Explanation:
Okay, so I need to factor the expression ab + 5b + 2a + 10. Hmm, factoring can sometimes be tricky, but let me try to remember the methods. I think grouping might be useful here. Let me see...

First, let me write down the expression again to make sure I have it right: ab + 5b + 2a + 10. Alright. Now, factoring by grouping usually involves grouping terms in pairs and then factoring out common factors from each pair. Let me try that.

Looking at the terms, I have ab, 5b, 2a, and 10. Let me group the first two terms together and the last two terms together. So, (ab + 5b) + (2a + 10). Now, let me factor out the greatest common factor (GCF) from each group.

Starting with the first group, ab + 5b. Both terms have a 'b' in them, so I can factor out a 'b'. That would leave me with b(a + 5). Let me check: b times a is ab, and b times 5 is 5b. Yep, that works.

Now the second group is 2a + 10. Both terms are divisible by 2, so factoring out a 2 gives me 2(a + 5). Let me verify: 2 times a is 2a, and 2 times 5 is 10. Perfect.

So now, the expression becomes: b(a + 5) + 2(a + 5). Wait a second, both terms have a common factor of (a + 5). That means I can factor that out. Let me do that.

Factoring out (a + 5) from both terms, I get: (a + 5)(b + 2). Let me check if that's correct by expanding it back.

Expanding (a + 5)(b + 2): first multiply a by b, which is ab. Then a times 2 is 2a. Then 5 times b is 5b. Finally, 5 times 2 is 10. So combining those terms: ab + 2a + 5b + 10. Which is the same as the original expression ab + 5b + 2a + 10. The order of the terms is different, but addition is commutative, so it's the same.

Therefore, the factored form is (a + 5)(b + 2). Let me just make sure there isn't another way to factor it or if I missed something.

Alternatively, could I have grouped the terms differently? Let me try grouping ab with 2a and 5b with 10. So, (ab + 2a) + (5b + 10). Factoring out a from the first group: a(b + 2). Factoring out 5 from the second group: 5(b + 2). Then combining them: (a + 5)(b + 2). Same result. So regardless of how I group them, I end up with the same factors.

Another check: maybe factor out something else? Let me see. The original expression is ab + 5b + 2a + 10. If I try to factor out a 'b' from the first two terms and a '2' from the last two, which is what I did first, and then factor out the common (a + 5).

Alternatively, maybe factor out something else? Let me think. Suppose I tried to factor out a different term. For example, if I look for common factors across all terms. The terms are ab, 5b, 2a, 10. The only common factor across all terms is... Well, there's no common factor in all four terms. For example, a is only in the first and third terms, b is in the first and second, and 5 and 2 are in different terms. So factoring out a single common factor from all terms isn't possible. Hence, grouping is the way to go.

Another way to think about it: maybe factor by rearranging terms? Let me try rearranging the original expression. Let's see, ab + 2a + 5b + 10. Then group (ab + 2a) + (5b + 10) as before. Which gives a(b + 2) + 5(b + 2) = (a + 5)(b + 2). Same result.

Alternatively, if I rearrange as ab + 5b + 2a + 10, maybe factor b from the first two terms and 2 from the last two. Which is what I did first.

So regardless of the grouping, the result is the same. Therefore, I think the factoring is correct.

Wait, just to make sure, let me plug in some numbers for a and b and check if both the original expression and the factored form give the same result. Let me choose a = 1 and b = 1.

Original expression: ab + 5b + 2a + 10 = (1)(1) + 5(1) + 2(1) + 10 = 1 + 5 + 2 + 10 = 18.

Factored form: (a + 5)(b + 2) = (1 + 5)(1 + 2) = 6 * 3 = 18. Same result.

Another test: a = 2, b = 3.

Original: (2)(3) + 5(3) + 2(2) + 10 = 6 + 15 + 4 + 10 = 35.

Factored: (2 + 5)(3 + 2) = 7 * 5 = 35. Same again.

Another test: a = 0, b = 0.

Original: 0 + 0 + 0 + 10 = 10.

Factored: (0 + 5)(0 + 2) = 5 * 2 = 10. Correct.

Another test: a = -5, b = -2.

Original: (-5)(-2) + 5(-2) + 2(-5) + 10 = 10 -10 -10 +10 = 0.

Factored: (-5 + 5)(-2 + 2) = 0 * 0 = 0. Correct.

So all these tests work. Therefore, I can be confident that the factoring is correct.

Therefore, the expression factors to (a + 5)(b + 2).

**Final Answer**
The factored form of the expression is \boxed{(a + 5)(b + 2)}.

To factor the expression $ ab + 5b + 2a + 10 $, we can use the method of **factoring by grouping**. This method is particularly useful when the expression has four terms and can be split into two pairs that share a common factor.

---

### Step 1: Group the terms

We group the expression into two pairs:

$$
(ab + 5b) + (2a + 10)
$$

---

### Step 2: Factor out the greatest common factor (GCF) from each group

- From the first group $ ab + 5b $, the GCF is $ b $, so we factor it out:

$$
b(a + 5)
$$

- From the second group $ 2a + 10 $, the GCF is $ 2 $, so we factor it out:

$$
2(a + 5)
$$

Now the expression becomes:

$$
b(a + 5) + 2(a + 5)
$$

---

### Step 3: Factor out the common binomial factor

Both terms now have a common factor of $ (a + 5) $. We factor that out:

$$
(a + 5)(b + 2)
$$

---

### Step 4: Verify the result

To ensure the factoring is correct, we can expand the factored form:

$$
(a + 5)(b + 2) = ab + 2a + 5b + 10
$$

This matches the original expression $ ab + 5b + 2a + 10 $, confirming that the factoring is correct.

---

### Final Answer

$$
\boxed{(a + 5)(b + 2)}
$$

[Assistant]
Explanation:
Okay, so I need to factor the expression ab + 5b + 2a + 10. Hmm, factoring can sometimes be tricky, but let me try to remember the methods. I think grouping might be useful here. Let me see...

First, let me write down the expression again to make sure I have it right: ab + 5b + 2a + 10. Alright. Now, factoring by grouping usually involves grouping terms in pairs and then factoring out common factors from each pair. Let me try that.

Looking at the terms, I have ab, 5b, 2a, and 10. Let me group the first two terms together and the last two terms together. So, (ab + 5b) + (2a + 10). Now, let me factor out the greatest common factor (GCF) from each group.

Starting with the first group, ab + 5b. Both terms have a 'b' in them, so I can factor out a 'b'. That would leave me with b(a + 5). Let me check: b times a is ab, and b times 5 is 5b. Yep, that works.

Now the second group is 2a + 10. Both terms are divisible by 2, so factoring out a 2 gives me 2(a + 5). Let me verify: 2 times a is 2a, and 2 times 5 is 10. Perfect.

So now, the expression becomes: b(a + 5) + 2(a + 5). Wait a second, both terms have a common factor of (a + 5). That means I can factor that out. Let me do that.

Factoring out (a + 5) from both terms, I get: (a + 5)(b + 2). Let me check if that's correct by expanding it back.

Expanding (a + 5)(b + 2): first multiply a by b, which is ab. Then a times 2 is 2a. Then 5 times b is 5b. Finally, 5 times 2 is 10. So combining those terms: ab + 2a + 5b + 10. Which is the same as the original expression ab + 5b + 2a + 10. The order of the terms is different, but addition is commutative, so it's the same.

Therefore, the factored form is (a + 5)(b + 2). Let me just make sure there isn't another way to factor it or if I missed something.

Alternatively, could I have grouped the terms differently? Let me try grouping ab with 2a and 5b with 10. So, (ab + 2a) + (5b + 10). Factoring out a from the first group: a(b + 2). Factoring out 5 from the second group: 5(b + 2). Then combining them: (a + 5)(b + 2). Same result. So regardless of how I group them, I end up with the same factors.

Another check: maybe factor out something else? Let me see. The original expression is ab + 5b + 2a + 10. If I try to factor out a 'b' from the first two terms and a '2' from the last two, which is what I did first, and then factor out the common (a + 5).

Alternatively, maybe factor out something else? Let me think. Suppose I tried to factor out a different term. For example, if I look for common factors across all terms. The terms are ab, 5b, 2a, 10. The only common factor across all terms is... Well, there's no common factor in all four terms. For example, a is only in the first and third terms, b is in the first and second, and 5 and 2 are in different terms. So factoring out a single common factor from all terms isn't possible. Hence, grouping is the way to go.

Another way to think about it: maybe factor by rearranging terms? Let me try rearranging the original expression. Let's see, ab + 2a + 5b + 10. Then group (ab + 2a) + (5b + 10) as before. Which gives a(b + 2) + 5(b + 2) = (a + 5)(b + 2). Same result.

Alternatively, if I rearrange as ab + 5b + 2a + 10, maybe factor b from the first two terms and 2 from the last two. Which is what I did first.

So regardless of the grouping, the result is the same. Therefore, I think the factoring is correct.

Wait, just to make sure, let me plug in some numbers for a and b and check if both the original expression and the factored form give the same result. Let me choose a = 1 and b = 1.

Original expression: ab + 5b + 2a + 10 = (1)(1) + 5(1) + 2(1) + 10 = 1 + 5 + 2 + 10 = 18.

Factored form: (a + 5)(b + 2) = (1 + 5)(1 + 2) = 6 * 3 = 18. Same result.

Another test: a = 2, b = 3.

Original: (2)(3) + 5(3) + 2(2) + 10 = 6 + 15 + 4 + 10 = 35.

Factored: (2 + 5)(3 + 2) = 7 * 5 = 35. Same again.

Another test: a = 0, b = 0.

Original: 0 + 0 + 0 + 10 = 10.

Factored: (0 + 5)(0 + 2) = 5 * 2 = 10. Correct.

Another test: a = -5, b = -2.

Original: (-5)(-2) + 5(-2) + 2(-5) + 10 = 10 -10 -10 +10 = 0.

Factored: (-5 + 5)(-2 + 2) = 0 * 0 = 0. Correct.

So all these tests work. Therefore, I can be confident that the factoring is correct.

Therefore, the expression factors to (a + 5)(b + 2).

**Final Answer**
The factored form of the expression is \boxed{(a + 5)(b + 2)}.

To factor the expression $ ab + 5b + 2a + 10 $, we can use the method of **factoring by grouping**. This method is particularly useful when the expression has four terms and can be split into two pairs that share a common factor.

