💡Okay, so I need to find the multiplicative inverse of 3 modulo 221. That means I need to find an integer x such that when I multiply it by 3, the result is congruent to 1 modulo 221. In other words, I need to solve the equation 3x ≡ 1 mod 221.Hmm, I remember that to find the inverse, the number and the modulus need to be coprime, which means their greatest common divisor (GCD) should be 1. So first, I should check if 3 and 221 are coprime. Let me see, 221 divided by 3 is about 73.666, so 3 doesn't divide 221 evenly. Let me check the prime factors of 221. 221 divided by 13 is 17, right? Because 13 times 17 is 221. So the prime factors are 13 and 17. Since 3 isn't a factor of either 13 or 17, the GCD of 3 and 221 is indeed 1. Okay, so an inverse should exist.Now, how do I find this inverse? I think the extended Euclidean algorithm is the way to go. I remember it's a method to find integers x and y such that ax + by = gcd(a, b). In this case, a is 3 and b is 221, and since their GCD is 1, I can use the algorithm to find x and y such that 3x + 221y = 1. The x here will be the multiplicative inverse of 3 modulo 221.Let me try to apply the extended Euclidean algorithm step by step. I'll start by dividing the larger number by the smaller one and keep track of the remainders.First step: Divide 221 by 3.221 ÷ 3 is 73 with a remainder of 2. So, I can write this as:221 = 3 * 73 + 2Now, I'll take the divisor 3 and divide it by the remainder 2.3 ÷ 2 is 1 with a remainder of 1. So:3 = 2 * 1 + 1Next, I take the divisor 2 and divide it by the remainder 1.2 ÷ 1 is 2 with a remainder of 0. So:2 = 1 * 2 + 0Since the remainder is now 0, the algorithm stops, and the last non-zero remainder is 1, which confirms that the GCD is indeed 1.Now, I need to work backwards to express 1 as a linear combination of 3 and 221. Starting from the second last equation:1 = 3 - 2 * 1But I know from the first equation that 2 = 221 - 3 * 73. So, I can substitute this into the equation above:1 = 3 - (221 - 3 * 73) * 1Let me simplify this:1 = 3 - 221 * 1 + 3 * 731 = 3 * (1 + 73) - 221 * 11 = 3 * 74 - 221 * 1So, this shows that 1 = 3 * 74 - 221 * 1. Therefore, x is 74 and y is -1. This means that 74 is the multiplicative inverse of 3 modulo 221 because when I multiply 3 by 74, I get 222, which is 1 more than 221, so 222 ≡ 1 mod 221.Wait, let me double-check that. 3 * 74 is 222. 222 divided by 221 is 1 with a remainder of 1. So yes, 222 ≡ 1 mod 221. That seems correct.But just to make sure I didn't make any mistakes in my calculations, let me go through the steps again.Starting with 221 divided by 3:221 = 3 * 73 + 2Then, 3 divided by 2:3 = 2 * 1 + 1Then, 2 divided by 1:2 = 1 * 2 + 0So, the GCD is 1, which is good. Then, expressing 1 as a combination:1 = 3 - 2 * 1But 2 is from the first equation, which is 221 - 3 * 73. So substituting:1 = 3 - (221 - 3 * 73) * 11 = 3 - 221 + 3 * 731 = 3 * 74 - 221Yes, that's correct. So x is 74, which is the inverse.I think that's solid. So, the inverse of 3 modulo 221 is 74.

