Explanation:
Okay, so I need to figure out how many inches on the map correspond to 53.25 miles in real life. The problem says that 3 inches on the map equals 10 miles in real life. Hmm, let me think about how to approach this.

First, maybe I should find out how many inches correspond to one mile. If 3 inches is 10 miles, then dividing both numbers by 10 should give me the inches per mile. Let me write that down:

3 inches / 10 miles = 0.3 inches per mile. Wait, is that right? Let me check. If 3 inches is 10 miles, then 1 mile would be 3 inches divided by 10, which is indeed 0.3 inches. Yeah, that seems correct.

Alternatively, maybe I can set up a proportion. Let me try that method too to verify. If 3 inches corresponds to 10 miles, then x inches corresponds to 53.25 miles. So the proportion would be:

3 inches / 10 miles = x inches / 53.25 miles

Cross-multiplying, that would be 10x = 3 * 53.25. Then solving for x would give me x = (3 * 53.25) / 10. Let me compute that.

First, 3 times 53.25. Let me calculate 53.25 * 3. 50*3 is 150, 3.25*3 is 9.75, so total is 150 + 9.75 = 159.75. Then divide by 10, so 159.75 / 10 = 15.975 inches. Hmm, so that would be 15.975 inches. But the question says to express it as a fraction. So I need to convert 15.975 into a fraction.

Wait, let me check if my first method was correct. If 1 mile is 0.3 inches, then 53.25 miles would be 53.25 * 0.3 inches. Let me compute that. 53 * 0.3 is 15.9, and 0.25 * 0.3 is 0.075, so total is 15.9 + 0.075 = 15.975 inches. Same result. So that seems consistent.

But the problem wants the answer as a fraction. So 15.975 inches. Let me convert that decimal to a fraction. First, note that 0.975 is the decimal part. Let me think. 0.975 can be written as 975/1000. Let me simplify that fraction. Let's divide numerator and denominator by 25: 975 ÷ 25 = 39, 1000 ÷ 25 = 40. So 39/40. Wait, 39 divided by 40 is 0.975. So 15.975 is 15 + 39/40, which is 15 39/40 inches. But maybe we can write that as an improper fraction. Let me do that.

15 * 40 = 600, plus 39 is 639. So 639/40 inches. Let me check if that reduces. Let's see, 639 and 40. The factors of 40 are 2^3 * 5. Let me check if 639 is divisible by 2? No, it's odd. Divisible by 5? Ends with 9, so no. So 639/40 is the simplified fraction. Wait, but let me check if I did the decimal conversion correctly.

Alternatively, maybe there's a better way to do this without converting to decimal. Let me try that approach. Since the original ratio is 3 inches to 10 miles, so the scale is 3/10 inches per mile. Therefore, for 53.25 miles, the map distance is (3/10) * 53.25 inches. Let me compute that.

But 53.25 is a decimal. Maybe I can write 53.25 as a fraction. 53.25 is equal to 53 and 1/4, which is 213/4. Let me confirm: 53 * 4 = 212, plus 1 is 213. So 53.25 = 213/4. Therefore, multiplying by 3/10:

(3/10) * (213/4) = (3 * 213) / (10 * 4) = 639 / 40. Which is the same as before. So 639/40 inches. So that's the fraction. Let me check if that reduces. As before, 639 and 40 have no common factors. 639 divided by 3 is 213, 40 divided by 3 is not an integer. 639 divided by 5? No. So yes, 639/40 is the simplified fraction.

Alternatively, maybe I can check my answer by converting 639/40 back to decimal. 40 goes into 639 how many times? 40*15=600, so 15 with remainder 39. So 15 + 39/40 = 15.975, which matches the decimal we had earlier. So that seems correct.

But let me just make sure I didn't make any mistakes in the initial steps. The problem says 3 inches on the map is 10 miles in real life. So the scale is 3 inches : 10 miles. So to find the map distance for 53.25 miles, we can set up a proportion:

3 inches / 10 miles = x inches / 53.25 miles

Cross-multiplying gives 10x = 3 * 53.25, so x = (3 * 53.25)/10. Which is exactly what I did. So that's correct.

Alternatively, maybe I can think in terms of unit conversion. The scale is 3 inches per 10 miles, so the conversion factor is 3 inches / 10 miles. Therefore, multiplying the real distance by this conversion factor gives the map distance. So:

Map distance = 53.25 miles * (3 inches / 10 miles) = (53.25 * 3)/10 inches. Which again is the same calculation. So that's consistent.

So, all methods lead to 639/40 inches. Let me just check if 639/40 can be simplified further. Let me check the greatest common divisor (GCD) of 639 and 40. The prime factors of 40 are 2^3 * 5. Let's check if 639 is divisible by 2? No. Divisible by 5? Ends with 9, so no. Divisible by 3? Let's check: 6 + 3 + 9 = 18, which is divisible by 3. So 639 ÷ 3 = 213. So 639 = 3 * 213. Then 40 is 2^3 * 5. So there are no common factors between numerator and denominator. Therefore, 639/40 is indeed in simplest terms.

Alternatively, if I wanted to write it as a mixed number, it would be 15 39/40 inches, but the question says "expressed as a fraction," so probably improper fraction is acceptable. Let me check the problem statement again: "expressed as a fraction." It doesn't specify whether it needs to be a mixed number or improper fraction, but in math problems, unless specified, improper fraction is usually acceptable. However, sometimes they prefer mixed numbers. But since 639/40 is already a fraction, that's probably the answer they want.

Wait, but let me check if there's another way to represent this fraction. For example, maybe simplifying earlier steps? Let me think. When I had 3 * 53.25 / 10. Maybe I can write 53.25 as a fraction earlier. Let me try that again.

Original problem: 3 inches = 10 miles. So 1 inch = 10/3 miles. Wait, no, that's the other way. Wait, if 3 inches is 10 miles, then 1 inch is 10/3 miles. But we need to find inches given miles. So maybe invert that. If 1 inch corresponds to 10/3 miles, then miles correspond to inches by dividing by 10/3, which is multiplying by 3/10. Which is the same as before. So that's consistent.

Alternatively, maybe I can think of the scale as a ratio. The scale is 3:10, meaning 3 units on map to 10 units real. So to find the map distance, we can use the ratio. So if real distance is 53.25 miles, then map distance is (3/10) * 53.25. Which again is the same calculation.

Alternatively, maybe I can use dimensional analysis. Let me write that out:

53.25 miles * (3 inches / 10 miles) = (53.25 * 3 / 10) inches. The miles unit cancels out, leaving inches. Which is again the same.

So all methods lead to the same result. Therefore, I think 639/40 inches is the correct answer. Let me just check once more the calculation of 3 * 53.25. 53.25 * 3. Let me do it step by step:

First, 50 * 3 = 150.

Then, 3.25 * 3. Let's break that down: 3 * 3 = 9, 0.25 * 3 = 0.75. So 9 + 0.75 = 9.75.

Adding that to 150 gives 150 + 9.75 = 159.75. Then divide by 10: 159.75 / 10 = 15.975. Which is 15 and 0.975. As before, 0.975 is 39/40. So 15 39/40 is 639/40. Yep.

Alternatively, if I wanted to write 53.25 as a fraction from the start, 53.25 is 53 + 0.25. 0.25 is 1/4, so 53.25 is 53 1/4, which is (53*4 + 1)/4 = (212 + 1)/4 = 213/4. Then multiplying by 3/10:

(213/4) * (3/10) = (213*3)/(4*10) = 639/40. Yep, same result.