---

### Step 1: Group the terms

We group the expression into two pairs:

$$
(ab + 5b) + (2a + 10)
$$

---

### Step 2: Factor out the greatest common factor (GCF) from each group

- From the first group $ ab + 5b $, the GCF is $ b $, so we factor it out:

$$
b(a + 5)
$$

- From the second group $ 2a + 10 $, the GCF is $ 2 $, so we factor it out:

$$
2(a + 5)
$$

Now the expression becomes:

$$
b(a + 5) + 2(a + 5)
$$

---

### Step 3: Factor out the common binomial factor

Both terms now have a common factor of $ (a + 5) $. We factor that out:

$$
(a + 5)(b + 2)
$$

---

### Step 4: Verify the result

To ensure the factoring is correct, we can expand the factored form:

$$
(a + 5)(b + 2) = ab + 2a + 5b + 10
$$

This matches the original expression $ ab + 5b + 2a + 10 $, confirming that the factoring is correct.

---

### Final Answer

$$
\boxed{(a + 5)(b + 2)}
$$

Solution: We have $ab +5b+2a+10 = ab + 5b+2a + 2\cdot 5$, so we have a straightforward application of Simon's Favorite Factoring Trick: \[ab + 5b+2a+10 = \boxed{(a+5)(b+2)}.\]


Explanation:
Okay, let me try to figure out this problem. So, Zach has three bags and a bunch of pencils. He needs to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. The question is asking for the greatest number of pencils he could have left over. Hmm, let me parse this again.

Wait, so he wants to put the greatest number of pencils possible into each bag, but also keep the number of pencils in each bag the same? That seems a bit confusing. Maybe it's saying that he wants to divide the pencils into three bags such that each bag has the same number of pencils, and he wants to maximize the number of pencils in each bag. But then, if he wants to maximize the number in each bag, he would just divide all the pencils equally among the three bags. But then there would be no leftovers. But the question says "the greatest number of pencils he could have left over." So maybe there's something else here.

Wait, maybe the problem is that he has a certain number of pencils, and he wants to distribute them into three bags such that each bag has the same number of pencils, but he wants to maximize the number in each bag. However, if he can't divide the total number of pencils evenly by three, then there will be leftovers. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is as large as possible, but with the constraint that when divided by three, there's some remainder. But the problem doesn't specify a total number of pencils. Wait, maybe I need to re-examine the problem.

Wait, the problem says: "Zach has three bags and a bunch of pencils to be placed into the bags. He is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. What is the greatest number of pencils he could have left over?"

Wait, maybe the problem is that he has a certain number of pencils, and he wants to divide them into three bags with the same number in each bag. But he wants to maximize the number in each bag. However, if the total number of pencils isn't divisible by three, then there will be leftovers. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is as large as possible, but with the remainder when divided by three being as large as possible? Wait, but how does that work?

Wait, maybe the problem is that he has a certain number of pencils, and he wants to divide them into three bags with the same number in each bag. But he wants to maximize the number in each bag. However, if the total number of pencils isn't divisible by three, then he can't have all bags with the same number. Therefore, he has to leave some pencils over. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is just one more than a multiple of three? Wait, no. Let me think again.

Suppose the total number of pencils is N. If he wants to divide them into three bags with the same number in each, then N must be divisible by three. If it's not, then he can't do that. But the problem says he is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. So maybe he is trying to divide the pencils into three bags with equal numbers, but he can't do it perfectly. Therefore, he has to leave some pencils over. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is as large as possible, but with the remainder when divided by three being as large as possible? Wait, but how does that work?

Wait, maybe the problem is that he has a certain number of pencils, and he wants to divide them into three bags with the same number in each. However, he can't do that perfectly, so he has to leave some pencils. But the question is asking for the maximum number of pencils that could be left over. But if he wants to maximize the number in each bag, then he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is just one more than a multiple of three? Wait, no. Let me think.

Suppose the total number of pencils is N. If he wants to divide them into three bags with the same number in each, then N must be divisible by three. If it's not, then he can't do that. However, if he wants to place the greatest number of pencils possible into each bag while keeping the number in each bag the same, maybe he is trying to find the maximum number of pencils per bag such that when multiplied by three, it's less than or equal to N. So, the maximum number of pencils per bag would be floor(N / 3). Then, the number of pencils left over would be N - 3 * floor(N / 3). But the problem is asking for the greatest number of pencils he could have left over. But if N is arbitrary, then the maximum number of leftovers would be 2, since when you divide by three, the remainder can be 0, 1, or 2. Therefore, the maximum number of pencils left over is 2. But that seems too straightforward. However, the problem says "the greatest number of pencils he could have left over." So maybe the answer is 2? But why is the problem phrased so confusingly?

Wait, maybe I'm misunderstanding the problem. Let me read it again:

"Zach has three bags and a bunch of pencils to be placed into the bags. He is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. What is the greatest number of pencils he could have left over?"

Wait, maybe the problem is that he wants to place the greatest number of pencils possible into each bag, but also keep the number of pencils in each bag the same. So, if he wants to maximize the number in each bag, he would want to have as many as possible, but since they have to be the same in each bag, he can't have different numbers. So, if he has N pencils, then the maximum number per bag is floor(N / 3), and the remainder is N mod 3. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number in each bag, then the remainder would be minimized? Wait, no. Wait, if he wants to maximize the number in each bag, he would want to have as many as possible, so the remainder would be as small as possible? Wait, no. Wait, if you have N pencils, and you want to divide them into three bags with the same number in each, then the maximum number per bag is floor(N / 3). The remainder is N - 3 * floor(N / 3). So, the remainder can be 0, 1, or 2. Therefore, the maximum remainder is 2. Therefore, the answer is 2? But why is the problem asking for that? Maybe there's a trick here.

Wait, but maybe the problem is not about dividing all the pencils, but about placing pencils into the bags such that each bag has the same number of pencils, but he wants to maximize the number in each bag. However, if he can't divide all pencils, then he leaves some over. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number in each bag, he would want to have as many as possible, so the remainder would be as small as possible? Wait, but the question is asking for the greatest number of pencils left over. So, if he wants to maximize the number in each bag, he would have the remainder be as small as possible. But the question is asking for the maximum possible remainder. So, maybe the problem is not about a specific N, but about what is the maximum possible remainder when dividing by three? Which is 2. Therefore, the answer is 2. But maybe there's more to it?

Wait, but maybe the problem is in the wording. "Place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same." Maybe "keeping the number of pencils in each bag the same" refers to something else? Like, maybe he wants to have the same number in each bag as some previous configuration? But the problem doesn't mention anything else. Maybe it's a translation issue or wording issue.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the maximum number of pencils per bag is floor(N / 3), and the remainder is N - 3 * floor(N / 3). But since the problem says "the greatest number of pencils he could have left over," maybe it's asking for the maximum possible remainder, which is 2. Therefore, the answer is 2. But maybe the problem is more complex?

Wait, let me think again. Suppose that the total number of pencils is not fixed. The problem says "a bunch of pencils," so maybe the number of pencils is arbitrary. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number of pencils left over, then he would want to have as many as possible. But if he wants to place the greatest number of pencils possible into each bag while keeping the number in each bag the same, then maybe he is trying to maximize the number in each bag, which would require that the total number of pencils is as large as possible. But since the number of pencils isn't fixed, maybe the problem is about the maximum possible remainder when dividing by three? Which is 2. Therefore, the answer is 2. But maybe there's a different interpretation.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to maximize the number in each bag, he would want N to be as large as possible. But since N isn't given, maybe the question is about the maximum possible remainder? Which is 2. Therefore, the answer is 2. But maybe the problem is in another way.

Wait, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the number of pencils left over would be N - 3 * k, where k is the number per bag. To maximize the number left over, he would need to maximize N - 3k. But since k is the maximum possible, which is floor(N / 3), then N - 3k is the remainder. Therefore, the maximum remainder is 2. Therefore, the answer is 2. So, maybe the answer is 2. But why is the problem considered difficult? Maybe I'm missing something.

Wait, let me check with examples. Suppose he has 4 pencils. Then, he can put 1 pencil in each bag, leaving 1 pencil. If he has 5 pencils, he can put 1 in each bag, leaving 2. If he has 6 pencils, he can put 2 in each, leaving 0. If he has 7 pencils, he can put 2 in each, leaving 1. If he has 8 pencils, he can put 2 in each, leaving 2. So, the maximum remainder is 2. Therefore, regardless of the total number of pencils, the maximum number of pencils left over is 2. Therefore, the answer is 2. Therefore, the answer is \boxed{2}.

But maybe the problem is different. Wait, the problem says "the greatest number of pencils he could have left over." So, if he wants to maximize the number left over, he would want to have as many as possible. But if he wants to place the greatest number of pencils possible into each bag, then he would want to minimize the number left over. Wait, this is conflicting. Wait, the problem says: "place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same." So, he wants to maximize the number in each bag, which would require that the number left over is minimized. However, the question is asking for the greatest number of pencils he could have left over. So, maybe there's a misunderstanding here.

Wait, maybe the problem is that he is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. So, maybe he is trying to do both: maximize the number in each bag and keep the number in each bag the same. But if he wants to maximize the number in each bag, then he would have to have as many as possible, but if the total number of pencils isn't divisible by three, then he can't have the same number in each bag. Therefore, he has to leave some pencils over. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number in each bag, he would want to have the total number of pencils be as large as possible, but with the remainder being as large as possible? Wait, that seems contradictory.

Wait, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the number of pencils left over would be N - 3k, where k is the number per bag. To maximize the number left over, he would need to maximize N - 3k. But since k is the maximum possible, which is floor(N / 3), then N - 3k is the remainder. Therefore, the maximum remainder is 2. Therefore, the answer is 2. Therefore, regardless of N, the maximum remainder is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to maximize the number left over, he would need to have N as large as possible. But since N isn't given, maybe the problem is about the maximum possible remainder when dividing by three, which is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is in the translation. Maybe "place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same." Maybe "keeping the number of pencils in each bag the same" refers to something else. Maybe he wants to have the same number in each bag as some previous number? But there's no previous number mentioned.

Alternatively, maybe the problem is that he wants to place pencils into the bags such that each bag has the same number of pencils, and he wants to maximize the number in each bag. Therefore, the number of pencils left over is N - 3k, where k is the number per bag. To maximize the number left over, he would need to maximize N - 3k. But since k is the maximum possible, which is floor(N / 3), then N - 3k is the remainder. Therefore, the maximum remainder is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to maximize the number left over, he would need to have N as large as possible. But since N isn't given, maybe the problem is asking for the maximum possible remainder, which is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the number of pencils left over is N - 3k. If he wants to maximize the number left over, he would need to have N as large as possible, but since N isn't given, maybe the problem is asking for the maximum possible remainder when dividing by three, which is 2. Therefore, the answer is 2.