💡Okay, so I have this problem here: the function is given as y equals (m plus 3) times x minus 2. And the question is asking, to make the function value y decrease as the independent variable x increases, what should be the range of values for m? The options are A: m is greater than or equal to -3, B: m is greater than -3, C: m is less than or equal to -3, and D: m is less than -3.Alright, let me try to figure this out step by step. First, I remember that in a linear function of the form y = kx + b, the coefficient k determines whether the function is increasing or decreasing. If k is positive, the function increases as x increases, and if k is negative, the function decreases as x increases.So in this case, the function is y = (m + 3)x - 2. Comparing this to the general form y = kx + b, I can see that k is (m + 3) and b is -2. Since we want the function value y to decrease as x increases, we need the coefficient k to be negative. That means (m + 3) has to be less than zero.Let me write that down: m + 3 < 0. To solve for m, I subtract 3 from both sides. So, m < -3. Hmm, that seems straightforward. So m has to be less than -3 for the function to decrease as x increases.Let me double-check that. If m is less than -3, then m + 3 is negative, which makes the coefficient of x negative. So, as x increases, y decreases. That makes sense. If m were equal to -3, then m + 3 would be zero, and the function would become y = -2, which is a constant function. It doesn't increase or decrease; it's just a horizontal line. So, in that case, y isn't decreasing as x increases—it's staying the same. Therefore, m can't be equal to -3; it has to be strictly less than -3.Looking back at the options, A is m greater than or equal to -3, which would include m = -3 and higher. But we just saw that m = -3 gives a constant function, not a decreasing one. So A is not correct. B is m greater than -3, which is also not what we want because m needs to be less than -3. So B is out.C is m less than or equal to -3. Well, m equal to -3 is not acceptable because it results in a constant function, not a decreasing one. So C is also incorrect. That leaves us with D: m less than -3, which is exactly what we found. So D must be the correct answer.Just to make sure I haven't missed anything, let me think about the graph of the function. If m is less than -3, then the slope of the line is negative, meaning it's going downwards from left to right. So as x increases, y decreases, which is exactly the behavior we want. If m were greater than -3, the slope would be positive, and the function would increase as x increases, which is the opposite of what we need. If m equals -3, the slope is zero, so the graph is a horizontal line, and y doesn't change as x increases. Therefore, only when m is less than -3 do we get the desired behavior.I think that covers all the possibilities. So, yeah, I'm confident that the correct answer is D: m < -3.

💡First, I need to understand what the floor function does. The floor function, denoted as (lfloor x rfloor), returns the largest integer that is less than or equal to the given number (x).Next, I'll apply this definition to the number 5.7. I need to find the greatest integer that is not greater than 5.7.The integers around 5.7 are 5 and 6. Since 5 is less than 5.7 and 6 is greater than 5.7, the largest integer that is less than or equal to 5.7 is 5.Therefore, (lfloor 5.7 rfloor = 5).

💡Alright, so I've got this problem about a raft and some animals that need to get across a river. Let me try to wrap my head around it step by step. First, let's list out all the animals and their weights:- Five baby mice, each weighing 70 grams.- Three moles, each weighing 90 grams.- Four hamsters, each weighing 120 grams.So, the total number of animals is 5 + 3 + 4, which is 12 animals. That's quite a lot! Now, the raft can't move without a rower, which means someone or something has to be on the raft to make it go across. But all the animals are just approaching the shore; none of them are initially on the raft. Hmm, that's a bit confusing. Does that mean we need one of the animals to act as the rower?Wait, the problem says the raft cannot move across the river without a rower. So, someone needs to be on the raft to row it. But all the animals are on the shore, not on the raft. So, maybe one of the animals has to get on the raft first to row it to the other side? But then, how do we get that animal back? Because if the raft is on the other side, we need someone to bring it back. This seems like a classic river crossing puzzle, similar to the fox, chicken, and grain problem, where you have to figure out how to transport items without leaving certain things unattended. In this case, though, it's about transporting animals with different weights, and