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 of an example.

Suppose he has N pencils. If he wants to divide them into three bags with the same number in each, then N must be divisible by three. If not, then he can't. 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 he would have to take the floor division of N by three. But then the remainder would be N mod 3. However, 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 minimize the remainder? Wait, no. Wait, 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 remainder is as small as possible. Wait, but the problem says "the greatest number of pencils he could have left over." So maybe the question is phrased in a way that he is trying to divide the pencils into three bags with equal numbers, but he can't, 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 remainder is as small as possible. However, the problem says "the greatest number of pencils he could have left over." So maybe the problem is not about a specific number of pencils, but rather, given that he wants to maximize the number in each bag, what is the maximum possible remainder?

Wait, but the remainder when dividing by three can be 0, 1, or 2. So the maximum remainder is 2. Therefore, the greatest number of pencils he could have left over is 2? But that seems too straightforward. Maybe I'm misunderstanding the problem.

Wait, let me check again. 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 maybe he wants to have the same number in each bag, and he wants to maximize that number. So if he has N pencils, then the maximum number he can put in each bag is floor(N / 3). But then the leftover pencils would be N - 3 * floor(N / 3). But the question is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftover pencils, then he would want N - 3 * floor(N / 3) to be as large as possible. But since floor(N / 3) is the maximum number of pencils per bag, then the leftover is N mod 3. Therefore, the maximum possible leftover is 2. So the answer is 2? But that seems too simple. Maybe the problem is different.

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. However, if he can't divide the pencils equally, he can't do that. Therefore, the question is asking for the maximum number of pencils that could be left over if he tries to do that. But if he wants to maximize the number in each bag, he would have to have as many as possible, so the remainder would be minimized. Wait, but the problem says "the greatest number of pencils he could have left over." So maybe the problem is not about a specific N, but rather, given that he wants to maximize the number in each bag, what is the maximum possible remainder? But the remainder can be at most 2. So maybe the answer is 2. But why would the problem be phrased that way?

Alternatively, maybe the problem is that he has 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. However, there's a constraint that he has to have some pencils left over. Wait, but 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. But the problem says "the greatest number of pencils he could have left over." Maybe the problem is in translation or wording. Maybe it's saying that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag, but there's some leftover pencils. But then, the maximum number of leftovers would be when the total number of pencils is just one more than a multiple of three? Wait, no. If he wants to maximize the number in each bag, he would want to have as many as possible, so the total number of pencils is 3k + r, where r is the remainder. If he wants to maximize k, then r can be 0, 1, or 2. But if he wants to maximize the number in each bag, he would want to have k as large as possible, which would mean that r is as small as possible. However, the problem is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftovers, he would have r as large as possible, which is 2. Therefore, the answer is 2. But why is the problem phrased so confusingly?

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, and he wants to maximize the number in each bag, but he also wants to have as many pencils as possible left over. Wait, that seems contradictory. If he wants to maximize the number in each bag, he would want to minimize the leftovers. But if he wants to maximize the leftovers, he would want to minimize the number in each bag. So maybe the problem is not well-phrased. Maybe there's a different interpretation.

Wait, let me check again. 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 maybe he wants to have the same number in each bag, and he wants to maximize that number. So if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to have the same number in each bag, then the number of pencils left over is N - 3 * floor(N / 3). But the problem is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftovers, he would need to maximize N - 3 * floor(N / 3). But since floor(N / 3) is the maximum number per bag, then the leftover is N mod 3. Therefore, the maximum possible value of N mod 3 is 2. Therefore, the answer is 2. But maybe the problem is in another context?

Wait, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. But the problem says "the greatest number of pencils he could have left over." Maybe the problem is not about a specific number of pencils, but rather, given that he wants to maximize the number in each bag, what is the maximum possible remainder? Which would be 2. Therefore, the answer is 2. But maybe the problem is different.