So I think that's solid. Therefore, the answer is 639/40 inches. Let me check if that fraction can be simplified further. As I thought before, 639 divided by 3 is 213, and 40 divided by 3 is not an integer. 639 divided by 2 is 319.5, not integer. Divided by 5? 639 divided by 5 is 127.8, not integer. So no, it can't be simplified further. Therefore, 639/40 is the answer.

But just to be thorough, let me check if 639 and 40 have any common factors. The prime factors of 40 are 2, 2, 2, 5. The prime factors of 639: Let's divide by 3 first. 6 + 3 + 9 = 18, which is divisible by 3. So 639 ÷ 3 = 213. Then 213 ÷ 3 = 71. So 639 factors into 3 * 3 * 71. 71 is a prime number. So the prime factors of 639 are 3² * 71. The prime factors of 40 are 2³ * 5. There are no common prime factors between numerator and denominator, so yes, 639/40 is in simplest terms.

Therefore, the answer is 639/40 inches. Let me just check once more with decimal conversion. 639 divided by 40. 40*15=600, 639-600=39. So 15.975, which is 15.975 inches. Which is the same as before. So that's correct.

Therefore, I think I can confidently say that the distance on the map is 639/40 inches.

**Final Answer**
The distance between the buildings on the map is \boxed{\dfrac{639}{40}} inches.

To determine the distance between two buildings on a map, we are given a scale: **3 inches on the map corresponds to 10 miles in real life**. We are also told that the **real-life distance** between the buildings is **53.25 miles**, and we are to find the **map distance in inches**, expressed as a **fraction**.

---

### Step 1: Understand the Scale

The scale tells us that:

$$
3 \text{ inches} = 10 \text{ miles}
$$

This means that **1 mile** corresponds to:

$$
\frac{3}{10} \text{ inches}
$$

So, to find the map distance for any real-life distance, we multiply the real-life distance by this conversion factor.

---

### Step 2: Convert Real-Life Distance to Map Distance

We are given:

$$
\text{Real-life distance} = 53.25 \text{ miles}
$$

Using the scale:

$$
\text{Map distance} = 53.25 \times \frac{3}{10}
$$

---

### Step 3: Perform the Multiplication

First, convert 53.25 to a fraction:

$$
53.25 = \frac{213}{4}
$$

Now multiply:

$$
\frac{213}{4} \times \frac{3}{10} = \frac{213 \times 3}{4 \times 10} = \frac{639}{40}
$$

---

### Step 4: Simplify the Fraction

Check if the numerator and denominator have any common factors:

- Prime factorization of 639: $3 \times 3 \times 71$
- Prime factorization of 40: $2^3 \times 5$

There are **no common factors** between 639 and 40, so the fraction is already in its **simplest form**.

---

### Final Answer

The distance between the buildings on the map is:

$$
\boxed{\dfrac{639}{40}} \text{ inches}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many inches on the map correspond to 53.25 miles in real life. The problem says that 3 inches on the map equals 10 miles in real life. Hmm, let me think about how to approach this.

First, maybe I should find out how many inches correspond to one mile. If 3 inches is 10 miles, then dividing both numbers by 10 should give me the inches per mile. Let me write that down:

3 inches / 10 miles = 0.3 inches per mile. Wait, is that right? Let me check. If 3 inches is 10 miles, then 1 mile would be 3 inches divided by 10, which is indeed 0.3 inches. Yeah, that seems correct.

Alternatively, maybe I can set up a proportion. Let me try that method too to verify. If 3 inches corresponds to 10 miles, then x inches corresponds to 53.25 miles. So the proportion would be:

3 inches / 10 miles = x inches / 53.25 miles

Cross-multiplying, that would be 10x = 3 * 53.25. Then solving for x would give me x = (3 * 53.25) / 10. Let me compute that.

First, 3 times 53.25. Let me calculate 53.25 * 3. 50*3 is 150, 3.25*3 is 9.75, so total is 150 + 9.75 = 159.75. Then divide by 10, so 159.75 / 10 = 15.975 inches. Hmm, so that would be 15.975 inches. But the question says to express it as a fraction. So I need to convert 15.975 into a fraction.

Wait, let me check if my first method was correct. If 1 mile is 0.3 inches, then 53.25 miles would be 53.25 * 0.3 inches. Let me compute that. 53 * 0.3 is 15.9, and 0.25 * 0.3 is 0.075, so total is 15.9 + 0.075 = 15.975 inches. Same result. So that seems consistent.

But the problem wants the answer as a fraction. So 15.975 inches. Let me convert that decimal to a fraction. First, note that 0.975 is the decimal part. Let me think. 0.975 can be written as 975/1000. Let me simplify that fraction. Let's divide numerator and denominator by 25: 975 ÷ 25 = 39, 1000 ÷ 25 = 40. So 39/40. Wait, 39 divided by 40 is 0.975. So 15.975 is 15 + 39/40, which is 15 39/40 inches. But maybe we can write that as an improper fraction. Let me do that.

15 * 40 = 600, plus 39 is 639. So 639/40 inches. Let me check if that reduces. Let's see, 639 and 40. The factors of 40 are 2^3 * 5. Let me check if 639 is divisible by 2? No, it's odd. Divisible by 5? Ends with 9, so no. So 639/40 is the simplified fraction. Wait, but let me check if I did the decimal conversion correctly.

Alternatively, maybe there's a better way to do this without converting to decimal. Let me try that approach. Since the original ratio is 3 inches to 10 miles, so the scale is 3/10 inches per mile. Therefore, for 53.25 miles, the map distance is (3/10) * 53.25 inches. Let me compute that.

But 53.25 is a decimal. Maybe I can write 53.25 as a fraction. 53.25 is equal to 53 and 1/4, which is 213/4. Let me confirm: 53 * 4 = 212, plus 1 is 213. So 53.25 = 213/4. Therefore, multiplying by 3/10:

(3/10) * (213/4) = (3 * 213) / (10 * 4) = 639 / 40. Which is the same as before. So 639/40 inches. So that's the fraction. Let me check if that reduces. As before, 639 and 40 have no common factors. 639 divided by 3 is 213, 40 divided by 3 is not an integer. 639 divided by 5? No. So yes, 639/40 is the simplified fraction.

Alternatively, maybe I can check my answer by converting 639/40 back to decimal. 40 goes into 639 how many times? 40*15=600, so 15 with remainder 39. So 15 + 39/40 = 15.975, which matches the decimal we had earlier. So that seems correct.

But let me just make sure I didn't make any mistakes in the initial steps. The problem says 3 inches on the map is 10 miles in real life. So the scale is 3 inches : 10 miles. So to find the map distance for 53.25 miles, we can set up a proportion:

3 inches / 10 miles = x inches / 53.25 miles

Cross-multiplying gives 10x = 3 * 53.25, so x = (3 * 53.25)/10. Which is exactly what I did. So that's correct.

Alternatively, maybe I can think in terms of unit conversion. The scale is 3 inches per 10 miles, so the conversion factor is 3 inches / 10 miles. Therefore, multiplying the real distance by this conversion factor gives the map distance. So:

Map distance = 53.25 miles * (3 inches / 10 miles) = (53.25 * 3)/10 inches. Which again is the same calculation. So that's consistent.

