answer：Okay, so I have this problem where I need to find the smallest possible value of |2015 m⁵ - 2014 n⁴|, where m and n are natural numbers. Hmm, that seems a bit tricky, but let me try to break it down.First, I know that m and n are natural numbers, which means they are positive integers. So, I need to find m and n such that when I compute 2015 times m to the fifth power and subtract 2014 times n to the fourth power, the absolute value of that result is as small as possible. Ideally, I would like this difference to be zero because that would give me the smallest possible value. But I'm not sure if that's achievable, so maybe I need to look for the next smallest value.Let me start by considering the equation 2015 m⁵ = 2014 n⁴. If I can find m and n that satisfy this equation, then the absolute value will be zero, which is the smallest possible. So, how can I find such m and n?I think I need to express m and n in terms of the prime factors of 2014 and 2015. Let me factorize these numbers. 2014: Let's see, 2014 divided by 2 is 1007. Then, 1007 divided by 19 is 53. So, 2014 = 2 × 19 × 53.2015: Similarly, 2015 divided by 5 is 403. Then, 403 divided by 13 is 31. So, 2015 = 5 × 13 × 31.Okay, so both 2014 and 2015 are products of three distinct prime numbers. That might be useful.Now, going back to the equation 2015 m⁵ = 2014 n⁴. Let me write m and n in terms of their prime factors. Let's assume m is some multiple of 2014 and 2015, and similarly for n. Maybe m = 2014^a × 2015^b and n = 2014^c × 2015^d, where a, b, c, d are non-negative integers.Substituting these into the equation:2015 × (2014^a × 2015^b)^5 = 2014 × (2014^c × 2015^d)^4Simplify both sides:Left side: 2015 × 2014^(5a) × 2015^(5b) = 2014^(5a) × 2015^(5b + 1)Right side: 2014 × 2014^(4c) × 2015^(4d) = 2014^(4c + 1) × 2015^(4d)For these two expressions to be equal, the exponents of the corresponding prime factors must be equal. So, we have two equations:1. 5a = 4c + 12. 5b + 1 = 4dNow, I need to find non-negative integers a, b, c, d that satisfy these equations.Starting with the first equation: 5a = 4c + 1I need to find integers a and c such that 5a is one more than a multiple of 4. Let me try small values for a:- a = 1: 5(1) = 5, which is 1 more than 4. So, 5 = 4c + 1 => 4c = 4 => c = 1. So, a = 1, c = 1.Wait, that works. So, a = 1, c = 1.Now, the second equation: 5b + 1 = 4dSimilarly, I need to find integers b and d such that 5b + 1 is a multiple of 4. Let's try small values for b:- b = 1: 5(1) + 1 = 6, which is not a multiple of 4.- b = 2: 5(2) + 1 = 11, not a multiple of 4.- b = 3: 5(3) + 1 = 16, which is a multiple of 4. So, 16 = 4d => d = 4.So, b = 3, d = 4.Okay, so we have a = 1, c = 1, b = 3, d = 4.Therefore, m = 2014^1 × 2015^3 and n = 2014^1 × 2015^4.Let me verify this:Compute 2015 m⁵:m⁵ = (2014 × 2015³)^5 = 2014^5 × 2015^152015 m⁵ = 2015 × 2014^5 × 2015^15 = 2014^5 × 2015^16Compute 2014 n⁴:n⁴ = (2014 × 2015⁴)^4 = 2014^4 × 2015^162014 n⁴ = 2014 × 2014^4 × 2015^16 = 2014^5 × 2015^16So, both 2015 m⁵ and 2014 n⁴ equal 2014^5 × 2015^16, which means their difference is zero. Therefore, |2015 m⁵ - 2014 n⁴| = 0.Wait, that's great! So, the smallest possible value is zero. But let me double-check if there are smaller values of m and n that could also satisfy this condition.Suppose I try smaller exponents. For example, if a = 0, then 5a = 0, so 0 = 4c + 1, which would require c = -1/4, which is not a natural number. So, a cannot be zero.Similarly, if b = 0, then 5b + 1 = 1 = 4d, which would require d = 1/4, which is not an integer. So, b cannot be zero either.Therefore, the smallest exponents that work are a = 1, c = 1, b = 3, d = 4, leading to m = 2014 × 2015³ and n = 2014 × 2015⁴.But wait, these are quite large numbers. Maybe there's a smaller m and n that also make the expression zero? Or perhaps not, given the exponents required.Alternatively, maybe I can find m and n such that 2015 m⁵ is very close to 2014 n⁴ without being exactly equal, resulting in a small positive difference. But since I already found that the difference can be zero, which is the smallest possible, I think that's the answer.However, just to be thorough, let me check if there are any smaller m and n that could make the expression zero. For example, if m and n are both 1:2015 × 1⁵ - 2014 × 1⁴ = 2015 - 2014 = 1. So, the absolute value is 1.If m = 2, n = 2:2015 × 32 - 2014 × 16 = 64480 - 32224 = 32256. That's much larger.If m = 1, n = 2:2015 × 1 - 2014 × 16 = 2015 - 32224 = -30209, absolute value 30209.If m = 2, n = 1:2015 × 32 - 2014 × 1 = 64480 - 2014 = 62466.So, clearly, m = 1, n = 1 gives a difference of 1, which is small, but not zero. So, is there a way to get a difference smaller than 1? Well, since m and n are natural numbers, the expression |2015 m⁵ - 2014 n⁴| must be an integer, so the smallest possible value is 0 or 1. Since we found that 0 is achievable, that's the minimum.Wait, but in my earlier reasoning, I found m and n that make the expression zero, but those m and n are very large. Is there a way to get a smaller m and n that also make the expression zero? Or is the zero difference only possible with those large m and n?Let me think. The equation 2015 m⁵ = 2014 n⁴ implies that m⁵/n⁴ = 2014/2015. Since 2014 and 2015 are coprime (they are consecutive integers), the fraction 2014/2015 is in its simplest form. Therefore, for m⁵/n⁴ to equal 2014/2015, m must be a multiple of 2014 and n must be a multiple of 2015, or vice versa?Wait, no, because m and n are in the exponents. Let me think differently. Since 2014 and 2015 are coprime, the only way for m⁵ to have a factor of 2014 and n⁴ to have a factor of 2015 is if m is a multiple of 2014 and n is a multiple of 2015. So, m = 2014^a × k and n = 2015^b × l, where k and l are integers not containing factors of 2014 or 2015, respectively.But this might complicate things. Alternatively, perhaps the minimal solution is indeed the one I found earlier, with m = 2014 × 2015³ and n = 2014 × 2015⁴, leading to the difference being zero.Alternatively, maybe I can find smaller exponents a, b, c, d that satisfy the equations 5a = 4c + 1 and 5b + 1 = 4d.For the first equation, 5a = 4c + 1. We found a = 1, c = 1. Is there a smaller a? If a = 0, then 0 = 4c + 1, which is impossible. So, a = 1 is the smallest.For the second equation, 5b + 1 = 4d. We found b = 3, d = 4. Let's see if there's a smaller b:- b = 0: 1 = 4d => d = 0.25, not integer.- b = 1: 6 = 4d => d = 1.5, not integer.- b = 2: 11 = 4d => d = 2.75, not integer.- b = 3: 16 = 4d => d = 4.So, b = 3 is indeed the smallest b that gives an integer d.Therefore, the minimal exponents are a = 1, c = 1, b = 3, d = 4, leading to m = 2014 × 2015³ and n = 2014 × 2015⁴.Thus, the smallest possible value of |2015 m⁵ - 2014 n⁴| is 0.But wait, earlier when I tried m = 1, n = 1, I got a difference of 1. So, is 0 actually achievable? Yes, because with the specific m and n I found, the difference is zero. So, the minimal value is indeed 0.I think that's the answer. The smallest possible value is 0.