ensuring the raft doesn't exceed its capacity.So, the key here is to figure out the minimum weight capacity of the raft so that all animals can be transported across the river in several trips. The raft can't move without a rower, so we need at least one animal on the raft at all times.Let me think about the weights. The lightest animals are the mice at 70 grams each, then the moles at 90 grams, and the heaviest are the hamsters at 120 grams. To minimize the raft's capacity, we probably want to pair heavier animals with lighter ones to balance the load.But wait, if we pair a heavy animal with a light one, the raft's capacity needs to be at least the sum of their weights. For example, a hamster and a mouse would require the raft to hold 120 + 70 = 190 grams. But maybe we can do better by pairing lighter animals together.If we pair two mice, that's 70 + 70 = 140 grams. That seems manageable. But if we pair a mole and a mouse, that's 90 + 70 = 160 grams. Hmm, so pairing two mice is lighter than pairing a mole and a mouse. So, maybe the minimum capacity is 140 grams?But let's think about the process. If the raft can only hold two mice at a time, that's 140 grams. But then, how do we get the other animals across? Because we need someone to row the raft back. So, if two mice go over, one has to come back to bring the raft back. That means each trip can only transport one mouse across, which isn't efficient.Alternatively, if the raft can hold more weight, maybe we can transport more animals at once, reducing the number of trips. But the question is asking for the minimum weight capacity, so we need to find the smallest possible raft that can still get all animals across, even if it takes multiple trips.Let me try to outline a possible strategy:1. First, send two mice across. Total weight: 140 grams. One mouse stays on the other side, and the other mouse brings the raft back. Now, one mouse is on the far side, and four mice are back on the original side.2. Next, send two moles across. Total weight: 90 + 90 = 180 grams. One mole stays, and the other mole brings the raft back. Now, one mole is on the far side, and two moles are back.3. Then, send two hamsters across. Total weight: 120 + 120 = 240 grams. One hamster stays, and the other hamster brings the raft back. Now, one hamster is on the far side, and three hamsters are back.But wait, this seems inefficient because each time we're only transporting one animal across, and the raft is being brought back by another. Maybe there's a better way.What if we pair a mouse and a mole together? That would be 70 + 90 = 160 grams. So, if we send a mouse and a mole across, that's 160 grams. Then, the mouse can bring the raft back. Now, one mole is on the far side, and the mouse is back.Then, we can send two mice across again. Total weight: 140 grams. One mouse stays, and the other brings the raft back. Now, two mice are on the far side, and three mice are back.This seems a bit better, but still, the raft's capacity is 160 grams in this case, which is higher than the 140 grams we initially thought.Wait, but maybe we can do better. If we have the raft's capacity at 140 grams, can we still transport all animals?Let's try:1. Two mice go across: 140 grams. One mouse stays, the other brings the raft back. Now, one mouse is on the far side, and four mice are back.2. Two mice go across again: 140 grams. One mouse stays, the other brings the raft back. Now, two mice are on the far side, and three mice are back.3. Repeat this process until all mice are across. But then, we still have moles and hamsters to transport, and we need someone to bring the raft back. If all mice are on the far side, we can't bring the raft back because there's no one left to row it.So, we need to have at least one animal on the original side to bring the raft back. That means we can't transport all mice first; we need to leave at least one mouse behind to act as a rower.Therefore, maybe the strategy is:1. Two mice go across: 140 grams. One mouse stays, the other brings the raft back. Now, one mouse is on the far side, and four mice are back.2. Two mice go across again: 140 grams. One mouse stays, the other brings the raft back. Now, two mice are on the far side, and three mice are back.3. Two mice go across again: 140 grams. One mouse stays, the other brings the raft back. Now, three mice are on the far side, and two mice are back.4. Now, we have two mice back. We can send one mouse and one mole across: 70 + 90 = 160 grams. But wait, the raft's capacity is only 140 grams. So, this won't work.Hmm, so if the raft can only hold 140 grams, we can't send a mouse and a mole together. That means we need to find another way to transport the moles and hamsters.Maybe we can use the mice to ferry the moles and hamsters one by one. But