So, all methods lead to 639/40 inches. Let me just check if 639/40 can be simplified further. Let me check the greatest common divisor (GCD) of 639 and 40. The prime factors of 40 are 2^3 * 5. Let's check if 639 is divisible by 2? No. Divisible by 5? Ends with 9, so no. Divisible by 3? Let's check: 6 + 3 + 9 = 18, which is divisible by 3. So 639 ÷ 3 = 213. So 639 = 3 * 213. Then 40 is 2^3 * 5. So there are no common factors between numerator and denominator. Therefore, 639/40 is indeed in simplest terms.

Alternatively, if I wanted to write it as a mixed number, it would be 15 39/40 inches, but the question says "expressed as a fraction," so probably improper fraction is acceptable. Let me check the problem statement again: "expressed as a fraction." It doesn't specify whether it needs to be a mixed number or improper fraction, but in math problems, unless specified, improper fraction is usually acceptable. However, sometimes they prefer mixed numbers. But since 639/40 is already a fraction, that's probably the answer they want.

Wait, but let me check if there's another way to represent this fraction. For example, maybe simplifying earlier steps? Let me think. When I had 3 * 53.25 / 10. Maybe I can write 53.25 as a fraction earlier. Let me try that again.

Original problem: 3 inches = 10 miles. So 1 inch = 10/3 miles. Wait, no, that's the other way. Wait, if 3 inches is 10 miles, then 1 inch is 10/3 miles. But we need to find inches given miles. So maybe invert that. If 1 inch corresponds to 10/3 miles, then miles correspond to inches by dividing by 10/3, which is multiplying by 3/10. Which is the same as before. So that's consistent.

Alternatively, maybe I can think of the scale as a ratio. The scale is 3:10, meaning 3 units on map to 10 units real. So to find the map distance, we can use the ratio. So if real distance is 53.25 miles, then map distance is (3/10) * 53.25. Which again is the same calculation.

Alternatively, maybe I can use dimensional analysis. Let me write that out:

53.25 miles * (3 inches / 10 miles) = (53.25 * 3 / 10) inches. The miles unit cancels out, leaving inches. Which is again the same.

So all methods lead to the same result. Therefore, I think 639/40 inches is the correct answer. Let me just check once more the calculation of 3 * 53.25. 53.25 * 3. Let me do it step by step:

First, 50 * 3 = 150.

Then, 3.25 * 3. Let's break that down: 3 * 3 = 9, 0.25 * 3 = 0.75. So 9 + 0.75 = 9.75.

Adding that to 150 gives 150 + 9.75 = 159.75. Then divide by 10: 159.75 / 10 = 15.975. Which is 15 and 0.975. As before, 0.975 is 39/40. So 15 39/40 is 639/40. Yep.

Alternatively, if I wanted to write 53.25 as a fraction from the start, 53.25 is 53 + 0.25. 0.25 is 1/4, so 53.25 is 53 1/4, which is (53*4 + 1)/4 = (212 + 1)/4 = 213/4. Then multiplying by 3/10:

(213/4) * (3/10) = (213*3)/(4*10) = 639/40. Yep, same result.

So I think that's solid. Therefore, the answer is 639/40 inches. Let me check if that fraction can be simplified further. As I thought before, 639 divided by 3 is 213, and 40 divided by 3 is not an integer. 639 divided by 2 is 319.5, not integer. Divided by 5? 639 divided by 5 is 127.8, not integer. So no, it can't be simplified further. Therefore, 639/40 is the answer.

But just to be thorough, let me check if 639 and 40 have any common factors. The prime factors of 40 are 2, 2, 2, 5. The prime factors of 639: Let's divide by 3 first. 6 + 3 + 9 = 18, which is divisible by 3. So 639 ÷ 3 = 213. Then 213 ÷ 3 = 71. So 639 factors into 3 * 3 * 71. 71 is a prime number. So the prime factors of 639 are 3² * 71. The prime factors of 40 are 2³ * 5. There are no common prime factors between numerator and denominator, so yes, 639/40 is in simplest terms.

Therefore, the answer is 639/40 inches. Let me just check once more with decimal conversion. 639 divided by 40. 40*15=600, 639-600=39. So 15.975, which is 15.975 inches. Which is the same as before. So that's correct.

Therefore, I think I can confidently say that the distance on the map is 639/40 inches.

**Final Answer**
The distance between the buildings on the map is \boxed{\dfrac{639}{40}} inches.

To determine the distance between two buildings on a map, we are given a scale: **3 inches on the map corresponds to 10 miles in real life**. We are also told that the **real-life distance** between the buildings is **53.25 miles**, and we are to find the **map distance in inches**, expressed as a **fraction**.

---

### Step 1: Understand the Scale

The scale tells us that:

$$
3 \text{ inches} = 10 \text{ miles}
$$

This means that **1 mile** corresponds to:

$$
\frac{3}{10} \text{ inches}
$$

So, to find the map distance for any real-life distance, we multiply the real-life distance by this conversion factor.

---

### Step 2: Convert Real-Life Distance to Map Distance

We are given:

$$
\text{Real-life distance} = 53.25 \text{ miles}
$$

Using the scale:

$$
\text{Map distance} = 53.25 \times \frac{3}{10}
$$

---

### Step 3: Perform the Multiplication

First, convert 53.25 to a fraction:

$$
53.25 = \frac{213}{4}
$$

Now multiply:

$$
\frac{213}{4} \times \frac{3}{10} = \frac{213 \times 3}{4 \times 10} = \frac{639}{40}
$$

---

### Step 4: Simplify the Fraction

Check if the numerator and denominator have any common factors:

- Prime factorization of 639: $3 \times 3 \times 71$
- Prime factorization of 40: $2^3 \times 5$

There are **no common factors** between 639 and 40, so the fraction is already in its **simplest form**.

---

### Final Answer

The distance between the buildings on the map is:

$$
\boxed{\dfrac{639}{40}} \text{ inches}
$$

Solution: If we let the distance between the buildings on the map be $d$, then $\frac{d}{53.25} = \frac{3}{10}$. Cross-multiplying and solving for $d$, we obtain $10d = 159.75 \Rightarrow d=15.975$ inches. Expressed as a fraction, $d = 15\frac{975}{1000} = \boxed{15\frac{39}{40}},$ or $\boxed{\frac{639}{40}}$ inches.


Explanation:
Okay, let's try to figure out how many students showed up to the archery class this week. The problem says that between 30 and 50 students come each week. Usually, they can split into groups of equal size, but this week they couldn't. Then, when Betty and Wilma join the students, they still can't split into equal groups. So we need to find a number between 30 and 50 that isn't divisible by any number except 1 and itself, but wait, no, actually, the problem says that they couldn't break up into multiple groups of equal size. So that means the number of students isn't divisible by any number other than 1 and itself? Wait, no. Wait, if you can't break them into multiple groups of equal size, that would mean that the number is a prime number? Because primes can't be divided into smaller groups except 1 and itself. But wait, if there are, say, 4 students, you can split them into two groups of 2. So if the number is composite, you can split into groups. If it's prime, you can't split into groups unless you have 1 group. But the problem says "multiple groups", so maybe they need more than one group. So if the number is prime, then you can't split into multiple groups of equal size. But if it's composite, you can. So Betty noticed that she couldn't break the students into multiple groups of equal size, so the number of students is prime. Then Wilma noticed that if she and Betty joined the students, meaning adding 2 people, the total number is still not divisible by any number except 1 and itself? Wait, but adding 2 people to the original number, which was prime, so the new number is prime + 2. But the problem says that even after adding Betty and Wilma, they still couldn't break into groups of equal size. So that means that the original number is prime, and original number + 2 is also prime? So we need a prime number between 30 and 50, such that when you add 2, it's also prime. So twin primes? But twin primes are pairs of primes that differ by 2. So the original number is a prime, and original number + 2 is also prime. So we need to find such primes between 30 and 50. Let me list primes between 30 and 50 first. Let me recall primes between 30 and 50:

Primes between 30 and 50 are: 31, 37, 41, 43, 47. Let me check:

31 is prime.