Alternatively, maybe the problem is that he has to divide the pencils into three bags with the same number in each, but he can't do it perfectly, so he has to leave some pencils. But he wants to maximize the number in each bag. Therefore, the maximum number in each bag would be floor(N / 3), and the leftover is N - 3 * floor(N / 3). But the question is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftover, he would need to have N as large as possible? But there's no upper limit on N. Wait, but if N is arbitrary, then the leftover can be 0, 1, or 2. So the maximum possible leftover is 2. Therefore, regardless of N, the maximum number of pencils left over is 2. Therefore, the answer is 2. But maybe the problem is expecting a different answer?

Wait, let me think again. Maybe the problem is that he wants to have the same number of pencils in each bag, and he wants to maximize the number in each bag. However, he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. But the problem says "the greatest number of pencils he could have left over." Maybe the problem is in the translation or wording. Maybe it's supposed to say 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. Wait, that still doesn't make sense. Maybe it's a translation issue.

Alternatively, maybe the problem is that he has 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. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be as small as possible. Therefore, the maximum number of leftovers would be 2. But if he wants to have some pencils left over, then the maximum number of leftovers is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he has to divide the pencils into three bags with the same number in each, but he can't do it perfectly, so he has to leave some pencils. But he wants to maximize the number in each bag. Therefore, the maximum number in each bag is floor(N / 3), and the leftover is N - 3 * floor(N / 3). But since N can be any number, the maximum possible leftover is 2. Therefore, the answer is 2.

But maybe the problem is different. Let me think of an example. Suppose he has 4 pencils. If he wants to divide them into three bags with the same number, 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 bag, leaving 0. If he has 7 pencils, he can put 2 in each bag, leaving 1. If he has 8 pencils, he can put 2 in each bag, leaving 2. So the maximum number of leftovers is 2. Therefore, regardless of how many pencils he has, the maximum number of pencils left over is 2. Therefore, the answer is 2.

But maybe the problem is expecting a different answer. Wait, maybe the problem is not about dividing into three bags with equal numbers, but something else. 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 maybe he wants to have the same number in each bag, and he wants to maximize that number. Therefore, if he has N pencils, the maximum number per bag is floor(N / 3). But the question is asking for the greatest number of pencils he could have left over. Which would be N - 3 * floor(N / 3). But since N can be any number, the maximum value of this expression is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. Therefore, the maximum number of leftovers would be 2.

Alternatively, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag, but he also wants to have as many pencils as possible left over. Wait, that would be contradictory. If he wants to maximize the number in each bag, he would want to minimize the leftovers. If he wants to maximize the leftovers, he would want to minimize the number in each bag. Therefore, the problem is confusingly worded.

But given the problem as stated, the most logical interpretation is that he wants to divide the pencils into three bags with equal numbers, and he wants to maximize the number in each bag. Therefore, the maximum number of pencils left over is 2. Therefore, the answer is 2.

But let me check if there's another way. Suppose that the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. Therefore, the maximum number of leftovers would be 2.

Alternatively, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag, but he can't do it perfectly. Therefore, the question is asking for the maximum number of pencils that could be left over. But since the maximum remainder when dividing by 3 is 2, the answer is 2.

Therefore, after all this thinking, I think the answer is 2. So 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 also keeping the number of pencils in each bag the same**

The question then 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**.

- If Zach wants to **place the greatest number of pencils possible into each of the three bags**, and also **keep the number of pencils in each bag the same**, then he is essentially trying to **divide N into three equal parts**.
- That is, he wants to find the **maximum integer value** of **k** such that **3k ≤ N**, and **N - 3k** is the number of pencils left over.

So, the number of pencils left over is:
$$
\text{Leftover} = N - 3 \times \left\lfloor \frac{N}{3} \right\rfloor
$$

This is equivalent to:
$$
\text{Leftover} = N \mod 3
$$

---

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

The **modulo operation** with 3 can yield only three possible remainders:
- 0 (if N is divisible by 3)
- 1
- 2

So, the **maximum possible number of pencils left over** is:
$$
\boxed{2}
$$

This occurs when **N mod 3 = 2**, i.e., when the total number of pencils is **one more than a multiple of 3** (e.g., 5, 8, 11, etc.).