I think that's the only way to interpret it. Therefore, the answer is 2. Therefore, the greatest number of pencils he could have left over is 2. Therefore, the answer is \boxed{2}.

**Final Answer**
\boxed{2}

To solve the problem, we need to understand the key constraints and what is being asked:

---

### **Understanding the Problem**

Zach has **three bags** and a **bunch of pencils**. He is told to:

- **Place the greatest number of pencils possible into each of the three bags**
- **While keeping the number of pencils in each bag the same**

The question asks: **What is the greatest number of pencils he could have left over?**

---

### **Breaking Down the Requirements**

Let’s denote the total number of pencils as $ N $, and the number of pencils in each bag as $ k $. Since there are three bags, the total number of pencils used is $ 3k $.

So, the number of pencils left over is:

$$
\text{Leftover} = N - 3k
$$

To **maximize the number of pencils in each bag**, we want to choose the **largest possible** $ k $ such that $ 3k \leq N $. That is, $ k = \left\lfloor \frac{N}{3} \right\rfloor $, the **floor division** of $ N $ by 3.

Then, the number of pencils left over is:

$$
\text{Leftover} = N - 3 \left\lfloor \frac{N}{3} \right\rfloor
$$

This is the **remainder** when $ N $ is divided by 3.

---

### **What is the Maximum Possible Remainder?**

When dividing any integer $ N $ by 3, the possible remainders are:

- 0 (if $ N $ is divisible by 3)
- 1
- 2

So, the **maximum possible remainder** is **2**.

This means that **no matter how many pencils Zach has**, the **maximum number of pencils he could have left over** after placing the greatest number of pencils possible into each of the three bags (while keeping the number in each bag the same) is **2**.

---

### **Conclusion**

The greatest number of pencils Zach could have left over is:

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, Zach has three bags and a bunch of pencils. He needs to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. The question is asking for the greatest number of pencils he could have left over. Hmm, let me parse this again.

Wait, so he wants to put the greatest number of pencils possible into each bag, but also keep the number of pencils in each bag the same? That seems a bit confusing. Maybe it's saying that he wants to divide the pencils into three bags such that each bag has the same number of pencils, and he wants to maximize the number of pencils in each bag. But then, if he wants to maximize the number in each bag, he would just divide all the pencils equally among the three bags. But then there would be no leftovers. But the question says "the greatest number of pencils he could have left over." So maybe there's something else here.

Wait, maybe the problem is that he has a certain number of pencils, and he wants to distribute them into three bags such that each bag has the same number of pencils, but he wants to maximize the number in each bag. However, if he can't divide the total number of pencils evenly by three, then there will be leftovers. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is as large as possible, but with the constraint that when divided by three, there's some remainder. But the problem doesn't specify a total number of pencils. Wait, maybe I need to re-examine the problem.

Wait, the problem says: "Zach has three bags and a bunch of pencils to be placed into the bags. He is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. What is the greatest number of pencils he could have left over?"

Wait, maybe the problem is that he has a certain number of pencils, and he wants to divide them into three bags with the same number in each bag. But he wants to maximize the number in each bag. However, if the total number of pencils isn't divisible by three, then there will be leftovers. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is as large as possible, but with the remainder when divided by three being as large as possible? Wait, but how does that work?

Wait, maybe the problem is that he has a certain number of pencils, and he wants to divide them into three bags with the same number in each bag. But he wants to maximize the number in each bag. However, if the total number of pencils isn't divisible by three, then he can't have all bags with the same number. Therefore, he has to leave some pencils over. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is just one more than a multiple of three? Wait, no. Let me think again.

Suppose the total number of pencils is N. If he wants to divide them into three bags with the same number in each, then N must be divisible by three. If it's not, then he can't do that. But the problem says he is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. So maybe he is trying to divide the pencils into three bags with equal numbers, but he can't do it perfectly. Therefore, he has to leave some pencils over. But the question is asking for the greatest number of pencils he could have left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is as large as possible, but with the remainder when divided by three being as large as possible? Wait, but how does that work?

Wait, maybe the problem is that he has a certain number of pencils, and he wants to divide them into three bags with the same number in each. However, he can't do that perfectly, so he has to leave some pencils. But the question is asking for the maximum number of pencils that could be left over. But if he wants to maximize the number in each bag, then he would want to have as many pencils as possible in each bag, which would mean that the total number of pencils is just one more than a multiple of three? Wait, no. Let me think.

Suppose the total number of pencils is N. If he wants to divide them into three bags with the same number in each, then N must be divisible by three. If it's not, then he can't do that. However, if he wants to place the greatest number of pencils possible into each bag while keeping the number in each bag the same, maybe he is trying to find the maximum number of pencils per bag such that when multiplied by three, it's less than or equal to N. So, the maximum number of pencils per bag would be floor(N / 3). Then, the number of pencils left over would be N - 3 * floor(N / 3). But the problem is asking for the greatest number of pencils he could have left over. But if N is arbitrary, then the maximum number of leftovers would be 2, since when you divide by three, the remainder can be 0, 1, or 2. Therefore, the maximum number of pencils left over is 2. But that seems too straightforward. However, the problem says "the greatest number of pencils he could have left over." So maybe the answer is 2? But why is the problem phrased so confusingly?

Wait, maybe I'm misunderstanding the problem. Let me read it again:

"Zach has three bags and a bunch of pencils to be placed into the bags. He is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. What is the greatest number of pencils he could have left over?"

Wait, maybe the problem is that he wants to place the greatest number of pencils possible into each bag, but also keep the number of pencils in each bag the same. So, if he wants to maximize the number in each bag, he would want to have as many as possible, but since they have to be the same in each bag, he can't have different numbers. So, if he has N pencils, then the maximum number per bag is floor(N / 3), and the remainder is N mod 3. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number in each bag, then the remainder would be minimized? Wait, no. Wait, if he wants to maximize the number in each bag, he would want to have as many as possible, so the remainder would be as small as possible? Wait, no. Wait, if you have N pencils, and you want to divide them into three bags with the same number in each, then the maximum number per bag is floor(N / 3). The remainder is N - 3 * floor(N / 3). So, the remainder can be 0, 1, or 2. Therefore, the maximum remainder is 2. Therefore, the answer is 2? But why is the problem asking for that? Maybe there's a trick here.

Wait, but maybe the problem is not about dividing all the pencils, but about placing pencils into the bags such that each bag has the same number of pencils, but he wants to maximize the number in each bag. However, if he can't divide all pencils, then he leaves some over. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number in each bag, he would want to have as many as possible, so the remainder would be as small as possible? Wait, but the question is asking for the greatest number of pencils left over. So, if he wants to maximize the number in each bag, he would have the remainder be as small as possible. But the question is asking for the maximum possible remainder. So, maybe the problem is not about a specific N, but about what is the maximum possible remainder when dividing by three? Which is 2. Therefore, the answer is 2. But maybe there's more to it?

Wait, but maybe the problem is in the wording. "Place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same." Maybe "keeping the number of pencils in each bag the same" refers to something else? Like, maybe he wants to have the same number in each bag as some previous configuration? But the problem doesn't mention anything else. Maybe it's a translation issue or wording issue.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the maximum number of pencils per bag is floor(N / 3), and the remainder is N - 3 * floor(N / 3). But since the problem says "the greatest number of pencils he could have left over," maybe it's asking for the maximum possible remainder, which is 2. Therefore, the answer is 2. But maybe the problem is more complex?

Wait, let me think again. Suppose that the total number of pencils is not fixed. The problem says "a bunch of pencils," so maybe the number of pencils is arbitrary. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number of pencils left over, then he would want to have as many as possible. But if he wants to place the greatest number of pencils possible into each bag while keeping the number in each bag the same, then maybe he is trying to maximize the number in each bag, which would require that the total number of pencils is as large as possible. But since the number of pencils isn't fixed, maybe the problem is about the maximum possible remainder when dividing by three? Which is 2. Therefore, the answer is 2. But maybe there's a different interpretation.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to maximize the number in each bag, he would want N to be as large as possible. But since N isn't given, maybe the question is about the maximum possible remainder? Which is 2. Therefore, the answer is 2. But maybe the problem is in another way.

Wait, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the number of pencils left over would be N - 3 * k, where k is the number per bag. To maximize the number left over, he would need to maximize N - 3k. But since k is the maximum possible, which is floor(N / 3), then N - 3k is the remainder. Therefore, the maximum remainder is 2. Therefore, the answer is 2. So, maybe the answer is 2. But why is the problem considered difficult? Maybe I'm missing something.

Wait, let me check with examples. Suppose he has 4 pencils. Then, he can put 1 pencil in each bag, leaving 1 pencil. If he has 5 pencils, he can put 1 in each bag, leaving 2. If he has 6 pencils, he can put 2 in each, leaving 0. If he has 7 pencils, he can put 2 in each, leaving 1. If he has 8 pencils, he can put 2 in each, leaving 2. So, the maximum remainder is 2. Therefore, regardless of the total number of pencils, the maximum number of pencils left over is 2. Therefore, the answer is 2. Therefore, the answer is \boxed{2}.

But maybe the problem is different. Wait, the problem says "the greatest number of pencils he could have left over." So, if he wants to maximize the number left over, he would want to have as many as possible. But if he wants to place the greatest number of pencils possible into each bag, then he would want to minimize the number left over. Wait, this is conflicting. Wait, the problem says: "place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same." So, he wants to maximize the number in each bag, which would require that the number left over is minimized. However, the question is asking for the greatest number of pencils he could have left over. So, maybe there's a misunderstanding here.

Wait, maybe the problem is that he is told to place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same. So, maybe he is trying to do both: maximize the number in each bag and keep the number in each bag the same. But if he wants to maximize the number in each bag, then he would have to have as many as possible, but if the total number of pencils isn't divisible by three, then he can't have the same number in each bag. Therefore, he has to leave some pencils over. But the question is asking for the greatest number of pencils he could have left over. So, if he wants to maximize the number in each bag, he would want to have the total number of pencils be as large as possible, but with the remainder being as large as possible? Wait, that seems contradictory.

Wait, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the number of pencils left over would be N - 3k, where k is the number per bag. To maximize the number left over, he would need to maximize N - 3k. But since k is the maximum possible, which is floor(N / 3), then N - 3k is the remainder. Therefore, the maximum remainder is 2. Therefore, the answer is 2. Therefore, regardless of N, the maximum remainder is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to maximize the number left over, he would need to have N as large as possible. But since N isn't given, maybe the problem is about the maximum possible remainder when dividing by three, which is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is in the translation. Maybe "place the greatest number of pencils possible into each of the three bags while also keeping the number of pencils in each bag the same." Maybe "keeping the number of pencils in each bag the same" refers to something else. Maybe he wants to have the same number in each bag as some previous number? But there's no previous number mentioned.