32 is not, 33 is divisible by 3, 34 even, 35 divisible by 5, 36 even, 37 prime, 38 even, 39 divisible by 3, 40 even, 41 prime, 42 even, 43 prime, 44 even, 45 divisible by 5, 46 even, 47 prime, 48 even, 49 is 7 squared, 50 even. So yes, primes are 31, 37, 41, 43, 47.

Now, check which of these primes, when you add 2, are also primes. Let's check:

31 + 2 = 33, which is not prime (divisible by 3 and 11). So no.

37 + 2 = 39, which is not prime (divisible by 3 and 13). No.

41 + 2 = 43, which is prime. So 41 and 43 are twin primes.

43 + 2 = 45, which is not prime. 45 is divisible by 5 and 9.

47 + 2 = 49, which is 7 squared, not prime. So only 41 and 43 are twin primes in this range. So if the original number is 41, then adding 2 gives 43, which is also prime. Similarly, if original number is 43, adding 2 gives 45, which is not prime. Wait, so only 41 is such that original number is prime and original number + 2 is also prime. Wait, but 43 + 2 is 45, which is not prime. So only 41? Wait, but 31 + 2 is 33, not prime. 37 + 2 is 39, not prime. So only 41? Wait, but 41 + 2 is 43, which is prime. So if the original number is 41, then adding 2 gives 43, which is also prime. So that would satisfy the condition. But then, if the original number is 41, then Betty and Wilma join, making total 43, which is still prime. Therefore, they can't split into groups. So that would be the answer? But let me check if there are other possibilities. Wait, but maybe the original number isn't prime? Wait, the problem says that Betty noticed that she could not break the students up into multiple groups of equal size. So that would mean that the number of students is not divisible by any number other than 1 and itself. So that would mean that the number is prime. Because if it's composite, then it can be divided into groups. For example, if there are 4 students, you can split into two groups of 2. If there are 6 students, you can split into 2 groups of 3, etc. So if the number is prime, you can't split into multiple groups. Therefore, the original number is prime. Then, when adding 2 people (Betty and Wilma), the total number is still not divisible by any number except 1 and itself. Therefore, original number is prime, original number + 2 is also prime. Therefore, the original number is 41, since 41 + 2 = 43 is prime. But let me check if there are other numbers. Wait, maybe the original number is not prime, but when you add 2, it's still not divisible by any number except 1 and itself? Wait, but if the original number is not prime, then it can be divided into groups. So the problem says that Betty noticed that she could not break the students up into multiple groups of equal size. Therefore, original number must be prime. Then Wilma noticed that if she and Betty joined, they still couldn't break into groups. So original number + 2 is also prime. Therefore, the answer is 41? But let me check if there are other primes between 30 and 50 where adding 2 gives another prime. For example, 3 and 5, but those are below 30. 5 and 7, same. So in the range 30-50, only 41 and 43 are twin primes. But 43 + 2 is 45, which is not prime. So only 41. Therefore, the answer is 41? Wait, but let me check if there's another possibility. Maybe the original number is not prime, but it's a prime power? Wait, but if it's a prime power, like 4, which is 2 squared. Then you can split into groups of 2. So 4 can be split into two groups of 2. So even though it's a prime power, it's composite. So if the number is composite, you can split into groups. Therefore, the original number must be prime. Therefore, only 41. Wait, but let me check if there's another number between 30 and 50 which is not prime, but when you add 2, it's also not prime? Wait, but the problem says that when Betty and Wilma joined, they still couldn't break into groups. So that would mean that original number + 2 is also not divisible into groups, meaning original number + 2 is also prime? Wait, but if original number is composite, then original number can be split into groups, so Betty noticed that she couldn't split, so original number must be prime. Then original number + 2 must also be prime. Therefore, only 41. Therefore, answer is 41. Wait, but let me check if there's another number. Let me think again. Suppose the original number is 31. Then adding 2 is 33, which is composite. So Wilma noticed that even with adding 2, they couldn't split into groups. But 33 can be split into groups of 3, 11, etc. So Wilma would have noticed that they could split into groups. Therefore, 31 is invalid. Similarly, 37 + 2 = 39, which is composite. So 39 can be split into groups of 3, 13, etc. So Wilma would have noticed that they could split. Therefore, invalid. 41 + 2 = 43, prime. So they can't split. 43 + 2 = 45, composite. So if original number is 43, then adding 2 gives 45, which is composite. Therefore, Wilma would have noticed that they could split into groups. Therefore, invalid. 47 + 2 = 49, which is composite. So same thing. Therefore, only 41 is the number where original number is prime, and original number + 2 is also prime. Therefore, answer is 41. Wait, but let me check if there's another way. Maybe the original number is not prime, but it's a prime power? Wait, but if it's a prime power, like 25 (which is 5 squared). Then you can split into groups of 5. So 25 students can be split into 5 groups of 5. Therefore, Betty would have been able to split them. Therefore, original number must be prime. Therefore, only 41. Therefore, answer is 41. Wait, but let me check if there's another number between 30 and 50 which is prime, and when adding 2, it's also prime. Wait, 41 and 43 are twin primes. But 43 + 2 is 45, which is not prime. So only 41. Therefore, answer is 41. Wait, but let me check if there's another prime between 30 and 50 where adding 2 gives another prime. For example, 3 and 5, but they are too small. 5 and 7, same. 11 and 13, same. 17 and 19, same. 29 and 31. 29 is below 30. So 31 and 33. 33 is not prime. So no. Then 37 and 39. 39 not prime. 41 and 43. 43 and 45. 45 not prime. 47 and 49. 49 not prime. So only 41 and 43 are twin primes in that range. But since adding 2 to 43 gives 45, which is not prime, only 41 is valid. Therefore, answer is 41. Therefore, the number of students is 41. Wait, but let me check if there's another possibility. Suppose that the original number is not prime, but when you add 2, it's also not prime. But Betty noticed that she couldn't split the students into groups. So original number must be prime. Therefore, no. Therefore, answer is 41. Wait, but let me check if 41 is between 30 and 50. Yes, 41 is between 30 and 50. So that's the answer. Therefore, the answer is 41. Wait, but let me check again. Let me check if 41 is the only number. Suppose someone says, what about 37? If original number is 37, which is prime. Then adding 2 gives 39, which is composite. Therefore, Wilma would have noticed that they could split into groups. Therefore, invalid. Similarly, 31 + 2 = 33, composite. 43 + 2 = 45, composite. 47 + 2 = 49, composite. So only 41. Therefore, answer is 41. Therefore, the number of students is 41. So I think that's the answer.

**Final Answer**
The number of students who showed up to the archery class this week is \boxed{41}.

To determine how many students showed up to the archery class this week, we need to analyze the conditions given:

---

### **Understanding the Problem**

- **Range of students:** Between 30 and 50.
- **Betty's observation:** She could **not** break the students into **multiple groups of equal size**.
- **Wilma's observation:** Even after **Betty and Wilma joined** the students (i.e., adding 2 more people), the group **still could not** be split into multiple groups of equal size.