---

### **Conclusion**

Even though Zach wants to **maximize the number of pencils in each bag**, the **maximum number of pencils that could be left over** is determined by the **remainder when the total number of pencils is divided by 3**. The largest such remainder is **2**.

---

### **Final Answer**

$$
\boxed{2}
$$

[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 of an example.

Suppose he has N pencils. If he wants to divide them into three bags with the same number in each, then N must be divisible by three. If not, then he can't. 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 he would have to take the floor division of N by three. But then the remainder would be N mod 3. However, 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 minimize the remainder? Wait, no. Wait, 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 remainder is as small as possible. Wait, but the problem says "the greatest number of pencils he could have left over." So maybe the question is phrased in a way that he is trying to divide the pencils into three bags with equal numbers, but he can't, 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 remainder is as small as possible. However, the problem says "the greatest number of pencils he could have left over." So maybe the problem is not about a specific number of pencils, but rather, given that he wants to maximize the number in each bag, what is the maximum possible remainder?

Wait, but the remainder when dividing by three can be 0, 1, or 2. So the maximum remainder is 2. Therefore, the greatest number of pencils he could have left over is 2? But that seems too straightforward. Maybe I'm misunderstanding the problem.

Wait, let me check again. 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 maybe he wants to have the same number in each bag, and he wants to maximize that number. So if he has N pencils, then the maximum number he can put in each bag is floor(N / 3). But then the leftover pencils would be N - 3 * floor(N / 3). But the question is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftover pencils, then he would want N - 3 * floor(N / 3) to be as large as possible. But since floor(N / 3) is the maximum number of pencils per bag, then the leftover is N mod 3. Therefore, the maximum possible leftover is 2. So the answer is 2? But that seems too simple. Maybe the problem is different.

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. However, if he can't divide the pencils equally, he can't do that. Therefore, the question is asking for the maximum number of pencils that could be left over if he tries to do that. But if he wants to maximize the number in each bag, he would have to have as many as possible, so the remainder would be minimized. Wait, but the problem says "the greatest number of pencils he could have left over." So maybe the problem is not about a specific N, but rather, given that he wants to maximize the number in each bag, what is the maximum possible remainder? But the remainder can be at most 2. So maybe the answer is 2. But why would the problem be phrased that way?

Alternatively, maybe the problem is that he has 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. However, there's a constraint that he has to have some pencils left over. Wait, but 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. But the problem says "the greatest number of pencils he could have left over." Maybe the problem is in translation or wording. Maybe it's saying that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag, but there's some leftover pencils. But then, the maximum number of leftovers would be when the total number of pencils is just one more than a multiple of three? Wait, no. If he wants to maximize the number in each bag, he would want to have as many as possible, so the total number of pencils is 3k + r, where r is the remainder. If he wants to maximize k, then r can be 0, 1, or 2. But if he wants to maximize the number in each bag, he would want to have k as large as possible, which would mean that r is as small as possible. However, the problem is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftovers, he would have r as large as possible, which is 2. Therefore, the answer is 2. But why is the problem phrased so confusingly?

Alternatively, maybe the problem is that he wants to have the same number of pencils in each bag, and he wants to maximize the number in each bag, but he also wants to have as many pencils as possible left over. Wait, that seems contradictory. If he wants to maximize the number in each bag, he would want to minimize the leftovers. But if he wants to maximize the leftovers, he would want to minimize the number in each bag. So maybe the problem is not well-phrased. Maybe there's a different interpretation.

Wait, let me check again. 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 maybe he wants to have the same number in each bag, and he wants to maximize that number. So if he has N pencils, then the maximum number per bag is floor(N / 3). However, if he wants to have the same number in each bag, then the number of pencils left over is N - 3 * floor(N / 3). But the problem is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftovers, he would need to maximize N - 3 * floor(N / 3). But since floor(N / 3) is the maximum number per bag, then the leftover is N mod 3. Therefore, the maximum possible value of N mod 3 is 2. Therefore, the answer is 2. But maybe the problem is in another context?