Alternatively, maybe the problem is that he wants to place pencils into the bags such that each bag has the same number of pencils, and he wants to maximize the number in each bag. Therefore, the number of pencils left over is N - 3k, where k is the number per bag. To maximize the number left over, he would need to maximize N - 3k. But since k is the maximum possible, which is floor(N / 3), then N - 3k is the remainder. Therefore, the maximum remainder is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to maximize the number left over, he would need to have N as large as possible. But since N isn't given, maybe the problem is asking for the maximum possible remainder, which is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, but he wants to maximize the number in each bag. Therefore, the number of pencils left over is N - 3k. If he wants to maximize the number left over, he would need to have N as large as possible, but since N isn't given, maybe the problem is asking for the maximum possible remainder when dividing by three, which is 2. Therefore, the answer is 2.

I think that's the only way to interpret it. Therefore, the answer is 2. Therefore, the greatest number of pencils he could have left over is 2. Therefore, the answer is \boxed{2}.

**Final Answer**
\boxed{2}

To solve the problem, we need to understand the key constraints and what is being asked:

---

### **Understanding the Problem**

Zach has **three bags** and a **bunch of pencils**. He is told to:

- **Place the greatest number of pencils possible into each of the three bags**
- **While keeping the number of pencils in each bag the same**

The question asks: **What is the greatest number of pencils he could have left over?**

---

### **Breaking Down the Requirements**

Let’s denote the total number of pencils as $ N $, and the number of pencils in each bag as $ k $. Since there are three bags, the total number of pencils used is $ 3k $.

So, the number of pencils left over is:

$$
\text{Leftover} = N - 3k
$$

To **maximize the number of pencils in each bag**, we want to choose the **largest possible** $ k $ such that $ 3k \leq N $. That is, $ k = \left\lfloor \frac{N}{3} \right\rfloor $, the **floor division** of $ N $ by 3.

Then, the number of pencils left over is:

$$
\text{Leftover} = N - 3 \left\lfloor \frac{N}{3} \right\rfloor
$$

This is the **remainder** when $ N $ is divided by 3.

---

### **What is the Maximum Possible Remainder?**

When dividing any integer $ N $ by 3, the possible remainders are:

- 0 (if $ N $ is divisible by 3)
- 1
- 2

So, the **maximum possible remainder** is **2**.

This means that **no matter how many pencils Zach has**, the **maximum number of pencils he could have left over** after placing the greatest number of pencils possible into each of the three bags (while keeping the number in each bag the same) is **2**.

---

### **Conclusion**

The greatest number of pencils Zach could have left over is:

$$
\boxed{2}
$$

Solution: If Zach has three or more pencils left over, then he can add another pencil to each bag.  Therefore, Zach can have at most $\boxed{2}$ pencils left over.


Explanation:
Okay, so I need to find the point on the curve that's given by the parametric equations when t equals 2. The parametric equations are x equals t cubed plus 7, and y equals negative 3t squared minus 6t minus 5. Alright, let me think about how to approach this.

First, I remember that parametric equations define x and y in terms of a third variable, which is called a parameter—in this case, t. To find the point on the curve at a specific value of t, I just need to substitute that value of t into both equations and calculate the corresponding x and y coordinates. That seems straightforward. Let me confirm that.

So, for t = 2, I need to compute x and y. Let me write down the equations again:

x = t³ + 7

y = -3t² - 6t - 5

Yes, so substituting t = 2 into both equations should give me the coordinates of the point. Let me start with the x-coordinate.

Calculating x when t = 2:

x = (2)³ + 7

Hmm, 2 cubed is 8, right? So 8 + 7 is 15. So x = 15. That seems simple enough.

Now for the y-coordinate. Let me plug t = 2 into the y equation:

y = -3*(2)² - 6*(2) - 5

Let me compute each term step by step. First, (2) squared is 4. Then multiply that by -3: -3*4 = -12.

Next term is -6*(2). Let me calculate that: -6*2 = -12.

Then the last term is -5. So putting it all together:

y = -12 - 12 - 5

Let me add those up. Starting with -12 -12 is -24, and then -24 -5 is -29. So y equals -29?

Wait, let me check that again to make sure I didn't make a mistake with the signs. The original equation is y = -3t² -6t -5. So when t = 2:

First term: -3*(2)^2 = -3*4 = -12

Second term: -6*(2) = -12

Third term: -5

So yes, adding them up: -12 -12 -5. That's definitely -29. Okay, so y is -29.

Therefore, the point on the curve when t = 2 is (15, -29). Let me just verify once more to be sure.

For x: 2³ is 8, plus 7 is 15. Correct.

For y: Let me compute each term again:

-3*(2)^2: 2 squared is 4, times -3 is -12.

-6*(2): that's -12.

Then -5. So total is -12 -12 -5. Let me add them in another order: -12 -12 is -24, then -24 -5 is -29. Yep, same result.

So I think that's correct. Therefore, the point is (15, -29). Wait, but maybe I should check if there's any trick or something? The problem says "a curve is parameterized by..." and asks for the point at t=2. It seems straightforward. Maybe I need to check if there's any restrictions on t or something? But the problem doesn't mention any, so I think substituting t=2 is the way to go.

Alternatively, maybe I can check by plugging t=2 into the parametric equations again. Let me write them again:

x(t) = t³ + 7. So x(2) = 8 + 7 = 15. Correct.

y(t) = -3t² -6t -5. So y(2) = -3*(4) -6*(2) -5 = -12 -12 -5 = -29. Yep, same as before.

So I think that's solid. Therefore, the answer should be (15, -29). Let me just make sure that I didn't mix up any signs. The y-component is negative, so all the terms are negative? Let me check:

Original equation: y = -3t² -6t -5. So all three terms are negative? Wait, no. Wait, the equation is written as -3t² -6t -5. So that's equivalent to (-3t²) + (-6t) + (-5). So all three terms are negative. So when t is positive, all terms are negative. So when t=2, all terms are negative, so adding them up gives a negative number. So -29 is correct.

Alternatively, if I had written the equation as y = - (3t² + 6t + 5), that would be the same thing. So maybe that's another way to think about it. So plugging t=2 into 3t² + 6t + 5 would be 3*(4) + 6*(2) + 5 = 12 + 12 + 5 = 29, so then y is -29. Yep, same result. So that's another way to check. So that seems to confirm it again.

Therefore, I think I can be confident that the point is (15, -29). I don't see any mistakes in my calculations. Let me just recap:

For x: t³ + 7 at t=2 is 8 + 7 = 15.

For y: -3t² -6t -5 at t=2 is -12 -12 -5 = -29.

Yes, that's correct. So the answer is (15, -29). Therefore, I think that's the point the curve passes through at t=2.

**Final Answer**
The point the curve passes through at $ t = 2 $ is \boxed{(15, -29)}.

To find the point on the curve that the parameterization passes through at $ t = 2 $, we are given the parametric equations:

$$
x = t^3 + 7, \quad y = -3t^2 - 6t - 5
$$

---

### Step 1: Substitute $ t = 2 $ into the equation for $ x $

$$
x = (2)^3 + 7 = 8 + 7 = 15
$$

---

### Step 2: Substitute $ t = 2 $ into the equation for $ y $

$$
y = -3(2)^2 - 6(2) - 5
$$

Break it down:

- $ (2)^2 = 4 $
- $ -3 \cdot 4 = -12 $
- $ -6 \cdot 2 = -12 $
- $ -5 $ remains as is

Now add the terms:

$$
y = -12 - 12 - 5 = -29
$$

---

### Final Result

The point on the curve when $ t = 2 $ is:

$$
\boxed{(15, -29)}
$$

[Assistant]
Explanation:
Okay, so I need to find the point on the curve that's given by the parametric equations when t equals 2. The parametric equations are x equals t cubed plus 7, and y equals negative 3t squared minus 6t minus 5. Alright, let me think about how to approach this.

First, I remember that parametric equations define x and y in terms of a third variable, which is called a parameter—in this case, t. To find the point on the curve at a specific value of t, I just need to substitute that value of t into both equations and calculate the corresponding x and y coordinates. That seems straightforward. Let me confirm that.

So, for t = 2, I need to compute x and y. Let me write down the equations again:

x = t³ + 7

y = -3t² - 6t - 5

Yes, so substituting t = 2 into both equations should give me the coordinates of the point. Let me start with the x-coordinate.

Calculating x when t = 2:

x = (2)³ + 7

Hmm, 2 cubed is 8, right? So 8 + 7 is 15. So x = 15. That seems simple enough.

Now for the y-coordinate. Let me plug t = 2 into the y equation:

y = -3*(2)² - 6*(2) - 5

Let me compute each term step by step. First, (2) squared is 4. Then multiply that by -3: -3*4 = -12.

Next term is -6*(2). Let me calculate that: -6*2 = -12.

Then the last term is -5. So putting it all together:

y = -12 - 12 - 5

Let me add those up. Starting with -12 -12 is -24, and then -24 -5 is -29. So y equals -29?

Wait, let me check that again to make sure I didn't make a mistake with the signs. The original equation is y = -3t² -6t -5. So when t = 2:

First term: -3*(2)^2 = -3*4 = -12

Second term: -6*(2) = -12

Third term: -5

So yes, adding them up: -12 -12 -5. That's definitely -29. Okay, so y is -29.

Therefore, the point on the curve when t = 2 is (15, -29). Let me just verify once more to be sure.

For x: 2³ is 8, plus 7 is 15. Correct.

For y: Let me compute each term again:

-3*(2)^2: 2 squared is 4, times -3 is -12.

-6*(2): that's -12.

Then -5. So total is -12 -12 -5. Let me add them in another order: -12 -12 is -24, then -24 -5 is -29. Yep, same result.

So I think that's correct. Therefore, the point is (15, -29). Wait, but maybe I should check if there's any trick or something? The problem says "a curve is parameterized by..." and asks for the point at t=2. It seems straightforward. Maybe I need to check if there's any restrictions on t or something? But the problem doesn't mention any, so I think substituting t=2 is the way to go.

Alternatively, maybe I can check by plugging t=2 into the parametric equations again. Let me write them again:

x(t) = t³ + 7. So x(2) = 8 + 7 = 15. Correct.

y(t) = -3t² -6t -5. So y(2) = -3*(4) -6*(2) -5 = -12 -12 -5 = -29. Yep, same as before.