---

### **Key Insight**

If a number **cannot** be split into **multiple groups of equal size**, it means that the number is **not divisible** by any number other than 1 and itself. That is, the number is **prime**.

So, the number of students **must be a prime number**. Furthermore, when **2 more people** (Betty and Wilma) are added, the new total must also **not be divisible** by any number other than 1 and itself — meaning the new total must also be **prime**.

Thus, we are looking for a **prime number between 30 and 50**, such that **adding 2 to it** still results in a **prime number**.

---

### **Step-by-Step Search**

Let’s list the **prime numbers between 30 and 50**:

- 31, 37, 41, 43, 47

Now, check which of these, when **2 is added**, still results in a **prime number**:

- 31 + 2 = 33 → **Not prime** (divisible by 3)
- 37 + 2 = 39 → **Not prime** (divisible by 3)
- 41 + 2 = 43 → **Prime**
- 43 + 2 = 45 → **Not prime** (divisible by 5)
- 47 + 2 = 49 → **Not prime** (7²)

Only **41** satisfies both conditions:
- 41 is a prime number.
- 41 + 2 = 43 is also a prime number.

---

### **Conclusion**

The only number between 30 and 50 that is **prime**, and when **2 is added**, still results in a **prime number**, is **41**.

Thus, the number of students who showed up to the archery class this week is:

$$
\boxed{41}
$$

[Assistant]
Explanation:
Okay, let's try to figure out how many students showed up to the archery class this week. The problem says that between 30 and 50 students come each week. Usually, they can split into groups of equal size, but this week they couldn't. Then, when Betty and Wilma join the students, they still can't split into equal groups. So we need to find a number between 30 and 50 that isn't divisible by any number except 1 and itself, but wait, no, actually, the problem says that they couldn't break up into multiple groups of equal size. So that means the number of students isn't divisible by any number other than 1 and itself? Wait, no. Wait, if you can't break them into multiple groups of equal size, that would mean that the number is a prime number? Because primes can't be divided into smaller groups except 1 and itself. But wait, if there are, say, 4 students, you can split them into two groups of 2. So if the number is composite, you can split into groups. If it's prime, you can't split into groups unless you have 1 group. But the problem says "multiple groups", so maybe they need more than one group. So if the number is prime, then you can't split into multiple groups of equal size. But if it's composite, you can. So Betty noticed that she couldn't break the students into multiple groups of equal size, so the number of students is prime. Then Wilma noticed that if she and Betty joined the students, meaning adding 2 people, the total number is still not divisible by any number except 1 and itself? Wait, but adding 2 people to the original number, which was prime, so the new number is prime + 2. But the problem says that even after adding Betty and Wilma, they still couldn't break into groups of equal size. So that means that the original number is prime, and original number + 2 is also prime? So we need a prime number between 30 and 50, such that when you add 2, it's also prime. So twin primes? But twin primes are pairs of primes that differ by 2. So the original number is a prime, and original number + 2 is also prime. So we need to find such primes between 30 and 50. Let me list primes between 30 and 50 first. Let me recall primes between 30 and 50:

Primes between 30 and 50 are: 31, 37, 41, 43, 47. Let me check:

31 is prime.

32 is not, 33 is divisible by 3, 34 even, 35 divisible by 5, 36 even, 37 prime, 38 even, 39 divisible by 3, 40 even, 41 prime, 42 even, 43 prime, 44 even, 45 divisible by 5, 46 even, 47 prime, 48 even, 49 is 7 squared, 50 even. So yes, primes are 31, 37, 41, 43, 47.

Now, check which of these primes, when you add 2, are also primes. Let's check:

31 + 2 = 33, which is not prime (divisible by 3 and 11). So no.

37 + 2 = 39, which is not prime (divisible by 3 and 13). No.

41 + 2 = 43, which is prime. So 41 and 43 are twin primes.

43 + 2 = 45, which is not prime. 45 is divisible by 5 and 9.

47 + 2 = 49, which is 7 squared, not prime. So only 41 and 43 are twin primes in this range. So if the original number is 41, then adding 2 gives 43, which is also prime. Similarly, if original number is 43, adding 2 gives 45, which is not prime. Wait, so only 41 is such that original number is prime and original number + 2 is also prime. Wait, but 43 + 2 is 45, which is not prime. So only 41? Wait, but 31 + 2 is 33, not prime. 37 + 2 is 39, not prime. So only 41? Wait, but 41 + 2 is 43, which is prime. So if the original number is 41, then adding 2 gives 43, which is also prime. So that would satisfy the condition. But then, if the original number is 41, then Betty and Wilma join, making total 43, which is still prime. Therefore, they can't split into groups. So that would be the answer? But let me check if there are other possibilities. Wait, but maybe the original number isn't prime? Wait, the problem says that Betty noticed that she could not break the students up into multiple groups of equal size. So that would mean that the number of students is not divisible by any number other than 1 and itself. So that would mean that the number is prime. Because if it's composite, then it can be divided into groups. For example, if there are 4 students, you can split into two groups of 2. If there are 6 students, you can split into 2 groups of 3, etc. So if the number is prime, you can't split into multiple groups. Therefore, the original number is prime. Then, when adding 2 people (Betty and Wilma), the total number is still not divisible by any number except 1 and itself. Therefore, original number is prime, original number + 2 is also prime. Therefore, the original number is 41, since 41 + 2 = 43 is prime. But let me check if there are other numbers. Wait, maybe the original number is not prime, but when you add 2, it's still not divisible by any number except 1 and itself? Wait, but if the original number is not prime, then it can be divided into groups. So the problem says that Betty noticed that she could not break the students up into multiple groups of equal size. Therefore, original number must be prime. Then Wilma noticed that if she and Betty joined, they still couldn't break into groups. So original number + 2 is also prime. Therefore, the answer is 41? But let me check if there are other primes between 30 and 50 where adding 2 gives another prime. For example, 3 and 5, but those are below 30. 5 and 7, same. So in the range 30-50, only 41 and 43 are twin primes. But 43 + 2 is 45, which is not prime. So only 41. Therefore, the answer is 41? Wait, but let me check if there's another possibility. Maybe the original number is not prime, but it's a prime power? Wait, but if it's a prime power, like 4, which is 2 squared. Then you can split into groups of 2. So 4 can be split into two groups of 2. So even though it's a prime power, it's composite. So if the number is composite, you can split into groups. Therefore, the original number must be prime. Therefore, only 41. Wait, but let me check if there's another number between 30 and 50 which is not prime, but when you add 2, it's also not prime? Wait, but the problem says that when Betty and Wilma joined, they still couldn't break into groups. So that would mean that original number + 2 is also not divisible into groups, meaning original number + 2 is also prime? Wait, but if original number is composite, then original number can be split into groups, so Betty noticed that she couldn't split, so original number must be prime. Then original number + 2 must also be prime. Therefore, only 41. Therefore, answer is 41. Wait, but let me check if there's another number. Let me think again. Suppose the original number is 31. Then adding 2 is 33, which is composite. So Wilma noticed that even with adding 2, they couldn't split into groups. But 33 can be split into groups of 3, 11, etc. So Wilma would have noticed that they could split into groups. Therefore, 31 is invalid. Similarly, 37 + 2 = 39, which is composite. So 39 can be split into groups of 3, 13, etc. So Wilma would have noticed that they could split. Therefore, invalid. 41 + 2 = 43, prime. So they can't split. 43 + 2 = 45, composite. So if original number is 43, then adding 2 gives 45, which is composite. Therefore, Wilma would have noticed that they could split into groups. Therefore, invalid. 47 + 2 = 49, which is composite. So same thing. Therefore, only 41 is the number where original number is prime, and original number + 2 is also prime. Therefore, answer is 41. Wait, but let me check if there's another way. Maybe the original number is not prime, but it's a prime power? Wait, but if it's a prime power, like 25 (which is 5 squared). Then you can split into groups of 5. So 25 students can be split into 5 groups of 5. Therefore, Betty would have been able to split them. Therefore, original number must be prime. Therefore, only 41. Therefore, answer is 41. Wait, but let me check if there's another number between 30 and 50 which is prime, and when adding 2, it's also prime. Wait, 41 and 43 are twin primes. But 43 + 2 is 45, which is not prime. So only 41. Therefore, answer is 41. Wait, but let me check if there's another prime between 30 and 50 where adding 2 gives another prime. For example, 3 and 5, but they are too small. 5 and 7, same. 11 and 13, same. 17 and 19, same. 29 and 31. 29 is below 30. So 31 and 33. 33 is not prime. So no. Then 37 and 39. 39 not prime. 41 and 43. 43 and 45. 45 not prime. 47 and 49. 49 not prime. So only 41 and 43 are twin primes in that range. But since adding 2 to 43 gives 45, which is not prime, only 41 is valid. Therefore, answer is 41. Therefore, the number of students is 41. Wait, but let me check if there's another possibility. Suppose that the original number is not prime, but when you add 2, it's also not prime. But Betty noticed that she couldn't split the students into groups. So original number must be prime. Therefore, no. Therefore, answer is 41. Wait, but let me check if 41 is between 30 and 50. Yes, 41 is between 30 and 50. So that's the answer. Therefore, the answer is 41. Wait, but let me check again. Let me check if 41 is the only number. Suppose someone says, what about 37? If original number is 37, which is prime. Then adding 2 gives 39, which is composite. Therefore, Wilma would have noticed that they could split into groups. Therefore, invalid. Similarly, 31 + 2 = 33, composite. 43 + 2 = 45, composite. 47 + 2 = 49, composite. So only 41. Therefore, answer is 41. Therefore, the number of students is 41. So I think that's the answer.