that would require the raft to hold at least the weight of a mole or a hamster plus a mouse.Wait, but the raft's capacity is only 140 grams. A mole is 90 grams, so if we send a mole and a mouse together, that's 160 grams, which exceeds the raft's capacity. So, that won't work.Alternatively, can we send a mole alone? A mole is 90 grams, which is less than 140 grams. So, yes, a mole can go alone, but then someone needs to bring the raft back. If a mole goes alone, it can't bring the raft back because it's on the far side. So, we need a mouse to bring the raft back.But if we have a mouse on the far side, it can bring the raft back. So, maybe:1. Two mice go across: 140 grams. One mouse stays, the other brings the raft back. Now, one mouse is on the far side, and four mice are back.2. A mole goes across alone: 90 grams. Now, one mole is on the far side, and the raft is there. But we need someone to bring the raft back. So, the mouse on the far side brings the raft back. Now, the mole is on the far side, and the mouse is back.3. Now, send two mice across again: 140 grams. One mouse stays, the other brings the raft back. Now, two mice are on the far side, and three mice are back.4. Send the mole across again: 90 grams. Now, the mole is on the far side, and the raft is there. The mouse brings the raft back. Now, the mole is on the far side, and the mouse is back.This seems like a loop where we're not making progress. We're just moving the mole back and forth without actually transporting it.Maybe we need a different approach. What if we use the mice to ferry the moles and hamsters individually, but in a way that doesn't exceed the raft's capacity.Wait, if the raft can hold 140 grams, and a hamster is 120 grams, which is less than 140 grams. So, a hamster can go alone. But then, someone needs to bring the raft back. If a hamster goes alone, it can't bring the raft back. So, we need a mouse to bring the raft back.So, the process would be:1. Two mice go across: 140 grams. One mouse stays, the other brings the raft back. Now, one mouse is on the far side, and four mice are back.2. A hamster goes across alone: 120 grams. Now, the hamster is on the far side, and the raft is there. The mouse brings the raft back. Now, the hamster is on the far side, and the mouse is back.3. Now, send two mice across again: 140 grams. One mouse stays, the other brings the raft back. Now, two mice are on the far side, and three mice are back.4. Send the hamster across again: 120 grams. Now, the hamster is on the far side, and the raft is there. The mouse brings the raft back. Now, the hamster is on the far side, and the mouse is back.Again, we're stuck in a loop. We're not actually transporting the hamster across; we're just moving it back and forth.This suggests that with a raft capacity of 140 grams, it's impossible to transport both the moles and hamsters because we can't get them across without someone bringing the raft back, and the raft's capacity isn't enough to carry both a heavy animal and a mouse together.So, maybe we need a higher raft capacity. Let's try 160 grams.With a 160-gram capacity, we can send a mouse and a mole together: 70 + 90 = 160 grams. That works.So, here's a possible strategy:1. Send a mouse and a mole across: 160 grams. The mouse stays, and the mole brings the raft back. Now, one mouse is on the far side, and the mole is back.2. Send two mice across: 140 grams. One mouse stays, and the other brings the raft back. Now, two mice are on the far side, and three mice are back.3. Send the mole and a mouse across again: 160 grams. The mole stays, and the mouse brings the raft back. Now, two mice and one mole are on the far side, and two mice are back.4. Repeat this process to transport the remaining moles and hamsters.But wait, we also have hamsters to transport. A hamster is 120 grams, so if we pair a hamster with a mouse, that's 120 + 70 = 190 grams, which exceeds the 160-gram capacity. So, we can't send a hamster and a mouse together.Alternatively, can we send a hamster alone? Yes, 120 grams is less than 160 grams. So, a hamster can go alone, but then someone needs to bring the raft back. If a hamster goes alone, it can't bring the raft back. So, we need a mouse to bring the raft back.So, the process would be:1. Send a mouse and a mole across: 160 grams. The mouse stays, and the mole brings the raft back. Now, one mouse is on the far side, and the mole is back.2. Send two mice across: 140 grams. One mouse stays, and the other brings the raft back. Now, two mice are on the far side, and two mice are back.3. Send a hamster across alone: 120 grams. Now, the hamster is on the far side, and the raft is there. The mouse brings the raft back. Now, the hamster is on the far side, and the mouse is back.4. Send the mole and a mouse across: 