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Let's solve each problem one by one.

---

### A-84: School Population Problem

**Problem:**
West High School’s population is 250 students fewer than twice the population of East High School. The two schools have a total of 2858 students. How many students attend East High School?

**Solution:**
1. Let the number of students at East High School be \( x \).
2. The number of students at West High School is \( 2x - 250 \).
3. The total number of students is 2858.

We can write the equation:
\[
x + (2x - 250) = 2858
\]
Simplify and solve for \( x \):
\[
x + 2x - 250 = 2858
\]
\[
3x - 250 = 2858
\]
\[
3x = 2858 + 250
\]
\[
3x = 3108
\]
\[
x = \frac{3108}{3} = 1036
\]
So, the number of students at East High School is **1036**.

To find the number of students at West High School:
\[
2x - 250 = 2(1036) - 250 = 2072 - 250 = 1822
\]
Thus, West High School has 1822 students, and East High School has 1036 students.

**Answer:**
There are 1036 students at East High School.

---

### A-99: Rectangle Perimeter Problem

**Problem:**
The length of a rectangle is three centimeters more than twice the width. The perimeter is 78 centimeters. Find the dimensions of the rectangle.

**Solution:**
1. Let the width of the rectangle be \( w \).
2. The length of the rectangle is \( 2w + 3 \).
3. The formula for the perimeter of a rectangle is \( P = 2(\text{length} + \text{width}) \).

We are given that the perimeter is 78 cm:
\[
2((2w + 3) + w) = 78
\]
Simplify:
\[
2(3w + 3) = 78
\]
\[
6w + 6 = 78
\]
\[
6w = 78 - 6 = 72
\]
\[
w = \frac{72}{6} = 12
\]
So, the width is **12 cm**.

Now, find the length:
\[
\text{length} = 2w + 3 = 2(12) + 3 = 24 + 3 = 27 \, \text{cm}
\]

**Answer:**
The width is 12 cm, and the length is 27 cm.

---

### A-100: Dog Show Problem

**Problem:**
Chantel noticed that there were three more Scotties than Schnauzers. She also realized that the number of Wirehaired Terriers was five less than twice the number of Schnauzers. If there were 78 dogs in total, how many Schnauzers were there?

**Solution:**
1. Let the number of Schnauzers be \( x \).
2. The number of Scotties is \( x + 3 \).
3. The number of Wirehaired Terriers is \( 2x - 5 \).

The total number of dogs is 78, so:
\[
x + (x + 3) + (2x - 5) = 78
\]
Simplify and solve for \( x \):
\[
x + x + 3 + 2x - 5 = 78
\]
\[
4x - 2 = 78
\]
\[
4x = 78 + 2 = 80
\]
\[
x = \frac{80}{4} = 20
\]
So, the number of Schnauzers is **20**.

Now, calculate the number of Scotties and Wirehaired Terriers:
- Scotties: \( 20 + 3 = 23 \)
- Wirehaired Terriers: \( 2(20) - 5 = 40 - 5 = 35 \)

**Answer:**
There are 20 Schnauzers, 23 Scotties, and 35 Wirehaired Terriers.

---

Let me know if you'd like more details or clarifications on any part! Here are 5 related questions to expand your understanding:

1. What would happen if the total number of students in A-84 were different?
2. How would the perimeter formula change if you knew the area of the rectangle in A-99?
3. If the number of Wirehaired Terriers in A-100 increased by 5, how would that affect the total number of dogs?
4. Can you derive the relationship between length and width in A-99 without the perimeter?
5. How would you solve the dog problem in A-100 if the ratio of Schnauzers to Scotties was given instead?

**Tip:** Always define variables clearly when solving word problems, as it helps simplify the process!

Solve the following problems:
1. West High School’s population is 250 students fewer than twice the population of East High School. The two schools have a total of 2858 students. How many students attend East High School?
2. The length of a rectangle is three centimeters more than twice the width. The perimeter is 78 centimeters. Find the dimensions of the rectangle.
3. At an annual dog show, Chantel noticed there were three more Scotties than Schnauzers. She also realized that the number of Wirehaired Terriers was five less than twice the number of Schnauzers. If there were 78 dogs in total, how many Schnauzers were there?

Learn how to solve algebra and geometry problems involving school population calculation, rectangle perimeter, and counting dog breeds at a dog show. Suitable for students in Grades 7-9.

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