A concert audience of 400 people consists of adults, students, and children. The ticket prices are $40 for adults, $20 for students, and $10 for children. The total amount of money taken in was $10,600. The numbers of children tickets sold is 200 less than the number of adults and student tickets in total. How many adults, students, and children are in attendance? Solve this problem in Augmented Matrix form

Let's denote the number of adults, students, and children in attendance by:

- $ a $ for adults,
- $ s $ for students,
- $ c $ for children.

We are given the following conditions:

1. **Total number of people**:  
   The total number of people in attendance is 400:
   \[
   a + s + c = 400
   \]

2. **Total ticket sales**:  
   The total amount of money taken in was $10,600. With ticket prices of $40 for adults, $20 for students, and $10 for children, the total revenue equation is:
   \[
   40a + 20s + 10c = 10600
   \]

3. **Relationship between children and others**:  
   The number of children is 200 less than the number of adults and students combined:
   \[
   c = (a + s) - 200
   \]

### Step 1: Write the system of equations

We have the system of equations:

1. $ a + s + c = 400 $
2. $ 40a + 20s + 10c = 10600 $
3. $ c = a + s - 200 $

We can rewrite the third equation as:
\[
a + s - c = 200
\]

So the system becomes:

1. $ a + s + c = 400 $
2. $ 40a + 20s + 10c = 10600 $
3. $ a + s - c = 200 $

### Step 2: Write the system as an augmented matrix

We can express this system as an augmented matrix:

\[
\begin{pmatrix}
1 & 1 & 1 & | & 400 \\
40 & 20 & 10 & | & 10600 \\
1 & 1 & -1 & | & 200
\end{pmatrix}
\]

### Step 3: Perform Gaussian elimination

Now, let's solve this system using Gaussian elimination. We will first convert the matrix into row-echelon form.

\[
\begin{pmatrix}
1 & 1 & 1 & | & 400 \\
40 & 20 & 10 & | & 10600 \\
1 & 1 & -1 & | & 200
\end{pmatrix}
\]

We can first eliminate the second row by subtracting 40 times the first row from the second row.

\[
R_2 \rightarrow R_2 - 40R_1
\]

\[
\begin{pmatrix}
1 & 1 & 1 & | & 400 \\
0 & -20 & -30 & | & -5400 \\
1 & 1 & -1 & | & 200
\end{pmatrix}
\]

Next, we eliminate the third row by subtracting the first row from the third row:

\[
R_3 \rightarrow R_3 - R_1
\]

\[
\begin{pmatrix}
1 & 1 & 1 & | & 400 \\
0 & -20 & -30 & | & -5400 \\
0 & 0 & -2 & | & -200
\end{pmatrix}
\]

### Step 4: Back substitution

Now that we have the matrix in row-echelon form, we can use back substitution to solve for $ a $, $ s $, and $ c $.

From the third row, we have:
\[
-2c = -200 \quad \Rightarrow \quad c = 100
\]

Substitute $ c = 100 $ into the second row:
\[
-20s - 30(100) = -5400 \quad \Rightarrow \quad -20s - 3000 = -5400
\]
\[
-20s = -2400 \quad \Rightarrow \quad s = 120
\]

Substitute $ s = 120 $ and $ c = 100 $ into the first row:
\[
a + 120 + 100 = 400 \quad \Rightarrow \quad a = 180
\]

### Final Answer:

- Number of adults: $ a = 180 $
- Number of students: $ s = 120 $
- Number of children: $ c = 100 $

Would you like further details or clarifications? Here are five related questions for practice:

1. What happens if the number of children is 150 less than the number of adults and students?
2. How would the equations change if the ticket price for children was $15 instead of $10?
3. What if the total number of people was 500, but the same revenue was collected?
4. How can we verify the solution by substitution into the original equations?
5. What method would you use if only 2 types of tickets were sold?

**Tip**: Always check your solution by substituting it back into the original system of equations to ensure it satisfies all conditions.

A concert audience of 400 people consists of adults, students, and children. The ticket prices are $40 for adults, $20 for students, and $10 for children. The total amount of money taken in was $10,600. The number of children tickets sold is 200 less than the number of adults and student tickets in total. How many adults, students, and children are in attendance? Solve this problem in Augmented Matrix form.

Learn how to determine the number of adults, students, and children in a concert audience of 400 people using a system of linear equations and augmented matrices. The problem involves ticket sales of $10,600 and a relationship between children and other attendees. Suitable for students in Grades 10-12.

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