Wait, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. But the problem says "the greatest number of pencils he could have left over." Maybe the problem is not about a specific number of pencils, but rather, given that he wants to maximize the number in each bag, what is the maximum possible remainder? Which would be 2. Therefore, the answer is 2. But maybe the problem is different.

Alternatively, maybe the problem is that he has to divide the pencils into three bags with the same number in each, but he can't do it perfectly, so he has to leave some pencils. But he wants to maximize the number in each bag. Therefore, the maximum number in each bag would be floor(N / 3), and the leftover is N - 3 * floor(N / 3). But the question is asking for the greatest number of pencils he could have left over. So if he wants to maximize the leftover, he would need to have N as large as possible? But there's no upper limit on N. Wait, but if N is arbitrary, then the leftover can be 0, 1, or 2. So the maximum possible leftover is 2. Therefore, regardless of N, the maximum number of pencils left over is 2. Therefore, the answer is 2. But maybe the problem is expecting a different answer?

Wait, let me think again. Maybe the problem is that he wants to have the same number of pencils in each bag, and he wants to maximize the number in each bag. However, he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. But the problem says "the greatest number of pencils he could have left over." Maybe the problem is in the translation or wording. Maybe it's supposed to say 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. Wait, that still doesn't make sense. Maybe it's a translation issue.

Alternatively, maybe the problem is that he has 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. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be as small as possible. Therefore, the maximum number of leftovers would be 2. But if he wants to have some pencils left over, then the maximum number of leftovers is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he has to divide the pencils into three bags with the same number in each, but he can't do it perfectly, so he has to leave some pencils. But he wants to maximize the number in each bag. Therefore, the maximum number in each bag is floor(N / 3), and the leftover is N - 3 * floor(N / 3). But since N can be any number, the maximum possible leftover is 2. Therefore, the answer is 2.

But maybe the problem is different. Let me think of an example. Suppose he has 4 pencils. If he wants to divide them into three bags with the same number, 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 bag, leaving 0. If he has 7 pencils, he can put 2 in each bag, leaving 1. If he has 8 pencils, he can put 2 in each bag, leaving 2. So the maximum number of leftovers is 2. Therefore, regardless of how many pencils he has, the maximum number of pencils left over is 2. Therefore, the answer is 2.

But maybe the problem is expecting a different answer. Wait, maybe the problem is not about dividing into three bags with equal numbers, but something else. 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 maybe he wants to have the same number in each bag, and he wants to maximize that number. Therefore, if he has N pencils, the maximum number per bag is floor(N / 3). But the question is asking for the greatest number of pencils he could have left over. Which would be N - 3 * floor(N / 3). But since N can be any number, the maximum value of this expression is 2. Therefore, the answer is 2.

Alternatively, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. Therefore, the maximum number of leftovers would be 2.

Alternatively, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag, but he also wants to have as many pencils as possible left over. Wait, that would be contradictory. If he wants to maximize the number in each bag, he would want to minimize the leftovers. If he wants to maximize the leftovers, he would want to minimize the number in each bag. Therefore, the problem is confusingly worded.

But given the problem as stated, the most logical interpretation is that he wants to divide the pencils into three bags with equal numbers, and he wants to maximize the number in each bag. Therefore, the maximum number of pencils left over is 2. Therefore, the answer is 2.

But let me check if there's another way. Suppose that the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag. However, there's a constraint that he has to have some pencils left over. Wait, but if he wants to maximize the number in each bag, he would want to have as many as possible, so the leftovers would be minimized. Therefore, the maximum number of leftovers would be 2.

Alternatively, maybe the problem is that he has to place pencils into the bags such that each bag has the same number, and he wants to maximize the number in each bag, but he can't do it perfectly. Therefore, the question is asking for the maximum number of pencils that could be left over. But since the maximum remainder when dividing by 3 is 2, the answer is 2.

Therefore, after all this thinking, I think the answer is 2. So 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 also keeping the number of pencils in each bag the same**

The question then 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**.