160 grams. The mole stays, and the mouse brings the raft back. Now, two mice, one mole, and one hamster are on the far side, and one mouse is back.5. Repeat this process to transport the remaining animals.This seems more feasible. We can continue this pattern, using the 160-gram capacity to transport a mouse and a mole together, and using the mouse to bring the raft back. For the hamsters, we can send them alone, and have a mouse bring the raft back each time.But wait, we have four hamsters, and each time we send one, we need a mouse to bring the raft back. So, we need to ensure we have enough mice left to bring the raft back after each trip.Let's see:- Initially, we have five mice, three moles, and four hamsters on the original side.- After sending two mice across and bringing one back, we have four mice back, one mouse on the far side.- Then, send a mole and a mouse across: now, two mice and one mole are on the far side, and three mice and two moles are back.- Bring the mole back: now, two mice are on the far side, and three mice and three moles are back.- Send two mice across again: now, four mice are on the far side, and one mouse is back.- Send a mole and a mouse across: now, five mice and one mole are on the far side, and two moles are back.- Bring the mole back: now, five mice and one mole are on the far side, and two moles are back.- Send two moles across: 90 + 90 = 180 grams, which exceeds the 160-gram capacity. So, we can't do that.Hmm, so we have a problem here. We can't send two moles together because that exceeds the raft's capacity. So, we need to find another way to transport the remaining moles.Maybe instead of sending two moles together, we can send them one by one, using the mouse to bring the raft back each time.So, after we have five mice and one mole on the far side, and two moles back:1. Send a mole and a mouse across: 160 grams. The mole stays, and the mouse brings the raft back. Now, five mice and two moles are on the far side, and one mole is back.2. Send the last mole and a mouse across: 160 grams. The mole stays, and the mouse brings the raft back. Now, five mice and three moles are on the far side, and no moles are back.3. Now, we have four hamsters to transport. Each time, send a hamster alone: 120 grams. After each trip, a mouse brings the raft back.So, first hamster goes across, mouse brings raft back.Second hamster goes across, mouse brings raft back.Third hamster goes across, mouse brings raft back.Fourth hamster goes across, mouse brings raft back.Now, all hamsters are on the far side, and the raft is back with a mouse.Finally, send the last mouse across: 70 grams. Now, all mice, moles, and hamsters are on the far side.Wait, but we have five mice, and we've already sent four across, so the last mouse is the one that was bringing the raft back. So, yes, it can go across.So, in total, with a raft capacity of 160 grams, we can transport all animals across the river. But is 160 grams the minimum?Earlier, we thought 140 grams might work, but we ran into issues where we couldn't transport the moles and hamsters without exceeding the capacity. So, maybe 160 grams is the minimum.But let's double-check. Is there a way to do it with a lower capacity, say 150 grams?If the raft can hold 150 grams, can we transport all animals?A mole is 90 grams, so we could send a mole and a mouse together: 90 + 70 = 160 grams, which exceeds 150 grams. So, that's not possible.Alternatively, send a mole alone: 90 grams. Then, someone needs to bring the raft back. If a mole goes alone, it can't bring the raft back. So, we need a mouse to bring it back. But then, the mouse is on the far side, and we need to get it back.Wait, let's try:1. Send two mice across: 140 grams. One mouse stays, the other brings the raft back. Now, one mouse is on the far side, and four mice are back.2. Send a mole across alone: 90 grams. Now, the mole is on the far side, and the raft is there. The mouse brings the raft back. Now, the mole is on the far side, and the mouse is back.3. Send two mice across again: 140 grams. One mouse stays, the other brings the raft back. Now, two mice are on the far side, and three mice are back.4. Send the mole across again: 90 grams. Now, the mole is on the far side, and the raft is there. The mouse brings the raft back. Now, the mole is on the far side, and the mouse is back.Again, we're stuck in a loop. We can't transport the mole without someone bringing the raft back, and with a 150-gram capacity, we can't send a mole and a mouse together because that's 160 grams.So, 150 grams isn't sufficient. Therefore, 160 grams seems to be the minimum capacity needed.But wait, let's think differently. Maybe we can use the hamsters to help row the raft. If a hamster can row, then we don't need to bring a mouse back every