answer：Alright, so I've got this problem here where Carl is on a vertex of a regular pentagon. Every minute, he randomly chooses an adjacent vertex and walks to it, each with a probability of 1/2. I need to find the probability that after 10 minutes, he ends up back where he started.Okay, let's break this down. First, a regular pentagon has five vertices, right? So, Carl is starting at one of these five vertices. Each minute, he can move to one of two adjacent vertices—either clockwise or counterclockwise. So, every move he makes is like a step either to the left or to the right on this pentagon.Now, after 10 minutes, he's made 10 moves. I need to figure out the probability that he's back at his starting point. Hmm, so he needs to have made an equal number of clockwise and counterclockwise moves, right? Because if he moves clockwise and counterclockwise the same number of times, he'll end up where he started.Wait, but actually, it's not just about equal numbers. Since it's a pentagon, moving five steps clockwise or counterclockwise brings you back to the starting point. So, the net movement after 10 steps should be a multiple of 5 steps in either direction. That is, the difference between the number of clockwise and counterclockwise moves should be a multiple of 5.Let me define this more formally. Let’s say Carl makes 'a' clockwise moves and 'b' counterclockwise moves. Then, a + b = 10, since he makes 10 moves in total. For him to end up at the starting point, the net movement should be 0 modulo 5. So, a - b ≡ 0 mod 5.That means, a ≡ b mod 5. Since a + b = 10, we can substitute b = 10 - a. So, a ≡ (10 - a) mod 5, which simplifies to 2a ≡ 10 mod 5. But 10 mod 5 is 0, so 2a ≡ 0 mod 5. This implies that a must be a multiple of 5/2. Wait, but a has to be an integer, so 2a must be a multiple of 5, meaning a must be a multiple of 5/2. But since a is an integer, the only way this works is if a is a multiple of 5.So, a can be 0, 5, or 10. Therefore, the number of clockwise moves Carl makes must be 0, 5, or 10. Similarly, the number of counterclockwise moves would be 10, 5, or 0, respectively.Now, the total number of possible paths Carl can take in 10 minutes is 2^10, since each minute he has two choices. That’s 1024 total paths.Next, I need to count how many of these paths result in Carl being back at the starting point. That is, the number of paths where he makes 0 clockwise and 10 counterclockwise moves, 5 clockwise and 5 counterclockwise moves, or 10 clockwise and 0 counterclockwise moves.For 0 clockwise and 10 counterclockwise moves, there's only 1 way: he moves counterclockwise every time. Similarly, for 10 clockwise and 0 counterclockwise moves, there's only 1 way.For 5 clockwise and 5 counterclockwise moves, the number of such paths is the number of ways to choose 5 moves out of 10 to be clockwise (the rest will be counterclockwise). That's given by the binomial coefficient C(10,5).Calculating C(10,5): 10! / (5! * 5!) = (10*9*8*7*6)/(5*4*3*2*1) = 252.So, the total number of valid paths is 1 (for 0 clockwise) + 252 (for 5 clockwise) + 1 (for 10 clockwise) = 254.Therefore, the probability is 254 / 1024. Simplifying this fraction: both numerator and denominator are divisible by 2, so 254/1024 = 127/512.Wait, let me double-check that. 254 divided by 2 is 127, and 1024 divided by 2 is 512. Yep, that's correct.So, the probability that Carl ends up back at his starting vertex after 10 minutes is 127/512.

answer：Okay, so I have this problem where I need to prove that for a differentiable function ( f(x) ) on the interval ([a, b]), there exists some interior point ( x_0 ) such that the derivative of ( f ) at ( x_0 ) equals the average rate of change of ( f ) over the interval. That is, ( f'(x_0) = frac{f(a) - f(b)}{a - b} ). Hmm, I remember something about this from my calculus class. It sounds like it's related to the Mean Value Theorem, which is also called Lagrange's theorem. But wait, isn't the Mean Value Theorem usually stated as ( f'(x_0) = frac{f(b) - f(a)}{b - a} )? That seems similar but not exactly the same. Oh, right, because ( frac{f(a) - f(b)}{a - b} ) is just the negative of ( frac{f(b) - f(a)}{b - a} ). So, they're essentially the same thing, just written differently. Alright, so I think the approach here is to use Rolle's Theorem, which is a specific case of the Mean Value Theorem. Rolle's Theorem states that if a function ( g ) is continuous on ([a, b]), differentiable on ((a, b)), and ( g(a) = g(b) ), then there exists some ( c ) in ((a, b)) such that ( g'(c) = 0 ). So, maybe I can construct a function ( g(x) ) that satisfies the conditions of Rolle's Theorem and relates to the original function ( f(x) ). Let me