- If Zach wants to **place the greatest number of pencils possible into each of the three bags**, and also **keep the number of pencils in each bag the same**, then he is essentially trying to **divide N into three equal parts**.
- That is, he wants to find the **maximum integer value** of **k** such that **3k ≤ N**, and **N - 3k** is the number of pencils left over.

So, the number of pencils left over is:
$$
\text{Leftover} = N - 3 \times \left\lfloor \frac{N}{3} \right\rfloor
$$

This is equivalent to:
$$
\text{Leftover} = N \mod 3
$$

---

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

The **modulo operation** with 3 can yield only three possible remainders:
- 0 (if N is divisible by 3)
- 1
- 2

So, the **maximum possible number of pencils left over** is:
$$
\boxed{2}
$$

This occurs when **N mod 3 = 2**, i.e., when the total number of pencils is **one more than a multiple of 3** (e.g., 5, 8, 11, etc.).

---

### **Conclusion**

Even though Zach wants to **maximize the number of pencils in each bag**, the **maximum number of pencils that could be left over** is determined by the **remainder when the total number of pencils is divided by 3**. The largest such remainder is **2**.

---

### **Final Answer**

$$
\boxed{2}
$$

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 just -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 make sure I didn't mix up any signs or miscalculate the exponents.

For x: t³ when t=2 is 8, plus 7 is 15. Correct.

For y: t squared is 4, times -3 is -12. Then -6t is -12, and then -5. So total is -12 -12 -5. Yes, that's -29. Okay, seems right.

Wait, but let me check if I copied the original equations correctly. The problem says:

(x, y) = (t³ + 7, -3t² - 6t - 5). Yes, that's what I used. So no issues there.

Alternatively, maybe I can check by plugging t=2 into the equations again.

x = 2³ + 7 = 8 + 7 = 15. Correct.

y = -3*(2)^2 -6*(2) -5. Let me compute each term again:

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

-6*(2) is -12.

Then -5. So total is -12 -12 -5. Let me add them step by step:

Start with -12, then subtract 12: -12 -12 = -24. Then subtract 5: -24 -5 = -29. Yep, same result.

So I think that's correct. Therefore, the point is (15, -29). Is there anything else I need to consider?

Wait, sometimes with parametric equations, there might be something else, like maybe the curve is defined only for certain t values? But the problem doesn't specify any restrictions on t, so I think t=2 is valid. So I think that's the answer.

Alternatively, maybe I can check by converting the parametric equations to a Cartesian equation? Although that might be more complicated, but just to see if there's another way.

But maybe that's overkill. Let me think. If I wanted to eliminate the parameter t, I could try solving for t from the x equation and substituting into the y equation. Let's see.

From x = t³ + 7, solving for t:

t³ = x - 7

t = cube root of (x - 7)

But then substituting into y would involve cube roots and might not be straightforward. Let me try.

But since we only need the point at t=2, which we already found, maybe it's not necessary. However, just for verification, if I plug x=15 into the equation t³ = x -7, then t³ = 15 -7 = 8, so t=2. Which matches. So that's consistent. Then, if I plug t=2 into y, we get y=-29. So that's consistent. Therefore, the point (15, -29) is indeed on the curve when t=2.

Alternatively, if I wanted to check if there's another point with the same x-coordinate but different t, but since the problem only asks for t=2, that's not necessary.

Therefore, I think my answer is correct. The point is (15, -29). Let me just write that in the required format.

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

To find the point on the curve defined by the parametric equations:

$$
(x, y) = (t^3 + 7, -3t^2 - 6t - 5)
$$

at $ t = 2 $, we substitute $ t = 2 $ into both the $ x $ and $ y $ components of the parametric equations.

---

### Step 1: Compute the $ x $-coordinate

$$
x = t^3 + 7
$$

Substitute $ t = 2 $:

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

---

### Step 2: Compute the $ y $-coordinate

$$
y = -3t^2 - 6t - 5
$$

Substitute $ t = 2 $:

$$
y = -3(2)^2 - 6(2) - 5 = -3(4) - 12 - 5 = -12 - 12 - 5 = -29
$$

---

### 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 just -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 make sure I didn't mix up any signs or miscalculate the exponents.