**Final Answer**
The number of students who showed up to the archery class this week is \boxed{41}.

To determine how many students showed up to the archery class this week, we need to analyze the conditions given:

---

### **Understanding the Problem**

- **Range of students:** Between 30 and 50.
- **Betty's observation:** She could **not** break the students into **multiple groups of equal size**.
- **Wilma's observation:** Even after **Betty and Wilma joined** the students (i.e., adding 2 more people), the group **still could not** be split into multiple groups of equal size.

---

### **Key Insight**

If a number **cannot** be split into **multiple groups of equal size**, it means that the number is **not divisible** by any number other than 1 and itself. That is, the number is **prime**.

So, the number of students **must be a prime number**. Furthermore, when **2 more people** (Betty and Wilma) are added, the new total must also **not be divisible** by any number other than 1 and itself — meaning the new total must also be **prime**.

Thus, we are looking for a **prime number between 30 and 50**, such that **adding 2 to it** still results in a **prime number**.

---

### **Step-by-Step Search**

Let’s list the **prime numbers between 30 and 50**:

- 31, 37, 41, 43, 47

Now, check which of these, when **2 is added**, still results in a **prime number**:

- 31 + 2 = 33 → **Not prime** (divisible by 3)
- 37 + 2 = 39 → **Not prime** (divisible by 3)
- 41 + 2 = 43 → **Prime**
- 43 + 2 = 45 → **Not prime** (divisible by 5)
- 47 + 2 = 49 → **Not prime** (7²)

Only **41** satisfies both conditions:
- 41 is a prime number.
- 41 + 2 = 43 is also a prime number.

---

### **Conclusion**

The only number between 30 and 50 that is **prime**, and when **2 is added**, still results in a **prime number**, is **41**.

Thus, the number of students who showed up to the archery class this week is:

$$
\boxed{41}
$$

Solution: Let $n$ be the number of students at class.  Betty noticed that $n$ has no divisors between 1 and itself, so $n$ is prime.  Wilma noticed that $n + 2$ is prime.  This means we are looking for the smaller of two primes that differ by 2 that are between 30 and 50.  The primes in that range are 31, 37, 41, 43, and 47.  Since 41 and 43 differ by 2, $n = \boxed{41}$.


Explanation:
Okay, so I need to solve this system of logarithmic equations:

First equation: log base 6 of (x - y) plus log base 6 of (x + y) equals 2.

Second equation: log base y of 5x equals 2.

And I need to find the value of x. Alright, let me start by recalling some logarithm properties.

For the first equation, since both terms have the same base, 6, I can use the logarithm property that says log_b(a) + log_b(c) = log_b(a*c). So combining those two logs into one:

log_6[(x - y)(x + y)] = 2.

Then, converting that logarithmic equation to its exponential form. Remember that log_b(A) = C is equivalent to b^C = A. So here, that would be:

6^2 = (x - y)(x + y).

Calculating 6 squared is 36, so:

(x - y)(x + y) = 36.

But (x - y)(x + y) is a difference of squares, so that simplifies to x² - y² = 36. Let me write that down:

Equation (1): x² - y² = 36.

Now moving on to the second equation: log base y of 5x equals 2. Again, converting that to exponential form. log_y(5x) = 2 means that y² = 5x. So:

Equation (2): y² = 5x.

So now I have two equations:

1) x² - y² = 36

2) y² = 5x

Hmm, maybe I can substitute equation (2) into equation (1). Let me try that. Since equation (2) gives me y² in terms of x, I can replace y² in equation (1) with 5x. Let me do that:

x² - (5x) = 36.

So that simplifies to x² - 5x - 36 = 0.

Now, solving this quadratic equation for x. Let me check if I can factor it. Let me think of two numbers that multiply to -36 and add up to -5. Let me see... factors of 36 are 1 & 36, 2 & 18, 3 & 12, 4 & 9, 6 & 6. Since the product is negative, one has to be positive and the other negative. Let me try 4 and -9: 4 * (-9) = -36, and 4 + (-9) = -5. Yes! Perfect. So the quadratic factors as:

(x + 4)(x - 9) = 0.

Therefore, the solutions are x = -4 or x = 9.

But wait, I need to check if these solutions are valid in the original equations, especially considering the domains of the logarithms.

First, let's recall that logarithms have certain domain restrictions. For log_6(x - y) and log_6(x + y) to be defined, the arguments must be positive. So:

x - y > 0 and x + y > 0.

Also, for log_y(5x) to be defined, the base y must be positive and not equal to 1, and the argument 5x must be positive. So:

y > 0, y ≠ 1, and 5x > 0 ⇒ x > 0.

So, from the second equation, log_y(5x) = 2, we have that y must be positive and not equal to 1, and x must be positive. Therefore, x must be positive. So x = -4 is invalid because x must be positive. Therefore, x = 9 is the only possible solution.

But let me check if x = 9 works with the other equations. Let me find y first. From equation (2): y² = 5x. If x = 9, then y² = 5*9 = 45. Therefore, y = sqrt(45) or y = -sqrt(45). But since y must be positive (from the domain of log_y(5x)), we take y = sqrt(45) = 3*sqrt(5).