So I think that's solid. Therefore, the answer should be (15, -29). Let me just make sure that I didn't mix up any signs. The y-component is negative, so all the terms are negative? Let me check:

Original equation: y = -3t² -6t -5. So all three terms are negative? Wait, no. Wait, the equation is written as -3t² -6t -5. So that's equivalent to (-3t²) + (-6t) + (-5). So all three terms are negative. So when t is positive, all terms are negative. So when t=2, all terms are negative, so adding them up gives a negative number. So -29 is correct.

Alternatively, if I had written the equation as y = - (3t² + 6t + 5), that would be the same thing. So maybe that's another way to think about it. So plugging t=2 into 3t² + 6t + 5 would be 3*(4) + 6*(2) + 5 = 12 + 12 + 5 = 29, so then y is -29. Yep, same result. So that's another way to check. So that seems to confirm it again.

Therefore, I think I can be confident that the point is (15, -29). I don't see any mistakes in my calculations. Let me just recap:

For x: t³ + 7 at t=2 is 8 + 7 = 15.

For y: -3t² -6t -5 at t=2 is -12 -12 -5 = -29.

Yes, that's correct. So the answer is (15, -29). Therefore, I think that's the point the curve passes through at t=2.

**Final Answer**
The point the curve passes through at $ t = 2 $ is \boxed{(15, -29)}.

To find the point on the curve that the parameterization passes through at $ t = 2 $, we are given the parametric equations:

$$
x = t^3 + 7, \quad y = -3t^2 - 6t - 5
$$

---

### Step 1: Substitute $ t = 2 $ into the equation for $ x $

$$
x = (2)^3 + 7 = 8 + 7 = 15
$$

---

### Step 2: Substitute $ t = 2 $ into the equation for $ y $

$$
y = -3(2)^2 - 6(2) - 5
$$

Break it down:

- $ (2)^2 = 4 $
- $ -3 \cdot 4 = -12 $
- $ -6 \cdot 2 = -12 $
- $ -5 $ remains as is

Now add the terms:

$$
y = -12 - 12 - 5 = -29
$$

---

### Final Result

The point on the curve when $ t = 2 $ is:

$$
\boxed{(15, -29)}
$$

Solution: At $t = 2,$ $(x,y) = (2^3 + 7, -3 \cdot 2^2 - 6 \cdot 2 - 5) = \boxed{(15,-29)}.$


Explanation:
Okay, so I need to solve the equation 2/3 equals 4 divided by (x minus 5). Let me write that down first to make sure I have it right: 2/3 = 4/(x - 5). Alright, so I need to find the value of x that makes this equation true.

Hmm, let me think. This looks like a proportion, where two fractions are equal. So maybe I can use cross-multiplication here? I remember that if a/b = c/d, then a*d = b*c. Let me try that.

So applying cross-multiplication to 2/3 = 4/(x - 5), that would mean 2 times (x - 5) equals 3 times 4. Let me write that out: 2*(x - 5) = 3*4.

Calculating the right side first, 3 times 4 is 12. So now the equation is 2*(x - 5) = 12. Okay, now I need to solve for x. Let me distribute the 2 on the left side. That would be 2*x minus 2*5, which is 2x - 10. So now the equation is 2x - 10 = 12.

Alright, now I need to isolate x. Let me add 10 to both sides to get rid of the -10. So adding 10 to both sides gives me 2x = 12 + 10. Let me calculate that: 12 + 10 is 22. So now it's 2x = 22.

To solve for x, I should divide both sides by 2. Dividing both sides by 2 gives me x = 22/2. Simplifying that, 22 divided by 2 is 11. So x equals 11.

Wait, let me check if that makes sense. Let me plug x = 11 back into the original equation to verify.

Original equation: 2/3 = 4/(x - 5). Substitute x with 11: 4/(11 - 5) = 4/6. Simplify 4/6 by dividing numerator and denominator by 2: 2/3. Which is equal to the left side of the equation. Perfect, that checks out.

But just to make sure I didn't make any mistakes in my steps. Let me go through them again.

Starting with cross-multiplication: 2*(x - 5) = 3*4. That's correct because cross-multiplying gives a*d = b*c. Then 2x - 10 = 12. Then adding 10 to both sides: 2x = 22. Dividing by 2: x = 11. Yep, that seems right.

Another thing to check is if there are any restrictions on x. The original equation has a denominator of (x - 5), so x cannot be 5 because that would make the denominator zero, which is undefined. Since our solution is x = 11, which is not 5, that's okay. So no issues there.

Is there another way to approach this problem? Maybe by manipulating the equation step by step without cross-multiplying? Let me try that.

Starting with 2/3 = 4/(x - 5). Maybe I can first solve for x - 5. Let me rewrite the equation as 4/(x - 5) = 2/3. Then, if I take reciprocal of both sides, I get (x - 5)/4 = 3/2. Wait, is that right? Taking reciprocal of both sides would flip both fractions. So reciprocal of 4/(x - 5) is (x - 5)/4, and reciprocal of 2/3 is 3/2. So yes, that gives (x - 5)/4 = 3/2.

Then, solving for x - 5, I can multiply both sides by 4. So (x - 5) = (3/2)*4. Calculating that, 3/2 times 4 is 6. So x - 5 = 6. Then adding 5 to both sides gives x = 6 + 5 = 11. Same answer.

So that method also gives x = 11. So that's reassuring.

Alternatively, maybe I can solve it by first simplifying the equation. Let me see.

Original equation: 2/3 = 4/(x - 5). Let me think about what this means. The fraction 2/3 is equal to 4 divided by (x - 5). So if I can write 4 divided by something equals 2/3, then that something must be 4 divided by (2/3). Let me think.

If 4 divided by (x - 5) equals 2/3, then (x - 5) equals 4 divided by (2/3). Dividing by a fraction is the same as multiplying by its reciprocal, so 4 divided by (2/3) is 4*(3/2) = 12/2 = 6. Therefore, x - 5 = 6, so x = 6 + 5 = 11.

Same result again. So regardless of the method, I get x = 11.

I think that's solid. Let me just check once more with substitution.

Left side: 2/3. Right side: 4/(11 - 5) = 4/6 = 2/3. Yep, same.

Is there any chance I could have messed up the cross-multiplication? Let me think. If I have a/b = c/d, cross-multiplying gives ad = bc. So in this case, a is 2, b is 3, c is 4, d is (x - 5). Therefore, 2*(x - 5) = 3*4. Which is exactly what I did. So that's correct.

Another common mistake might be to incorrectly distribute or solve the equation. For example, if someone forgot to multiply both terms inside the parentheses by 2, they might have done 2x - 5 = 12, which would lead to a wrong answer. But I didn't make that mistake.

Also, when adding 10 to both sides, making sure that 12 + 10 is 22. That's correct.

So, all steps seem to check out.

Therefore, I can confidently say that the solution is x = 11.

**Final Answer**
The solution is \boxed{11}.

To solve the equation:

$$
\frac{2}{3} = \frac{4}{x - 5}
$$

we aim to find the value of $ x $ that makes the equation true.

---

### Step 1: Cross-Multiply

Since the equation is a proportion, we can use **cross-multiplication** to eliminate the fractions:

$$
2(x - 5) = 3 \cdot 4
$$

---

### Step 2: Simplify Both Sides

On the left side, distribute the 2:

$$
2x - 10 = 12
$$

---

### Step 3: Solve for $ x $

Add 10 to both sides:

$$
2x = 22
$$

Now divide both sides by 2:

$$
x = 11
$$

---

### Step 4: Verify the Solution

Substitute $ x = 11 $ back into the original equation:

$$
\frac{2}{3} = \frac{4}{11 - 5} = \frac{4}{6} = \frac{2}{3}
$$

Both sides are equal, confirming that the solution is correct.

---

### Step 5: Check for Restrictions

The original equation has a denominator $ x - 5 $, so we must ensure that $ x \neq 5 $ to avoid division by zero. Since $ x = 11 $, this condition is satisfied.

---

### Final Answer

$$
\boxed{11}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation 2/3 equals 4 divided by (x minus 5). Let me write that down first to make sure I have it right: 2/3 = 4/(x - 5). Alright, so I need to find the value of x that makes this equation true.

Hmm, let me think. This looks like a proportion, where two fractions are equal. So maybe I can use cross-multiplication here? I remember that if a/b = c/d, then a*d = b*c. Let me try that.

So applying cross-multiplication to 2/3 = 4/(x - 5), that would mean 2 times (x - 5) equals 3 times 4. Let me write that out: 2*(x - 5) = 3*4.

Calculating the right side first, 3 times 4 is 12. So now the equation is 2*(x - 5) = 12. Okay, now I need to solve for x. Let me distribute the 2 on the left side. That would be 2*x minus 2*5, which is 2x - 10. So now the equation is 2x - 10 = 12.

Alright, now I need to isolate x. Let me add 10 to both sides to get rid of the -10. So adding 10 to both sides gives me 2x = 12 + 10. Let me calculate that: 12 + 10 is 22. So now it's 2x = 22.

To solve for x, I should divide both sides by 2. Dividing both sides by 2 gives me x = 22/2. Simplifying that, 22 divided by 2 is 11. So x equals 11.

Wait, let me check if that makes sense. Let me plug x = 11 back into the original equation to verify.

Original equation: 2/3 = 4/(x - 5). Substitute x with 11: 4/(11 - 5) = 4/6. Simplify 4/6 by dividing numerator and denominator by 2: 2/3. Which is equal to the left side of the equation. Perfect, that checks out.

But just to make sure I didn't make any mistakes in my steps. Let me go through them again.

Starting with cross-multiplication: 2*(x - 5) = 3*4. That's correct because cross-multiplying gives a*d = b*c. Then 2x - 10 = 12. Then adding 10 to both sides: 2x = 22. Dividing by 2: x = 11. Yep, that seems right.

Another thing to check is if there are any restrictions on x. The original equation has a denominator of (x - 5), so x cannot be 5 because that would make the denominator zero, which is undefined. Since our solution is x = 11, which is not 5, that's okay. So no issues there.

Is there another way to approach this problem? Maybe by manipulating the equation step by step without cross-multiplying? Let me try that.

Starting with 2/3 = 4/(x - 5). Maybe I can first solve for x - 5. Let me rewrite the equation as 4/(x - 5) = 2/3. Then, if I take reciprocal of both sides, I get (x - 5)/4 = 3/2. Wait, is that right? Taking reciprocal of both sides would flip both fractions. So reciprocal of 4/(x - 5) is (x - 5)/4, and reciprocal of 2/3 is 3/2. So yes, that gives (x - 5)/4 = 3/2.