think about how to define such a function. Perhaps I can define ( g(x) ) as the difference between ( f(x) ) and a linear function that connects the points ( (a, f(a)) ) and ( (b, f(b)) ). That way, ( g(a) ) and ( g(b) ) would both be zero, satisfying Rolle's condition. Let me write that out. If I define ( g(x) = f(x) - left( frac{f(b) - f(a)}{b - a}(x - a) + f(a) right) ), then at ( x = a ), ( g(a) = f(a) - left( 0 + f(a) right) = 0 ). Similarly, at ( x = b ), ( g(b) = f(b) - left( frac{f(b) - f(a)}{b - a}(b - a) + f(a) right) = f(b) - (f(b) - f(a) + f(a)) = f(b) - f(b) = 0 ). Great, so ( g(a) = g(b) = 0 ), which satisfies Rolle's Theorem. Now, since ( f(x) ) is differentiable on ([a, b]), ( g(x) ) is also differentiable on ([a, b]) because it's just a linear combination of differentiable functions. By Rolle's Theorem, there must exist some ( x_0 ) in ((a, b)) such that ( g'(x_0) = 0 ). Let's compute ( g'(x) ). The derivative of ( g(x) ) is ( g'(x) = f'(x) - frac{f(b) - f(a)}{b - a} ). Setting this equal to zero at ( x_0 ) gives ( f'(x_0) - frac{f(b) - f(a)}{b - a} = 0 ), which simplifies to ( f'(x_0) = frac{f(b) - f(a)}{b - a} ). Wait a second, the problem statement says ( f'(x_0) = frac{f(a) - f(b)}{a - b} ). But ( frac{f(b) - f(a)}{b - a} ) is the same as ( frac{f(a) - f(b)}{a - b} ) because both numerator and denominator are multiplied by -1. So, they're equivalent. Therefore, I've shown that there exists some ( x_0 ) in ((a, b)) such that ( f'(x_0) = frac{f(a) - f(b)}{a - b} ), which is exactly what was to be proven. I think that makes sense. I constructed a function that fits the criteria of Rolle's Theorem, applied the theorem, and then related the result back to the original function ( f(x) ). It all checks out. Just to recap, the key steps were:1. Define an auxiliary function ( g(x) ) that subtracts a linear term from ( f(x) ) to make ( g(a) = g(b) = 0 ).2. Apply Rolle's Theorem to ( g(x) ) to find a point where the derivative is zero.3. Compute the derivative of ( g(x) ) and set it to zero to find the desired condition on ( f'(x_0) ).This seems solid. I don't see any gaps or mistakes in the logic. Maybe I should double-check the computation of ( g(b) ) to ensure it's indeed zero. Let's compute ( g(b) ) again:[g(b) = f(b) - left( frac{f(b) - f(a)}{b - a}(b - a) + f(a) right) = f(b) - (f(b) - f(a) + f(a)) = f(b) - f(b) = 0.]Yep, that's correct. So, everything holds up. I think I'm confident with this proof now.

answer：Okay, so I need to figure out how to compute the tangent of 45 degrees using geometric principles, similar to how we compute the tangent of 60 degrees. Hmm, I remember that for 60 degrees, we used an equilateral triangle and some properties of 30-60-90 triangles. Maybe I can do something similar for 45 degrees.First, I should recall what a 45-degree angle represents in geometry. I think it's part of a 45-45-90 triangle, which is an isosceles right triangle. In such a triangle, the two legs are equal, and the angles opposite those legs are both 45 degrees. That makes sense because the sum of angles in a triangle is 180 degrees, so if two angles are 45 each, the third must be 90 degrees.Now, in a 45-45-90 triangle, the sides have a specific ratio. I believe it's 1:1:√2, where the legs are equal, and the hypotenuse is √2 times the length of each leg. So, if I consider a unit circle, where the radius is 1, the hypotenuse of the triangle would be 1. That means each leg would be 1 divided by √2, right? Because if the hypotenuse is √2 times the leg, then the leg is hypotenuse divided by √2.Wait, let me write that down to make sure I'm not mixing things up. If the hypotenuse is 1, then each leg is 1/√2. That seems correct because (1/√2)^2 + (1/√2)^2 = 1/2 + 1/2 = 1, which matches the Pythagorean theorem for a right triangle with hypotenuse 1.So, in the unit circle, the point corresponding to 45 degrees would have coordinates (1/√2, 1/√2). That makes sense because both the x and y coordinates are equal in a 45-degree angle, reflecting the symmetry of the isosceles right triangle.Now, tangent of an angle in the unit circle is defined as the y-coordinate divided by the x-coordinate. So, tan(45°) would be (1/√2) divided by (1/√2). Let me compute that: (1/√2) / (1/√2) = 1. So, tan(45°) is 1.But wait, I should make sure I'm not missing anything here. Is there another way to think about this? Maybe using similar triangles or some other geometric principle?I remember that in a right triangle, tangent is the ratio of the opposite side to the adjacent side. In the case of a 45-45-90 triangle, both the opposite and adjacent sides are equal. So, if both sides are equal, their ratio would be 1. That aligns with what I