Now, check the first equation's domain conditions: x - y > 0 and x + y > 0. Let's compute x - y:

x - y = 9 - 3√5. Let me approximate √5 ≈ 2.236, so 3√5 ≈ 6.708. Therefore, 9 - 6.708 ≈ 2.292, which is positive.

x + y = 9 + 3√5 ≈ 9 + 6.708 ≈ 15.708, which is also positive. So both conditions are satisfied.

Additionally, check that y ≠ 1. Since y = 3√5 ≈ 6.708, which is not 1, so that's okay.

Therefore, x = 9 is valid.

But just to be thorough, let me check if substituting x = 9 and y = 3√5 into the original equations works.

First equation: log_6(x - y) + log_6(x + y) = log_6[(x - y)(x + y)] = log_6(x² - y²). From equation (1), x² - y² = 36, so log_6(36). Since 6² = 36, log_6(36) = 2. Which matches the first equation.

Second equation: log_y(5x) = log_{3√5}(5*9) = log_{3√5}(45). Let me check if this equals 2.

If log_{3√5}(45) = 2, then (3√5)^2 should equal 45. Let's compute (3√5)^2: 3²*(√5)^2 = 9*5 = 45. Yes! So that works.

Therefore, x = 9 is indeed the correct solution.

But just to make sure there are no other solutions, let me check if x = -4 could be a solution even though it's negative. If x = -4, then from equation (2): y² = 5*(-4) = -20. But y² can't be negative, so that's impossible. Therefore, x = -4 is invalid.

Therefore, the only valid solution is x = 9.

Wait, but let me think again. When I solved the quadratic equation, I got x = -4 and x = 9. But since x must be positive, x = -4 is rejected. So yes, only x = 9.

Therefore, the answer is 9.

But let me check once more if there's any other possible mistake. For example, when I combined the logs, did I do that correctly?

Original first equation: log_6(x - y) + log_6(x + y) = 2.

Yes, combining them gives log_6[(x - y)(x + y)] = 2. Then converting to exponential form: 6^2 = (x - y)(x + y) = x² - y² = 36. Correct.

Second equation: log_y(5x) = 2 ⇒ y² = 5x. Correct.

Substituting into first equation: x² - 5x = 36 ⇒ x² -5x -36 = 0. Factored as (x + 4)(x - 9) = 0. Correct.

So all steps are correct.

Therefore, I think x = 9 is the correct answer.

**Final Answer**
The value of $ x $ is \boxed{9}.

To solve the system of equations:

$$
\log_6 (x - y) + \log_6 (x + y) = 2 \quad \text{(1)}
$$
$$
\log_y (5x) = 2 \quad \text{(2)}
$$

we proceed step by step using logarithmic identities and algebraic manipulation.

---

### Step 1: Simplify the First Equation

Using the logarithmic identity:

$$
\log_b a + \log_b c = \log_b (ac)
$$

we combine the two logarithms in equation (1):

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)]
$$

So the equation becomes:

$$
\log_6 [(x - y)(x + y)] = 2
$$

Now convert this to exponential form:

$$
(x - y)(x + y) = 6^2 = 36
$$

Recognizing the difference of squares:

$$
x^2 - y^2 = 36 \quad \text{(Equation A)}
$$

---

### Step 2: Simplify the Second Equation

From equation (2):

$$
\log_y (5x) = 2
$$

Convert to exponential form:

$$
y^2 = 5x \quad \text{(Equation B)}
$$

---

### Step 3: Substitute Equation B into Equation A

Substitute $ y^2 = 5x $ into $ x^2 - y^2 = 36 $:

$$
x^2 - 5x = 36
$$

Rewriting:

$$
x^2 - 5x - 36 = 0
$$

---

### Step 4: Solve the Quadratic Equation

Factor the quadratic:

$$
x^2 - 5x - 36 = (x + 4)(x - 9) = 0
$$

So the solutions are:

$$
x = -4 \quad \text{or} \quad x = 9
$$

---

### Step 5: Check for Validity of Solutions

We must ensure that the logarithms are defined, which imposes the following conditions:

- $ x > 0 $ (since $ \log_y (5x) $ requires $ 5x > 0 $)
- $ y > 0 $ and $ y \ne 1 $ (since $ \log_y $ requires a valid base)
- $ x - y > 0 $ and $ x + y > 0 $ (since the arguments of the logs in equation (1) must be positive)

#### Case 1: $ x = -4 $

- $ y^2 = 5x = -20 $, which is not possible (no real $ y $ satisfies this)
- So $ x = -4 $ is **invalid**

#### Case 2: $ x = 9 $

- From equation (2): $ y^2 = 5x = 45 \Rightarrow y = \sqrt{45} = 3\sqrt{5} $
- Check domain conditions:
  - $ x = 9 > 0 $
  - $ y = 3\sqrt{5} > 0 $, and $ y \ne 1 $
  - $ x - y = 9 - 3\sqrt{5} \approx 9 - 6.708 = 2.292 > 0 $
  - $ x + y = 9 + 3\sqrt{5} > 0 $

All conditions are satisfied.

---

### Step 6: Verify the Solution

Check equation (1):

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)] = \log_6 (x^2 - y^2)
$$

From equation (A): $ x^2 - y^2 = 36 $, so:

$$
\log_6 (36) = 2 \quad \text{✓}
$$

Check equation (2):

$$
\log_y (5x) = \log_{3\sqrt{5}} (45)
$$

Check if $ (3\sqrt{5})^2 = 45 $:

$$
(3\sqrt{5})^2 = 9 \cdot 5 = 45 \quad \text{✓}
$$

So the solution satisfies both equations.

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to solve this system of logarithmic equations:

First equation: log base 6 of (x - y) plus log base 6 of (x + y) equals 2.

Second equation: log base y of 5x equals 2.

And I need to find the value of x. Alright, let me start by recalling some logarithm properties.

For the first equation, since both terms have the same base, 6, I can use the logarithm property that says log_b(a) + log_b(c) = log_b(a*c). So combining those two logs into one:

log_6[(x - y)(x + y)] = 2.

Then, converting that logarithmic equation to its exponential form. Remember that log_b(A) = C is equivalent to b^C = A. So here, that would be:

6^2 = (x - y)(x + y).

Calculating 6 squared is 36, so:

(x - y)(x + y) = 36.

But (x - y)(x + y) is a difference of squares, so that simplifies to x² - y² = 36. Let me write that down:

Equation (1): x² - y² = 36.

Now moving on to the second equation: log base y of 5x equals 2. Again, converting that to exponential form. log_y(5x) = 2 means that y² = 5x. So:

Equation (2): y² = 5x.

So now I have two equations:

1) x² - y² = 36

2) y² = 5x

Hmm, maybe I can substitute equation (2) into equation (1). Let me try that. Since equation (2) gives me y² in terms of x, I can replace y² in equation (1) with 5x. Let me do that:

x² - (5x) = 36.

So that simplifies to x² - 5x - 36 = 0.

Now, solving this quadratic equation for x. Let me check if I can factor it. Let me think of two numbers that multiply to -36 and add up to -5. Let me see... factors of 36 are 1 & 36, 2 & 18, 3 & 12, 4 & 9, 6 & 6. Since the product is negative, one has to be positive and the other negative. Let me try 4 and -9: 4 * (-9) = -36, and 4 + (-9) = -5. Yes! Perfect. So the quadratic factors as:

(x + 4)(x - 9) = 0.

Therefore, the solutions are x = -4 or x = 9.

