was \(\frac{1}{4}\) the number of children. Solve the system $$ \left\\{\begin{array}{l} c+a=115 \\ a=\frac{1}{4} c \end{array}\right. $$ to find \(c,\) the number of children and \(a,\) the number of adults.

Step-by-Step Solution

Answer

There are 92 children and 23 adults.

Step 1: Define the variables

Let \(c\) represent the number of children and \(a\) represent the number of adults. The problem states the total number of children and adults who went on the field trip was 115, which gives us the equation: \text{\( (1)\) } c + a = 115

Step 2: Express one variable in terms of the other

The problem also states that the number of adults was \( Four \) fractional \the number of children. This gives us the second equation: \text{\( (2) \)} a = \frac{1}{4} c

Step 3: Substitute the second equation into the first equation

To solve the system, substitute the expression for \(a\) from the second equation (2) into the first equation (1): \( c + \frac{1}{4} c = 115 \)

Step 4: Combine like terms

Combine the terms involving \(c\) on the left-hand side: \( 1c + \frac{1}{4}c \) \( \frac{5}{4}c = 115 \)

Step 5: Solve for \(c\)

Multiply both sides of the equation by \( \frac {4}{5} \) to solve for \(c\): \( c = 92\)

Step 6: Find the value of \(a\)

Use the second original equation to find \(a\). Substitute \(c = 92\) into \( a = \frac {1}{4} c\): \( a = \frac {1}{4} \times 92 \) \text{so:} \( a = 23 \).

Key Concepts

elementary algebra

Elementary algebra forms the foundation of all higher-level mathematics. It helps in understanding how to manipulate mathematical symbols and solve equations. In this exercise, we worked with variables and basic arithmetic operations. When solving word problems:

• Identify the unknowns. We defined the number of children as \( c \) and adults as \( a \).
• Set up equations based on the problem's context. From the problem, we know their total is 115 and developed the equation: \( c + a = 115 \).
• Replace one variable in one equation when given a relationship between variables, such as \( a = \frac{1}{4} c \).

Practicing with elementary algebra prepares students for more complex algebraic concepts.

substitution method

The substitution method is a technique used to solve systems of equations. Here's how it was used in our problem:

• First, express one of the variables in terms of the other from one of the given equations. We expressed \( a \) in terms of \( c \) as: \( a = \frac{1}{4} c \).
• Next, substitute that expression into the other equation. We replaced \( a \) in the first equation \( c + a = 115 \) with \( \frac{1}{4} c \).
• This results in a single equation in one variable: \( c + \frac{1}{4} c = 115 \).
• Combine like terms and solve for the variable, \( c \): \( \frac{5}{4} c = 115 \).
• Once you find the value of one variable, substitute it back into the original equation to find the other variable. We found \( a \) by substituting \( c \) back into: \( a = \frac{1}{4} \times 92 = 23 \).

This method ensures accuracy and clarity when solving systems of linear equations.

linear equations

Linear equations are equations between two variables that produce a straight line when graphed. In our problem, we dealt with the system of linear equations:

• \( c + a = 115 \)
• \( a = \frac{1}{4} c \)

A linear equation follows the standard form \( ax + by = c \), where \( a \) and \( b \) are coefficients, and \( c \) is a constant.
Here's how we solved the system:

• We combined information from both equations, transforming the second one to fit into the first.
• Often, solving a system can involve graphing the equations and finding their intersection, but here, algebraic methods were sufficient.
• Understanding how to manipulate and combine these equations is a fundamental skill in algebra, paving the way for solving real-life problems.

Linear equations are one of math's basic tools for describing relationships between quantities.