Problem 40 - Algebra A Combined Function

Question

A charity fundraiser consisted of a spaghetti supper where a total of 387 people were fed. They charged \(\$ 6.80\) for adults and half price for children. If they took in \(\$ 2444.60\), find how many adults and how many children attended the supper.

Step-by-Step Solution

Answer

332 adults and 55 children attended the supper.

Step 1: Define Variables

Let's define the variables for this problem. Let \( a \) represent the number of adults, and \( c \) represent the number of children. Based on the problem, adults were charged \( \\(6.80 \), and children were charged half, which is \( \\)3.40 \).

Step 2: Create Equations

We have two conditions given: one is the total number of people and the other is the total money collected. First, the equation for the total number of people is \( a + c = 387 \). Second, the equation for the total revenue is \( 6.80a + 3.40c = 2444.60 \).

Step 3: Simplify the Revenue Equation

To simplify the calculations, note that \( 6.80a + 3.40c = 2444.60 \) can be divided by 3.40:\[2a + c = 719\]This simplifies our computations.

Step 4: Use Substitution or Elimination

We now have two equations:1. \( a + c = 387 \)2. \( 2a + c = 719 \)We can use the elimination method. Subtract the first equation from the second:\[(2a + c) - (a + c) = 719 - 387\]Simplifying gives:\[ a = 332 \]

Step 5: Solve for the Other Variable

Now, substitute \( a = 332 \) back into the first equation:\[332 + c = 387\]Solving for \( c \), we subtract 332 from both sides:\[ c = 55 \]

Step 6: Verify the Solution

Finally, verify by checking the revenue:- Revenue from adults is \( 332 \times 6.80 = 2257.60 \)- Revenue from children is \( 55 \times 3.40 = 187.00 \)- Total revenue: \( 2257.60 + 187.00 = 2444.60 \)This confirms our calculations are correct.

Key Concepts

Linear Equations

In this problem, linear equations help us describe the relationship between variables. We define two linear equations based on the information given:

• One equation for the total number of people: \( a + c = 387 \)
• Another for the total money collected: \( 6.80a + 3.40c = 2444.60 \)

Linear equations like these reflect a straight-line relationship between variables.
The key is that each variable represents an unknown quantity, here the number of adults and children. Solving linear equations helps us find these values.

Substitution Method

The substitution method is a powerful technique for solving systems of equations. Although in this exercise the elimination method was used, it is beneficial to know how substitution works.
The idea is straightforward:

• Solve one of the equations for one variable in terms of the others.
• Then substitute that expression into the other equation.

For instance, from the equation \( a + c = 387 \), you can express \( c \) as \( c = 387 - a \) and substitute this into the other equation. This method is particularly handy when one equation is easily isolated.

Elimination Method

This exercise uses the elimination method, a preferred choice when equations are set up nicely to cancel out a variable.
Consider the problem at hand:

• We have the simplified equations: \( a + c = 387 \) and \( 2a + c = 719 \).
• By subtracting the first equation from the second, we eliminate \( c \) and easily solve for \( a \).

The trick with elimination is aligning terms so that adding or subtracting the equations removes one variable. This leaves you one equation with one variable, making it straightforward to solve.

System of Equations

In algebra, a system of equations is a set of two or more equations with the same variables.
To solve a system of equations means finding a value for each variable that turns both equations into true statements.

• For our problem, we are dealing with \( a + c = 387 \) and \( 2a + c = 719 \).
• The solutions \( a = 332 \) and \( c = 55 \) satisfy both equations, representing the correct number of adults and children.

Solving systems of equations is a crucial algebraic skill, applicable in many real-world scenarios. Each method, substitution or elimination, provides a way to find these solutions.