For x: t³ when t=2 is 8, plus 7 is 15. Correct.

For y: t squared is 4, times -3 is -12. Then -6t is -12, and then -5. So total is -12 -12 -5. Yes, that's -29. Okay, seems right.

Wait, but let me check if I copied the original equations correctly. The problem says:

(x, y) = (t³ + 7, -3t² - 6t - 5). Yes, that's what I used. So no issues there.

Alternatively, maybe I can check by plugging t=2 into the equations again.

x = 2³ + 7 = 8 + 7 = 15. Correct.

y = -3*(2)^2 -6*(2) -5. Let me compute each term again:

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

-6*(2) is -12.

Then -5. So total is -12 -12 -5. Let me add them step by step:

Start with -12, then subtract 12: -12 -12 = -24. Then subtract 5: -24 -5 = -29. Yep, same result.

So I think that's correct. Therefore, the point is (15, -29). Is there anything else I need to consider?

Wait, sometimes with parametric equations, there might be something else, like maybe the curve is defined only for certain t values? But the problem doesn't specify any restrictions on t, so I think t=2 is valid. So I think that's the answer.

Alternatively, maybe I can check by converting the parametric equations to a Cartesian equation? Although that might be more complicated, but just to see if there's another way.

But maybe that's overkill. Let me think. If I wanted to eliminate the parameter t, I could try solving for t from the x equation and substituting into the y equation. Let's see.

From x = t³ + 7, solving for t:

t³ = x - 7

t = cube root of (x - 7)

But then substituting into y would involve cube roots and might not be straightforward. Let me try.

But since we only need the point at t=2, which we already found, maybe it's not necessary. However, just for verification, if I plug x=15 into the equation t³ = x -7, then t³ = 15 -7 = 8, so t=2. Which matches. So that's consistent. Then, if I plug t=2 into y, we get y=-29. So that's consistent. Therefore, the point (15, -29) is indeed on the curve when t=2.

Alternatively, if I wanted to check if there's another point with the same x-coordinate but different t, but since the problem only asks for t=2, that's not necessary.

Therefore, I think my answer is correct. The point is (15, -29). Let me just write that in the required format.

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

To find the point on the curve defined by the parametric equations:

$$
(x, y) = (t^3 + 7, -3t^2 - 6t - 5)
$$

at $ t = 2 $, we substitute $ t = 2 $ into both the $ x $ and $ y $ components of the parametric equations.

---

### Step 1: Compute the $ x $-coordinate

$$
x = t^3 + 7
$$

Substitute $ t = 2 $:

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

---

### Step 2: Compute the $ y $-coordinate

$$
y = -3t^2 - 6t - 5
$$

Substitute $ t = 2 $:

$$
y = -3(2)^2 - 6(2) - 5 = -3(4) - 12 - 5 = -12 - 12 - 5 = -29
$$

---

### 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.

Okay, 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 try that.

So 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 all methods lead to 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:

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

Simplify both sides:

$$
2(x - 5) = 12
$$

---

### Step 2: Distribute and Solve for $ x $

Distribute the 2 on the left-hand side:

$$
2x - 10 = 12
$$

Now, isolate $ x $ by adding 10 to both sides:

$$
2x = 22
$$

Divide both sides by 2:

$$
x = 11
$$

---

### Step 3: Verify the Solution

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

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

Both sides are equal, so the solution is correct.

---

### Step 4: 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.

Okay, 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 try that.

So 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 all methods lead to 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:

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

Simplify both sides:

$$
2(x - 5) = 12
$$

---

### Step 2: Distribute and Solve for $ x $

Distribute the 2 on the left-hand side:

$$
2x - 10 = 12
$$

Now, isolate $ x $ by adding 10 to both sides:

$$
2x = 22
$$

Divide both sides by 2:

$$
x = 11
$$

---

### Step 3: Verify the Solution

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

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

Both sides are equal, so the solution is correct.

---

### Step 4: 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 $200 + 2d = \boxed{350}$.