Then, solving for x - 5, I can multiply both sides by 4. So (x - 5) = (3/2)*4. Calculating that, 3/2 times 4 is 6. So x - 5 = 6. Then adding 5 to both sides gives x = 6 + 5 = 11. Same answer.

So that method also gives x = 11. So that's reassuring.

Alternatively, maybe I can solve it by first simplifying the equation. Let me see.

Original equation: 2/3 = 4/(x - 5). Let me think about what this means. The fraction 2/3 is equal to 4 divided by (x - 5). So if I can write 4 divided by something equals 2/3, then that something must be 4 divided by (2/3). Let me think.

If 4 divided by (x - 5) equals 2/3, then (x - 5) equals 4 divided by (2/3). Dividing by a fraction is the same as multiplying by its reciprocal, so 4 divided by (2/3) is 4*(3/2) = 12/2 = 6. Therefore, x - 5 = 6, so x = 6 + 5 = 11.

Same result again. So regardless of the method, I get x = 11.

I think that's solid. Let me just check once more with substitution.

Left side: 2/3. Right side: 4/(11 - 5) = 4/6 = 2/3. Yep, same.

Is there any chance I could have messed up the cross-multiplication? Let me think. If I have a/b = c/d, cross-multiplying gives ad = bc. So in this case, a is 2, b is 3, c is 4, d is (x - 5). Therefore, 2*(x - 5) = 3*4. Which is exactly what I did. So that's correct.

Another common mistake might be to incorrectly distribute or solve the equation. For example, if someone forgot to multiply both terms inside the parentheses by 2, they might have done 2x - 5 = 12, which would lead to a wrong answer. But I didn't make that mistake.

Also, when adding 10 to both sides, making sure that 12 + 10 is 22. That's correct.

So, all steps seem to check out.

Therefore, I can confidently say that the solution is x = 11.

**Final Answer**
The solution is \boxed{11}.

To solve the equation:

$$
\frac{2}{3} = \frac{4}{x - 5}
$$

we aim to find the value of $ x $ that makes the equation true.

---

### Step 1: Cross-Multiply

Since the equation is a proportion, we can use **cross-multiplication** to eliminate the fractions:

$$
2(x - 5) = 3 \cdot 4
$$

---

### Step 2: Simplify Both Sides

On the left side, distribute the 2:

$$
2x - 10 = 12
$$

---

### Step 3: Solve for $ x $

Add 10 to both sides:

$$
2x = 22
$$

Now divide both sides by 2:

$$
x = 11
$$

---

### Step 4: Verify the Solution

Substitute $ x = 11 $ back into the original equation:

$$
\frac{2}{3} = \frac{4}{11 - 5} = \frac{4}{6} = \frac{2}{3}
$$

Both sides are equal, confirming that the solution is correct.

---

### Step 5: Check for Restrictions

The original equation has a denominator $ x - 5 $, so we must ensure that $ x \neq 5 $ to avoid division by zero. Since $ x = 11 $, this condition is satisfied.

---

### Final Answer

$$
\boxed{11}
$$

Solution: Multiplying both sides by $x-5$ and by 3 gives $2(x-5) = 4(3)$.  Expanding the left side gives $2x-10 = 12$.  Adding 10 to both sides gives $2x = 22$ and dividing by 2 gives $x = \boxed{11}$.


Explanation:
Okay, so I need to find the volume of a cone. The formula given is V equals one-third times B times h. Alright, let me make sure I remember that correctly. Yeah, I think that's right. The volume of a cone is one-third the area of the base times the height.

The problem says the area of the base is 30 square units. So B is 30. And the height is 6.5 units. So h is 6.5. So I just need to plug those numbers into the formula. Let me write that down.

V = (1/3) * B * h
V = (1/3) * 30 * 6.5

Hmm, let me compute that. First, maybe I can simplify the numbers. Let me do 1/3 of 30 first. Because 30 divided by 3 is 10. So that would make it 10 times 6.5. Wait, is that right? Let me check. If I take 1/3 of 30, that's 10, yes. Then multiply by 6.5. So 10 times 6.5.

What's 10 times 6.5? Well, 10 times 6 is 60, and 10 times 0.5 is 5, so adding those together gives 65. So the volume should be 65 cubic units? Let me verify that again to make sure I didn't make a mistake.

Alternatively, maybe I should do the multiplication step by step. Let me try another way. Let's compute 30 times 6.5 first, and then divide by 3.

30 times 6.5. Hmm, 30 times 6 is 180, and 30 times 0.5 is 15. So adding those together, 180 + 15 = 195. Then divide by 3. 195 divided by 3. Let me do that division. 3 goes into 19 five times (15), remainder 4. 3 goes into 4 once (3), remainder 1. Bring down the 5, making 15. 3 goes into 15 five times. So that's 65. So same result. So 65.

So both ways I get 65. So that seems consistent. Therefore, the volume is 65 cubic units.

Wait, let me check if I interpreted the formula correctly. The formula is V = (1/3)Bh. Yes, that's the formula for the volume of a cone. The base area times height times one-third. So since they gave me the base area directly, I don't need to calculate the base area from radius or anything. They already told me B is 30. So that's straightforward.

Another way to think about it: if the base area was, say, πr², but since they gave me B, I don't need to worry about that. So yeah, just plug in the numbers.

So 30 times 6.5 is 195, divided by 3 is 65. Yep. So I think that's correct.

Wait, just to make sure, maybe I can convert 6.5 into a fraction. 6.5 is the same as 13/2. So let me try that.

So V = (1/3) * 30 * (13/2). Let me compute that.

First, multiply 1/3 and 30. 30 divided by 3 is 10. Then multiply by 13/2. So 10 * 13 / 2. Which is (130)/2 = 65. Yep, same answer.

So that's another way to do it. So regardless of the method, I get 65.

Therefore, I think the volume is 65 cubic units.

I don't see any mistakes in my calculations. All the different methods give me the same result. So I think that's the correct answer.

**Final Answer**
The volume of the cone is \boxed{65} cubic units.

To find the volume of a cone, we use the formula:

$$
V = \frac{1}{3}Bh
$$

where:
- $ V $ is the volume,
- $ B $ is the area of the base,
- $ h $ is the height of the cone.

---

### Step 1: Identify the Given Values

From the problem:
- The area of the base, $ B = 30 $ square units,
- The height, $ h = 6.5 $ units.

---

### Step 2: Plug the Values into the Formula

$$
V = \frac{1}{3} \times 30 \times 6.5
$$

---

### Step 3: Simplify the Expression

First, compute $ \frac{1}{3} \times 30 $:

$$
\frac{1}{3} \times 30 = 10
$$

Now multiply by 6.5:

$$
10 \times 6.5 = 65
$$

---

### Step 4: Final Answer

Thus, the volume of the cone is:

$$
\boxed{65}
$$

cubic units.

[Assistant]
Explanation:
Okay, so I need to find the volume of a cone. The formula given is V equals one-third times B times h. Alright, let me make sure I remember that correctly. Yeah, I think that's right. The volume of a cone is one-third the area of the base times the height.

The problem says the area of the base is 30 square units. So B is 30. And the height is 6.5 units. So h is 6.5. So I just need to plug those numbers into the formula. Let me write that down.

V = (1/3) * B * h
V = (1/3) * 30 * 6.5

Hmm, let me compute that. First, maybe I can simplify the numbers. Let me do 1/3 of 30 first. Because 30 divided by 3 is 10. So that would make it 10 times 6.5. Wait, is that right? Let me check. If I take 1/3 of 30, that's 10, yes. Then multiply by 6.5. So 10 times 6.5.

What's 10 times 6.5? Well, 10 times 6 is 60, and 10 times 0.5 is 5, so adding those together gives 65. So the volume should be 65 cubic units? Let me verify that again to make sure I didn't make a mistake.

Alternatively, maybe I should do the multiplication step by step. Let me try another way. Let's compute 30 times 6.5 first, and then divide by 3.

30 times 6.5. Hmm, 30 times 6 is 180, and 30 times 0.5 is 15. So adding those together, 180 + 15 = 195. Then divide by 3. 195 divided by 3. Let me do that division. 3 goes into 19 five times (15), remainder 4. 3 goes into 4 once (3), remainder 1. Bring down the 5, making 15. 3 goes into 15 five times. So that's 65. So same result. So 65.

So both ways I get 65. So that seems consistent. Therefore, the volume is 65 cubic units.

Wait, let me check if I interpreted the formula correctly. The formula is V = (1/3)Bh. Yes, that's the formula for the volume of a cone. The base area times height times one-third. So since they gave me the base area directly, I don't need to calculate the base area from radius or anything. They already told me B is 30. So that's straightforward.

Another way to think about it: if the base area was, say, πr², but since they gave me B, I don't need to worry about that. So yeah, just plug in the numbers.

So 30 times 6.5 is 195, divided by 3 is 65. Yep. So I think that's correct.

Wait, just to make sure, maybe I can convert 6.5 into a fraction. 6.5 is the same as 13/2. So let me try that.

So V = (1/3) * 30 * (13/2). Let me compute that.

First, multiply 1/3 and 30. 30 divided by 3 is 10. Then multiply by 13/2. So 10 * 13 / 2. Which is (130)/2 = 65. Yep, same answer.

So that's another way to do it. So regardless of the method, I get 65.

Therefore, I think the volume is 65 cubic units.

I don't see any mistakes in my calculations. All the different methods give me the same result. So I think that's the correct answer.

**Final Answer**
The volume of the cone is \boxed{65} cubic units.

To find the volume of a cone, we use the formula:

$$
V = \frac{1}{3}Bh
$$

where:
- $ V $ is the volume,
- $ B $ is the area of the base,
- $ h $ is the height of the cone.

---

### Step 1: Identify the Given Values

From the problem:
- The area of the base, $ B = 30 $ square units,
- The height, $ h = 6.5 $ units.

---

### Step 2: Plug the Values into the Formula

$$
V = \frac{1}{3} \times 30 \times 6.5
$$

---

### Step 3: Simplify the Expression

First, compute $ \frac{1}{3} \times 30 $:

$$
\frac{1}{3} \times 30 = 10
$$

Now multiply by 6.5:

$$
10 \times 6.5 = 65
$$

---

### Step 4: Final Answer

Thus, the volume of the cone is:

$$
\boxed{65}
$$

cubic units.