time.But the problem doesn't specify that only mice can row; it just says the raft cannot move without a rower. So, any animal can row, as long as it's on the raft.So, if we send a hamster and a mouse across, the hamster can row back. But wait, the hamster is heavier, so if we send a hamster and a mouse, that's 120 + 70 = 190 grams, which exceeds our assumed 160-gram capacity.But if the raft's capacity is 160 grams, we can't send both together. So, that doesn't help.Alternatively, if we send a mole and a hamster together: 90 + 120 = 210 grams, which is way over.So, no, that doesn't help either.Therefore, it seems that 160 grams is indeed the minimum capacity needed to transport all animals across the river, using the strategy of sending a mouse and a mole together, and sending hamsters alone, with mice bringing the raft back each time.But wait, let's see if we can do it with a lower capacity by using a different pairing.What if we pair a mole and a hamster? 90 + 120 = 210 grams, which is way too much.Alternatively, pair two moles: 90 + 90 = 180 grams, which is over 160.So, no, that doesn't help.Alternatively, pair a mouse and a hamster: 70 + 120 = 190 grams, which is over 160.So, no, that doesn't help either.Therefore, the only way is to send a mouse and a mole together at 160 grams, and send hamsters alone at 120 grams, with mice bringing the raft back.So, I think 160 grams is the minimum capacity needed.But wait, earlier I thought 140 grams might work, but it didn't because we couldn't transport the moles and hamsters without exceeding the capacity. So, 160 grams seems necessary.But let's see if there's a more efficient way.What if we send two moles together? That's 180 grams, which is over 160, so no.Alternatively, send a mole and two mice: 90 + 70 + 70 = 230 grams, which is way over.No, that's not helpful.Alternatively, send a mole and a hamster: 90 + 120 = 210 grams, which is over.So, no.Therefore, the only feasible way is to send a mouse and a mole together at 160 grams, and send hamsters alone at 120 grams, with mice bringing the raft back.Thus, the minimum weight capacity the raft must have is 160 grams.But wait, let's double-check.If the raft is 160 grams, can we transport all animals?Yes, as per the strategy above.But is there a way to do it with a lower capacity?If we have a raft capacity of 150 grams, can we do it?As we saw earlier, no, because we can't send a mole and a mouse together, and we can't send two moles together.So, 160 grams is indeed the minimum.Therefore, the minimum weight capacity the raft must have is 160 grams.

💡Okay, so I need to find the mass percentage of chlorine in the produced bleach, which is assumed to be pure sodium hypochlorite (NaClO). The problem states that 100 grams of bleach is produced. First, I think I need to understand what mass percentage means. Mass percentage is the ratio of the mass of a particular component in a mixture to the total mass of the mixture, multiplied by 100 to get a percentage. So, in this case, I need to find out how much of the 100 grams of bleach is chlorine.To do this, I should probably start by finding the molar masses of the elements involved in sodium hypochlorite. Sodium (Na), chlorine (Cl), and oxygen (O). I remember that the atomic mass of sodium is about 23 grams per mole, chlorine is about 35.5 grams per mole, and oxygen is about 16 grams per mole. So, the molar mass of NaClO would be the sum of these: 23 + 35.5 + 16, which equals 74.5 grams per mole.Now, if I have 100 grams of NaClO, I can find out how many moles of NaClO that is by dividing the mass by the molar mass. So, 100 grams divided by 74.5 grams per mole is approximately 1.342 moles of NaClO.Since each mole of NaClO contains one mole of chlorine, the number of moles of chlorine in 100 grams of NaClO is also 1.342 moles.Next, I need to find the mass of chlorine in those 1.342 moles. Since the molar mass of chlorine is 35.5 grams per mole, multiplying that by 1.342 moles gives me the mass of chlorine: 35.5 * 1.342 ≈ 47.6 grams.Finally, to find the mass percentage of chlorine in the bleach, I divide the mass of chlorine by the total mass of the bleach and multiply by 100. So, 47.6 grams divided by 100 grams times 100% gives me approximately 47.6%.Wait, that seems a bit high. Let me double-check my calculations. Maybe I made a mistake in the molar mass or the mole calculation. Let me recalculate the molar mass of NaClO: 23 (Na) + 35.5 (Cl) + 16 (O) = 74.5 grams per mole. That seems correct.Then, 100 grams divided by 74.5 grams per mole is indeed approximately 1.342 moles. Chlorine's molar mass is 35.5 grams per mole, so 1.342 * 35.5 ≈ 47.6 grams. Divided by 100 grams gives 47.6%. Hmm, maybe it's correct after all.I guess the mass percentage of chlorine in the bleach is about 47.6%.