But wait, I need to check if these solutions are valid in the original equations, especially considering the domains of the logarithms.

First, let's recall that logarithms have certain domain restrictions. For log_6(x - y) and log_6(x + y) to be defined, the arguments must be positive. So:

x - y > 0 and x + y > 0.

Also, for log_y(5x) to be defined, the base y must be positive and not equal to 1, and the argument 5x must be positive. So:

y > 0, y ≠ 1, and 5x > 0 ⇒ x > 0.

So, from the second equation, log_y(5x) = 2, we have that y must be positive and not equal to 1, and x must be positive. Therefore, x must be positive. So x = -4 is invalid because x must be positive. Therefore, x = 9 is the only possible solution.

But let me check if x = 9 works with the other equations. Let me find y first. From equation (2): y² = 5x. If x = 9, then y² = 5*9 = 45. Therefore, y = sqrt(45) or y = -sqrt(45). But since y must be positive (from the domain of log_y(5x)), we take y = sqrt(45) = 3*sqrt(5).

Now, check the first equation's domain conditions: x - y > 0 and x + y > 0. Let's compute x - y:

x - y = 9 - 3√5. Let me approximate √5 ≈ 2.236, so 3√5 ≈ 6.708. Therefore, 9 - 6.708 ≈ 2.292, which is positive.

x + y = 9 + 3√5 ≈ 9 + 6.708 ≈ 15.708, which is also positive. So both conditions are satisfied.

Additionally, check that y ≠ 1. Since y = 3√5 ≈ 6.708, which is not 1, so that's okay.

Therefore, x = 9 is valid.

But just to be thorough, let me check if substituting x = 9 and y = 3√5 into the original equations works.

First equation: log_6(x - y) + log_6(x + y) = log_6[(x - y)(x + y)] = log_6(x² - y²). From equation (1), x² - y² = 36, so log_6(36). Since 6² = 36, log_6(36) = 2. Which matches the first equation.

Second equation: log_y(5x) = log_{3√5}(5*9) = log_{3√5}(45). Let me check if this equals 2.

If log_{3√5}(45) = 2, then (3√5)^2 should equal 45. Let's compute (3√5)^2: 3²*(√5)^2 = 9*5 = 45. Yes! So that works.

Therefore, x = 9 is indeed the correct solution.

But just to make sure there are no other solutions, let me check if x = -4 could be a solution even though it's negative. If x = -4, then from equation (2): y² = 5*(-4) = -20. But y² can't be negative, so that's impossible. Therefore, x = -4 is invalid.

Therefore, the only valid solution is x = 9.

Wait, but let me think again. When I solved the quadratic equation, I got x = -4 and x = 9. But since x must be positive, x = -4 is rejected. So yes, only x = 9.

Therefore, the answer is 9.

But let me check once more if there's any other possible mistake. For example, when I combined the logs, did I do that correctly?

Original first equation: log_6(x - y) + log_6(x + y) = 2.

Yes, combining them gives log_6[(x - y)(x + y)] = 2. Then converting to exponential form: 6^2 = (x - y)(x + y) = x² - y² = 36. Correct.

Second equation: log_y(5x) = 2 ⇒ y² = 5x. Correct.

Substituting into first equation: x² - 5x = 36 ⇒ x² -5x -36 = 0. Factored as (x + 4)(x - 9) = 0. Correct.

So all steps are correct.

Therefore, I think x = 9 is the correct answer.

**Final Answer**
The value of $ x $ is \boxed{9}.

To solve the system of equations:

$$
\log_6 (x - y) + \log_6 (x + y) = 2 \quad \text{(1)}
$$
$$
\log_y (5x) = 2 \quad \text{(2)}
$$

we proceed step by step using logarithmic identities and algebraic manipulation.

---

### Step 1: Simplify the First Equation

Using the logarithmic identity:

$$
\log_b a + \log_b c = \log_b (ac)
$$

we combine the two logarithms in equation (1):

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)]
$$

So the equation becomes:

$$
\log_6 [(x - y)(x + y)] = 2
$$

Now convert this to exponential form:

$$
(x - y)(x + y) = 6^2 = 36
$$

Recognizing the difference of squares:

$$
x^2 - y^2 = 36 \quad \text{(Equation A)}
$$

---

### Step 2: Simplify the Second Equation

From equation (2):

$$
\log_y (5x) = 2
$$

Convert to exponential form:

$$
y^2 = 5x \quad \text{(Equation B)}
$$

---

### Step 3: Substitute Equation B into Equation A

Substitute $ y^2 = 5x $ into $ x^2 - y^2 = 36 $:

$$
x^2 - 5x = 36
$$

Rewriting:

$$
x^2 - 5x - 36 = 0
$$

---

### Step 4: Solve the Quadratic Equation

Factor the quadratic:

$$
x^2 - 5x - 36 = (x + 4)(x - 9) = 0
$$

So the solutions are:

$$
x = -4 \quad \text{or} \quad x = 9
$$

---

### Step 5: Check for Validity of Solutions

We must ensure that the logarithms are defined, which imposes the following conditions:

- $ x > 0 $ (since $ \log_y (5x) $ requires $ 5x > 0 $)
- $ y > 0 $ and $ y \ne 1 $ (since $ \log_y $ requires a valid base)
- $ x - y > 0 $ and $ x + y > 0 $ (since the arguments of the logs in equation (1) must be positive)

#### Case 1: $ x = -4 $

- $ y^2 = 5x = -20 $, which is not possible (no real $ y $ satisfies this)
- So $ x = -4 $ is **invalid**

#### Case 2: $ x = 9 $

- From equation (2): $ y^2 = 5x = 45 \Rightarrow y = \sqrt{45} = 3\sqrt{5} $
- Check domain conditions:
  - $ x = 9 > 0 $
  - $ y = 3\sqrt{5} > 0 $, and $ y \ne 1 $
  - $ x - y = 9 - 3\sqrt{5} \approx 9 - 6.708 = 2.292 > 0 $
  - $ x + y = 9 + 3\sqrt{5} > 0 $

All conditions are satisfied.

---

### Step 6: Verify the Solution

Check equation (1):

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)] = \log_6 (x^2 - y^2)
$$

From equation (A): $ x^2 - y^2 = 36 $, so:

$$
\log_6 (36) = 2 \quad \text{✓}
$$

Check equation (2):

$$
\log_y (5x) = \log_{3\sqrt{5}} (45)
$$

Check if $ (3\sqrt{5})^2 = 45 $:

$$
(3\sqrt{5})^2 = 9 \cdot 5 = 45 \quad \text{✓}
$$

So the solution satisfies both equations.

---

### Final Answer

$$
\boxed{9}
$$

Solution: Working on the first equation, we have from the difference of squares factorization that $\log_6 (x-y) + \log_6 (x+y) = \log_6 (x^2-y^2) = 2$, so $x^2 - y^2 = 6^2 = 36$. Using the change of base formula, the second equation becomes $$\frac{\log(5x)}{\log y} = 2 \Longrightarrow \log(5x) = 2\log y = \log y^2.$$Substituting that $y^2 = x^2 - 36$, it follows that $\log (x^2 - 36) = \log y^2 = 2\log y = \log 5x$. Since the logarithm is a one-to-one function, it follows that $x^2 - 36 = 5x$, so $x^2 - 5x - 36 = (x - 9)(x + 4) = 0$. Thus, $x = 9, - 4$, but the second does not work. Thus, our answer is $x = \boxed{9}$.