Solution: We are given that $B = 30$ and $h = 6.5$ and asked to find $\frac{1}{3}Bh$.  We find that \[\frac{1}{3}Bh = \frac{1}{3}(30)(6.5) = (10)(6.5) = \boxed{65}.\]


Explanation:
Okay, so Louis and Jack are sharing a pie, and I need to figure out how much of the pie they eat altogether. Let me think. Louis eats 1/5 of the pie, and Jack eats 2/3 of the pie. Hmm, so I guess I need to add those two fractions together to find the total amount they ate.

But wait, adding fractions requires them to have the same denominator, right? So first, I need to find a common denominator for 1/5 and 2/3. Let me recall how to do that. The denominators are 5 and 3. The least common denominator would be the least common multiple of 5 and 3. Let me think... 5 and 3 are both prime numbers, so their least common multiple is just 5 times 3, which is 15. Okay, so 15 is the common denominator.

Now, I need to convert both fractions to have 15 as the denominator. Starting with Louis's portion, which is 1/5. To convert 1/5 to fifteenths, I multiply both the numerator and the denominator by 3. Because 5 times 3 is 15. So, 1 times 3 is 3, and 5 times 3 is 15. So, 1/5 is equivalent to 3/15.

Next, Jack's portion is 2/3. To convert that to fifteenths, I need to multiply numerator and denominator by 5, since 3 times 5 is 15. So, 2 times 5 is 10, and 3 times 5 is 15. Therefore, 2/3 is equivalent to 10/15.

Now, adding the two converted fractions: 3/15 + 10/15. Since the denominators are the same, I can just add the numerators. 3 + 10 is 13. So, the sum is 13/15.

Wait, let me check if that makes sense. If Louis ate 1/5, which is 0.2 of the pie, and Jack ate 2/3, which is approximately 0.666... of the pie, then adding them together would be 0.2 + 0.666..., which is about 0.866... And 13 divided by 15 is approximately 0.866..., so that seems to check out.

But let me verify the fraction addition again. 1/5 + 2/3. Converting to 15ths: 3/15 + 10/15 = 13/15. Yes, that's correct. So altogether, they ate 13/15 of the pie.

Is there any chance I made a mistake in the common denominator? Let me think. 5 and 3 have LCM 15, so that's right. Converting 1/5 to 3/15 and 2/3 to 10/15. Adding them gives 13/15.

Alternatively, maybe I can think of it as decimals. 1 divided by 5 is 0.2, and 2 divided by 3 is approximately 0.666666... Adding them together: 0.2 + 0.666666... equals 0.866666..., which is the same as 13/15. Because 13 divided by 15 is 0.8666..., so that's consistent.

Another way to check is to subtract the total from 1 to see how much is left. If they ate 13/15, then the remaining pie is 1 - 13/15 = 2/15. Let me see if that makes sense. If Louis ate 1/5 (which is 3/15) and Jack ate 2/3 (which is 10/15), then together they ate 13/15, leaving 2/15. That seems reasonable.

Alternatively, if I didn't convert to common denominators, and just added 1/5 + 2/3, I might have made a mistake. For example, someone might incorrectly add the numerators and denominators separately, like 1+2 over 5+3, which would be 3/8. But that's wrong. But since I converted to common denominators, I think my answer is correct.

So, I think the answer is 13/15. Let me just write that in the box as instructed.

**Final Answer**
The total amount of the pie that Louis and Jack eat altogether is \boxed{\dfrac{13}{15}}.

To determine how much of the pie Louis and Jack eat altogether, we need to add the fractions of the pie that each of them consumes:

- Louis eats $\frac{1}{5}$ of the pie.
- Jack eats $\frac{2}{3}$ of the pie.

---

### Step 1: Find a Common Denominator

To add the fractions, we need a common denominator. The denominators are 5 and 3. The least common multiple (LCM) of 5 and 3 is 15.

---

### Step 2: Convert the Fractions

Convert both fractions to have a denominator of 15:

- $\frac{1}{5} = \frac{1 \times 3}{5 \times 3} = \frac{3}{15}$
- $\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

---

### Step 3: Add the Fractions

Now that both fractions have the same denominator, we can add them:

$$
\frac{3}{15} + \frac{10}{15} = \frac{13}{15}
$$

---

### Step 4: Interpret the Result

The sum $\frac{13}{15}$ represents the total portion of the pie that Louis and Jack ate together.

---

### Final Answer

$$
\boxed{\dfrac{13}{15}}
$$

[Assistant]
Explanation:
Okay, so Louis and Jack are sharing a pie, and I need to figure out how much of the pie they eat altogether. Let me think. Louis eats 1/5 of the pie, and Jack eats 2/3 of the pie. Hmm, so I guess I need to add those two fractions together to find the total amount they ate.

But wait, adding fractions requires them to have the same denominator, right? So first, I need to find a common denominator for 1/5 and 2/3. Let me recall how to do that. The denominators are 5 and 3. The least common denominator would be the least common multiple of 5 and 3. Let me think... 5 and 3 are both prime numbers, so their least common multiple is just 5 times 3, which is 15. Okay, so 15 is the common denominator.

Now, I need to convert both fractions to have 15 as the denominator. Starting with Louis's portion, which is 1/5. To convert 1/5 to fifteenths, I multiply both the numerator and the denominator by 3. Because 5 times 3 is 15. So, 1 times 3 is 3, and 5 times 3 is 15. So, 1/5 is equivalent to 3/15.

Next, Jack's portion is 2/3. To convert that to fifteenths, I need to multiply numerator and denominator by 5, since 3 times 5 is 15. So, 2 times 5 is 10, and 3 times 5 is 15. Therefore, 2/3 is equivalent to 10/15.

Now, adding the two converted fractions: 3/15 + 10/15. Since the denominators are the same, I can just add the numerators. 3 + 10 is 13. So, the sum is 13/15.

Wait, let me check if that makes sense. If Louis ate 1/5, which is 0.2 of the pie, and Jack ate 2/3, which is approximately 0.666... of the pie, then adding them together would be 0.2 + 0.666..., which is about 0.866... And 13 divided by 15 is approximately 0.866..., so that seems to check out.

But let me verify the fraction addition again. 1/5 + 2/3. Converting to 15ths: 3/15 + 10/15 = 13/15. Yes, that's correct. So altogether, they ate 13/15 of the pie.

Is there any chance I made a mistake in the common denominator? Let me think. 5 and 3 have LCM 15, so that's right. Converting 1/5 to 3/15 and 2/3 to 10/15. Adding them gives 13/15.

Alternatively, maybe I can think of it as decimals. 1 divided by 5 is 0.2, and 2 divided by 3 is approximately 0.666666... Adding them together: 0.2 + 0.666666... equals 0.866666..., which is the same as 13/15. Because 13 divided by 15 is 0.8666..., so that's consistent.

Another way to check is to subtract the total from 1 to see how much is left. If they ate 13/15, then the remaining pie is 1 - 13/15 = 2/15. Let me see if that makes sense. If Louis ate 1/5 (which is 3/15) and Jack ate 2/3 (which is 10/15), then together they ate 13/15, leaving 2/15. That seems reasonable.

Alternatively, if I didn't convert to common denominators, and just added 1/5 + 2/3, I might have made a mistake. For example, someone might incorrectly add the numerators and denominators separately, like 1+2 over 5+3, which would be 3/8. But that's wrong. But since I converted to common denominators, I think my answer is correct.

So, I think the answer is 13/15. Let me just write that in the box as instructed.

**Final Answer**
The total amount of the pie that Louis and Jack eat altogether is \boxed{\dfrac{13}{15}}.

To determine how much of the pie Louis and Jack eat altogether, we need to add the fractions of the pie that each of them consumes:

- Louis eats $\frac{1}{5}$ of the pie.
- Jack eats $\frac{2}{3}$ of the pie.

---

### Step 1: Find a Common Denominator

To add the fractions, we need a common denominator. The denominators are 5 and 3. The least common multiple (LCM) of 5 and 3 is 15.

---

### Step 2: Convert the Fractions

Convert both fractions to have a denominator of 15:

- $\frac{1}{5} = \frac{1 \times 3}{5 \times 3} = \frac{3}{15}$
- $\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

---

### Step 3: Add the Fractions

Now that both fractions have the same denominator, we can add them:

$$
\frac{3}{15} + \frac{10}{15} = \frac{13}{15}
$$

---

### Step 4: Interpret the Result

The sum $\frac{13}{15}$ represents the total portion of the pie that Louis and Jack ate together.

---

### Final Answer

$$
\boxed{\dfrac{13}{15}}
$$

Solution: The denominators $5$ and $3$ have a common multiple of $15$.  We use this to write $\frac{1}{5}\cdot \frac{3}{3}=\frac{3}{15}$ and $\frac{2}{3}\cdot \frac{5}{5}=\frac{10}{15}$.  Then, we can add the fractions by adding the numerators and keeping the denominator.  We have $$\frac{1}{5}+\frac{2}{3}=\frac{3}{15}+\frac{10}{15}=\frac{3+10}{15}=\boxed{\frac{13}{15}}.$$


Explanation:
Okay, so I need to find the sixth term of an arithmetic sequence where the fourth term is 200 and the eighth term is 500. Let me recall what an arithmetic sequence is. It's a sequence where each term increases by a constant difference. So, if I denote the first term as $ a $ and the common difference as $ d $, then the $ n $-th term of the sequence can be written as $ a_n = a + (n - 1)d $.

Alright, so the fourth term is given as 200. Let me write that down using the formula. For the fourth term, $ n = 4 $, so:

$ a_4 = a + (4 - 1)d = a + 3d = 200 $.

Similarly, the eighth term is 500. So for $ n = 8 $:

$ a_8 = a + (8 - 1)d = a + 7d = 500 $.

Now I have two equations:

1. $ a + 3d = 200 $
2. $ a + 7d = 500 $

I need to solve these two equations to find $ a $ and $ d $, and then use them to find the sixth term. Let me think about how to solve these equations. Since both equations have $ a $ and $ d $, I can subtract the first equation from the second to eliminate $ a $.

Subtracting equation 1 from equation 2:

$ (a + 7d) - (a + 3d) = 500 - 200 $

Simplify the left side:

$ a + 7d - a - 3d = 4d $

And the right side is:

$ 300 $

So, $ 4d = 300 $. Therefore, solving for $ d $:

$ d = 300 / 4 = 75 $.