💡Alright, so I've got this problem here: Given that (3x + 2y = 7) and (2x + 3y = 8), find the value of (13x^2 + 22xy + 13y^2).Hmm, okay. Let me try to figure this out step by step. I'm a bit nervous because I'm not super confident with algebra, but I'll give it a shot.First, I see that we have two equations with two variables, (x) and (y). That usually means I can solve for one variable in terms of the other and then substitute. But before I jump into solving for (x) and (y), I notice that the expression we need to find, (13x^2 + 22xy + 13y^2), looks kind of complicated. Maybe there's a way to simplify it or relate it to the given equations.Let me write down the given equations:1. (3x + 2y = 7)2. (2x + 3y = 8)And the expression we need is (13x^2 + 22xy + 13y^2). Hmm, I wonder if this expression can be rewritten in terms of the given equations. Maybe if I square the given equations or something like that?Let me try squaring both equations and see what happens.First, square equation 1:((3x + 2y)^2 = 7^2)Which expands to:(9x^2 + 12xy + 4y^2 = 49)Okay, that's one part. Now, square equation 2:((2x + 3y)^2 = 8^2)Expanding that:(4x^2 + 12xy + 9y^2 = 64)Alright, now I have two new equations:3. (9x^2 + 12xy + 4y^2 = 49)4. (4x^2 + 12xy + 9y^2 = 64)Hmm, if I add these two equations together, maybe something will cancel out or combine nicely.Adding equation 3 and equation 4:(9x^2 + 12xy + 4y^2 + 4x^2 + 12xy + 9y^2 = 49 + 64)Combine like terms:( (9x^2 + 4x^2) + (12xy + 12xy) + (4y^2 + 9y^2) = 113 )Which simplifies to:(13x^2 + 24xy + 13y^2 = 113)Wait a minute, the expression we need is (13x^2 + 22xy + 13y^2). That's almost the same as what I just got, except the middle term is (24xy) instead of (22xy). So, there's a difference of (2xy).Hmm, so (13x^2 + 24xy + 13y^2 = 113), but we need (13x^2 + 22xy + 13y^2). That means if I subtract (2xy) from both sides, I get:(13x^2 + 22xy + 13y^2 = 113 - 2xy)But wait, that introduces a new term, (-2xy), which I don't know the value of. So, that might not help directly.Maybe I need to find the value of (xy) first. If I can find (xy), then I can plug it into this equation and get the desired expression.How can I find (xy)? Well, I have the original two equations:1. (3x + 2y = 7)2. (2x + 3y = 8)I can solve this system of equations for (x) and (y). Let me try that.One method to solve for (x) and (y) is the elimination method. Let's try to eliminate one variable. For example, I can try to eliminate (x) by making the coefficients of (x) the same in both equations.Multiply equation 1 by 2:(6x + 4y = 14)Multiply equation 2 by 3:(6x + 9y = 24)Now, subtract the first new equation from the second new equation:((6x + 9y) - (6x + 4y) = 24 - 14)Simplify:(0x + 5y = 10)So, (5y = 10), which means (y = 2).Now that I have (y = 2), I can plug this back into one of the original equations to find (x). Let's use equation 1:(3x + 2y = 7)Substitute (y = 2):(3x + 2*2 = 7)(3x + 4 = 7)Subtract 4 from both sides:(3x = 3)Divide both sides by 3:(x = 1)Okay, so (x = 1) and (y = 2). Now, let's find (xy):(xy = 1*2 = 2)Great, so (xy = 2). Now, going back to the earlier equation where I had:(13x^2 + 22xy + 13y^2 = 113 - 2xy)Substitute (xy = 2):(13x^2 + 22xy + 13y^2 = 113 - 2*2)(13x^2 + 22xy + 13y^2 = 113 - 4)(13x^2 + 22xy + 13y^2 = 109)Wait, that's different from the initial thought where I thought it was 113. Did I make a mistake somewhere?Let me double-check my steps.First, squaring the equations:1. ((3x + 2y)^2 = 49) gives (9x^2 + 12xy + 4y^2 = 49)2. ((2x + 3y)^2 = 64) gives (4x^2 + 12xy + 9y^2 = 64)Adding them together:(9x^2 + 4x^2 = 13x^2)(12xy + 12xy = 24xy)(4y^2 + 9y^2 = 13y^2)So, (13x^2 + 24xy + 13y^2 = 113)Then, since we need (13x^2 + 22xy + 13y^2), we subtract (2xy):(13x^2 + 22xy + 13y^2 = 113 - 2xy)We found (xy = 2), so:(113 - 4 = 109)But wait, earlier, I thought the answer was 113. That must mean I made a mistake in my initial assumption.Wait, let me think again. Maybe I don't need to subtract (2xy). Maybe there's another way to express (13x^2 + 22xy + 13y^2).Alternatively, perhaps I can compute (13x^2 + 22xy + 13y^2) directly by plugging in (x = 1) and (y = 2).Let's try that.Compute each term:(13x^2 = 13*(1)^2 = 13*1 = 13)(22xy = 22*1*2 = 44)(13y^2 = 13*(2)^2 = 13*4 = 52)Now, add them up:(13 + 44 + 52 = 109)So, that confirms it: (13x^2 + 22xy + 13y^2 = 109)But wait, earlier, when I added the squares, I got 113, and then subtracted 4 to get 109. So, that seems consistent.But in my initial thought, I thought the answer was 113. That must have been a mistake. So, the correct answer is 109.Wait, but let me check again. Maybe I made a mistake in adding the equations.Wait, when I added the two squared equations:(9x^2 + 12xy + 4y^2 = 49)and(4x^2 + 12xy + 9y^2 = 64)Adding them:(9x^2 + 4x^2 = 13x^2)(12xy + 12xy = 24xy)(4y^2 + 9y^2 = 13y^2)So, total is (13x^2 + 24xy + 13y^2 = 113)Then, since the desired expression is (13x^2 + 22xy + 13y^2), which is 2xy less than 113.So, (13x^2 + 22xy + 13y^2 = 113 - 2xy)We found (xy = 2), so (113 - 4 = 109)Therefore, the correct value is 109.But wait, in the initial problem, the user wrote:"Given that (3x + 2y = 7) and (2x + 3y = 8), find the value of (13x^2 + 22xy + 13y^2)."And in the thought process, I initially thought it was 113, but after solving, it's 109.So, perhaps the initial thought was wrong, and the correct answer is 109.Alternatively, maybe there's another way to express (13x^2 + 22xy + 13y^2) without having to subtract.Wait, let me think. Maybe I can factor the expression (13x^2 + 22xy + 13y^2).Let me see:(13x^2 + 22xy + 13y^2)Hmm, this looks symmetric in (x) and (y). Maybe it's a perfect square or something similar.Let me try to write it as:((ax + by)^2 + (cx + dy)^2)But I'm not sure. Alternatively, maybe it's related to the sum of squares of the original equations.Wait, earlier, I saw that:((3x + 2y)^2 + (2x + 3y)^2 = 13x^2 + 24xy + 13y^2 = 113)But our desired expression is (13x^2 + 22xy + 13y^2), which is 2xy less.So, perhaps, (13x^2 + 22xy + 13y^2 = (3x + 2y)^2 + (2x + 3y)^2 - 2xy)But since we know (xy = 2), then:(113 - 2*2 = 113 - 4 = 109)So, that's consistent.Alternatively, maybe I can compute (13x^2 + 22xy + 13y^2) directly by plugging in (x = 1) and (y = 2), as I did earlier, which also gives 109.Therefore, the correct answer is 109.Wait, but in the initial thought process, I thought it was 113. That must have been a mistake. So, the correct answer is 109.But let me double-check everything to make sure I didn't make any errors.First, solving for (x) and (y):Equation 1: (3x + 2y = 7)Equation 2: (2x + 3y = 8)Using elimination:Multiply equation 1 by 2: (6x + 4y = 14)Multiply equation 2 by 3: (6x + 9y = 24)Subtract the first from the second: (5y = 10) → (y = 2)Substitute (y = 2) into equation 1: (3x + 4 = 7) → (3x = 3) → (x = 1)So, (x = 1), (y = 2). Correct.Then, (xy = 1*2 = 2). Correct.Then, (13x^2 + 22xy + 13y^2 = 13*1 + 22*2 + 13*4 = 13 + 44 + 52 = 109). Correct.Alternatively, using the squared equations:((3x + 2y)^2 = 49) → (9x^2 + 12xy + 4y^2 = 49)((2x + 3y)^2 = 64) → (4x^2 + 12xy + 9y^2 = 64)Adding them: (13x^2 + 24xy + 13y^2 = 113)Subtract (2xy): (13x^2 + 22xy + 13y^2 = 113 - 2xy = 113 - 4 = 109). Correct.So, both methods confirm that the value is 109.Therefore, the initial thought of 113 was incorrect because it didn't account for the difference in the middle term. The correct answer is 109.