Okay, so the common difference $ d $ is 75. Now I can plug this back into one of the equations to find $ a $. Let me use equation 1:

$ a + 3d = 200 $

Substituting $ d = 75 $:

$ a + 3*75 = 200 $

Calculate 3*75:

3*75 is 225. So:

$ a + 225 = 200 $

Subtract 225 from both sides:

$ a = 200 - 225 = -25 $

So the first term $ a $ is -25. Let me check if this makes sense with the second equation. Let's plug $ a = -25 $ and $ d = 75 $ into equation 2:

$ a + 7d = -25 + 7*75 $

Calculate 7*75: 7*70 is 490, 7*5 is 35, so total is 490 + 35 = 525. Then:

-25 + 525 = 500. Which matches the given eighth term. Great, that checks out.

Now, I need to find the sixth term. Let me use the formula again. The sixth term is $ a_6 = a + (6 - 1)d = a + 5d $.

We already know $ a = -25 $ and $ d = 75 $, so:

$ a_6 = -25 + 5*75 $

Calculate 5*75: 5*70 is 350, 5*5 is 25, so total is 350 + 25 = 375.

Then, adding to -25:

-25 + 375 = 350.

Wait, so the sixth term is 350? Let me verify this another way to make sure I didn't make a mistake.

Alternatively, since we know the fourth term is 200 and the eighth term is 500, maybe I can find the common difference by looking at the difference between the eighth term and the fourth term. The eighth term is four terms after the fourth term. So, the difference between term 8 and term 4 is 4d. Which we already calculated as 300, so 4d = 300, so d = 75. That's consistent with before.

So, from term 4 to term 6 is two terms. So, the sixth term is term 4 plus 2d. Let me check that:

Term 4: 200

Term 5: 200 + d = 200 + 75 = 275

Term 6: 275 + 75 = 350. Yep, same result. So that seems correct.

Alternatively, term 6 is halfway between term 4 and term 8? Wait, term 4 is 200, term 8 is 500. The difference between term 8 and term 4 is 300 over 4 terms. So, the average of term 4 and term 8 would be (200 + 500)/2 = 700/2 = 350. Since term 6 is exactly in the middle of term 4 and term 8 (since 4 to 6 is two terms, and 6 to 8 is two terms), so the sixth term is the average of term 4 and term 8. That's another way to see it. So that gives me 350 again. So that's a good check.

Therefore, the sixth term is 350. So I think that's the answer. Let me just recap to make sure I didn't skip any steps or make any calculation errors.

First, set up equations for term 4 and term 8. Solved for d by subtracting equations, found d = 75. Then found a = -25. Then calculated term 6 as a + 5d = -25 + 375 = 350. Cross-checked by moving from term 4 to term 6 with two steps of d, and also by averaging term 4 and term 8 since term 6 is halfway. All methods give 350, so I think that's solid.

I don't see any mistakes here. So the answer should be 350.

**Final Answer**
The sixth term is \boxed{350}.

To solve the problem, we are given that the fourth term of an arithmetic sequence is 200 and the eighth term is 500. We are asked to find the sixth term.

---

### Step 1: Use the General Formula for an Arithmetic Sequence

The general formula for the $ n $-th term of an arithmetic sequence is:

$$
a_n = a + (n - 1)d
$$

Where:
- $ a $ is the first term,
- $ d $ is the common difference,
- $ n $ is the term number.

---

### Step 2: Write Equations for the Given Terms

We are told:
- $ a_4 = 200 $, so:
  $$
  a + 3d = 200 \quad \text{(1)}
  $$
- $ a_8 = 500 $, so:
  $$
  a + 7d = 500 \quad \text{(2)}
  $$

---

### Step 3: Solve the System of Equations

Subtract equation (1) from equation (2):

$$
(a + 7d) - (a + 3d) = 500 - 200
$$
$$
4d = 300 \Rightarrow d = 75
$$

Now substitute $ d = 75 $ into equation (1) to find $ a $:

$$
a + 3(75) = 200 \Rightarrow a + 225 = 200 \Rightarrow a = -25
$$

---

### Step 4: Find the Sixth Term

Now that we know $ a = -25 $ and $ d = 75 $, we can find the sixth term:

$$
a_6 = a + 5d = -25 + 5(75) = -25 + 375 = 350
$$

---

### Step 5: Cross-Check

We can also verify this by noting that the sixth term is exactly halfway between the fourth and eighth terms. The average of 200 and 500 is:

$$
\frac{200 + 500}{2} = \frac{700}{2} = 350
$$

This confirms our result.

---

### Final Answer

$$
\boxed{350}
$$

[Assistant]
Explanation:
Okay, so I need to find the sixth term of an arithmetic sequence where the fourth term is 200 and the eighth term is 500. Let me recall what an arithmetic sequence is. It's a sequence where each term increases by a constant difference. So, if I denote the first term as $ a $ and the common difference as $ d $, then the $ n $-th term of the sequence can be written as $ a_n = a + (n - 1)d $.

Alright, so the fourth term is given as 200. Let me write that down using the formula. For the fourth term, $ n = 4 $, so:

$ a_4 = a + (4 - 1)d = a + 3d = 200 $.

Similarly, the eighth term is 500. So for $ n = 8 $:

$ a_8 = a + (8 - 1)d = a + 7d = 500 $.

Now I have two equations:

1. $ a + 3d = 200 $
2. $ a + 7d = 500 $

I need to solve these two equations to find $ a $ and $ d $, and then use them to find the sixth term. Let me think about how to solve these equations. Since both equations have $ a $ and $ d $, I can subtract the first equation from the second to eliminate $ a $.

Subtracting equation 1 from equation 2:

$ (a + 7d) - (a + 3d) = 500 - 200 $

Simplify the left side:

$ a + 7d - a - 3d = 4d $

And the right side is:

$ 300 $

So, $ 4d = 300 $. Therefore, solving for $ d $:

$ d = 300 / 4 = 75 $.

Okay, so the common difference $ d $ is 75. Now I can plug this back into one of the equations to find $ a $. Let me use equation 1:

$ a + 3d = 200 $

Substituting $ d = 75 $:

$ a + 3*75 = 200 $

Calculate 3*75:

3*75 is 225. So:

$ a + 225 = 200 $

Subtract 225 from both sides:

$ a = 200 - 225 = -25 $

So the first term $ a $ is -25. Let me check if this makes sense with the second equation. Let's plug $ a = -25 $ and $ d = 75 $ into equation 2:

$ a + 7d = -25 + 7*75 $

Calculate 7*75: 7*70 is 490, 7*5 is 35, so total is 490 + 35 = 525. Then:

-25 + 525 = 500. Which matches the given eighth term. Great, that checks out.

Now, I need to find the sixth term. Let me use the formula again. The sixth term is $ a_6 = a + (6 - 1)d = a + 5d $.

We already know $ a = -25 $ and $ d = 75 $, so:

$ a_6 = -25 + 5*75 $

Calculate 5*75: 5*70 is 350, 5*5 is 25, so total is 350 + 25 = 375.

Then, adding to -25:

-25 + 375 = 350.

Wait, so the sixth term is 350? Let me verify this another way to make sure I didn't make a mistake.

Alternatively, since we know the fourth term is 200 and the eighth term is 500, maybe I can find the common difference by looking at the difference between the eighth term and the fourth term. The eighth term is four terms after the fourth term. So, the difference between term 8 and term 4 is 4d. Which we already calculated as 300, so 4d = 300, so d = 75. That's consistent with before.

So, from term 4 to term 6 is two terms. So, the sixth term is term 4 plus 2d. Let me check that:

Term 4: 200

Term 5: 200 + d = 200 + 75 = 275

Term 6: 275 + 75 = 350. Yep, same result. So that seems correct.

Alternatively, term 6 is halfway between term 4 and term 8? Wait, term 4 is 200, term 8 is 500. The difference between term 8 and term 4 is 300 over 4 terms. So, the average of term 4 and term 8 would be (200 + 500)/2 = 700/2 = 350. Since term 6 is exactly in the middle of term 4 and term 8 (since 4 to 6 is two terms, and 6 to 8 is two terms), so the sixth term is the average of term 4 and term 8. That's another way to see it. So that gives me 350 again. So that's a good check.

Therefore, the sixth term is 350. So I think that's the answer. Let me just recap to make sure I didn't skip any steps or make any calculation errors.

First, set up equations for term 4 and term 8. Solved for d by subtracting equations, found d = 75. Then found a = -25. Then calculated term 6 as a + 5d = -25 + 375 = 350. Cross-checked by moving from term 4 to term 6 with two steps of d, and also by averaging term 4 and term 8 since term 6 is halfway. All methods give 350, so I think that's solid.

I don't see any mistakes here. So the answer should be 350.

**Final Answer**
The sixth term is \boxed{350}.

To solve the problem, we are given that the fourth term of an arithmetic sequence is 200 and the eighth term is 500. We are asked to find the sixth term.

---

### Step 1: Use the General Formula for an Arithmetic Sequence

The general formula for the $ n $-th term of an arithmetic sequence is:

$$
a_n = a + (n - 1)d
$$

Where:
- $ a $ is the first term,
- $ d $ is the common difference,
- $ n $ is the term number.

---

### Step 2: Write Equations for the Given Terms

We are told:
- $ a_4 = 200 $, so:
  $$
  a + 3d = 200 \quad \text{(1)}
  $$
- $ a_8 = 500 $, so:
  $$
  a + 7d = 500 \quad \text{(2)}
  $$

---

### Step 3: Solve the System of Equations

Subtract equation (1) from equation (2):

$$
(a + 7d) - (a + 3d) = 500 - 200
$$
$$
4d = 300 \Rightarrow d = 75
$$

Now substitute $ d = 75 $ into equation (1) to find $ a $:

$$
a + 3(75) = 200 \Rightarrow a + 225 = 200 \Rightarrow a = -25
$$

---

### Step 4: Find the Sixth Term

Now that we know $ a = -25 $ and $ d = 75 $, we can find the sixth term:

$$
a_6 = a + 5d = -25 + 5(75) = -25 + 375 = 350
$$

---

### Step 5: Cross-Check

We can also verify this by noting that the sixth term is exactly halfway between the fourth and eighth terms. The average of 200 and 500 is:

$$
\frac{200 + 500}{2} = \frac{700}{2} = 350
$$

This confirms our result.

---

### Final Answer

$$
\boxed{350}
$$

Solution: The sixth term is exactly halfway between the fourth and the eighth in the arithmetic sequence, so it is the average of the two terms. Therefore, the sixth term is $(200 + 500)/2 = \boxed{350}$.  We also could have found the common difference by noting that there are four steps between the fourth term and the eighth term.  So, if $d$ is the common difference, we have $4d = 500-200 = 300$. Therefore, we find $d=75$.  The sixth term is two steps after the fourth, or $200 + 2d = \boxed{350}$.