found earlier.Let me also consider the unit circle approach again. The unit circle has a radius of 1, and any point on the circle can be represented as (cosθ, sinθ), where θ is the angle from the positive x-axis. For θ = 45°, both cos(45°) and sin(45°) are equal to √2/2, which is the same as 1/√2. Therefore, tan(45°) = sin(45°)/cos(45°) = (√2/2)/(√2/2) = 1.Hmm, so both methods give me the same result. That makes me more confident that tan(45°) is indeed 1.I wonder if there's a way to visualize this. If I draw a 45-45-90 triangle inside the unit circle, the legs would both be 1/√2, and the hypotenuse would be 1. The tangent would be the slope of the terminal side of the angle, which in this case is a line at 45 degrees from the x-axis. The slope of a 45-degree line is 1, which again confirms that tan(45°) is 1.Is there any other perspective or method I can use to verify this? Maybe using the properties of special triangles or some trigonometric identities?I know that tan(θ) = sin(θ)/cos(θ), and for θ = 45°, both sin and cos are equal, so their ratio is 1. That's straightforward.Alternatively, I can think about the unit circle and the coordinates. Since both coordinates are equal at 45 degrees, their ratio must be 1. That's another way to see it.I guess another way is to consider the graph of the tangent function. I know that tan(0°) is 0, and as the angle increases towards 90°, tan(θ) increases without bound. At 45°, which is halfway between 0° and 90°, the value of tan(θ) should be 1, which is the midpoint in a sense between 0 and infinity.Wait, is that accurate? I mean, the tangent function isn't linear, so the midpoint in angle doesn't correspond to the midpoint in value. But in this specific case, it just so happens that tan(45°) is 1, which is a nice, clean value.I think I've covered several different ways to approach this problem: using the properties of 45-45-90 triangles, the unit circle definition, the slope of the terminal side, and even considering the behavior of the tangent function. All these methods consistently lead me to the conclusion that tan(45°) is 1.I don't see any flaws in my reasoning. Each method reinforces the others, and the result is consistent across different geometric principles. So, I feel confident that tan(45°) is indeed 1.

answer：Okay, so I have this problem about two ellipses, C₁ and C₂. The equations are given as:C₁: (x²/a²) + y² = 1, where a > 1.C₂: (x²/4) + y² = 1.They mention that the eccentricities of these ellipses, e₁ and e₂ respectively, satisfy the relationship e₂ = √3 e₁. I need to find the value of 'a' from the given options.First, I should recall what eccentricity means for an ellipse. Eccentricity (e) measures how "stretched" an ellipse is. For an ellipse, it's defined as e = c/a, where 'c' is the distance from the center to a focus, and 'a' is the semi-major axis length.So, for both ellipses, I need to find their eccentricities and then set up the equation e₂ = √3 e₁ to solve for 'a'.Let me start with ellipse C₂ because its equation is fully given, and I can compute its eccentricity directly.Ellipse C₂: (x²/4) + y² = 1.In standard form, an ellipse is (x²/a²) + (y²/b²) = 1, where 'a' is the semi-major axis and 'b' is the semi-minor axis. If a > b, the major axis is along the x-axis; if b > a, it's along the y-axis.Looking at C₂: (x²/4) + y² = 1.Here, a² = 4, so a = 2, and b² = 1, so b = 1. Since a > b, the major axis is along the x-axis.Now, the formula for eccentricity e = c/a, where c is the distance from the center to a focus. Also, c² = a² - b².So for C₂:c₂² = a₂² - b₂² = 4 - 1 = 3.Therefore, c₂ = √3.Thus, eccentricity e₂ = c₂ / a₂ = √3 / 2.Okay, so e₂ is √3 / 2.Now, moving on to ellipse C₁: (x²/a²) + y² = 1.Similarly, this is in standard form. Here, a² is under x², so the major axis is along the x-axis since a > 1 (given). So, semi-major axis is 'a', and semi-minor axis is 1.Again, the eccentricity e₁ = c₁ / a₁, where c₁ is the distance from center to focus.But first, I need to find c₁. Using the relationship c₁² = a₁² - b₁².Here, a₁ = a, and b₁ = 1.So, c₁² = a² - 1.Therefore, c₁ = √(a² - 1).Thus, eccentricity e₁ = c₁ / a = √(a² - 1) / a.Now, according to the problem, e₂ = √3 e₁.We already found e₂ = √3 / 2.So, substituting:√3 / 2 = √3 * e₁Divide both sides by √3:1/2 = e₁So, e₁ = 1/2.But we also have e₁ = √(a² - 1) / a.Therefore:√(a² - 1) / a = 1/2Let me solve for 'a'.Multiply both sides by 'a':√(a² - 1) = a / 2Now, square both sides to eliminate the square root:(√(a² - 1))² = (a / 2)²Simplify:a² - 1 = a² / 4Multiply both sides by 4 to eliminate the denominator:4(a² - 1) = a²Expand the left side:4a² - 4 = a²Subtract a² from both sides:3a² - 4 = 0Add 4 to both sides:3a² = 4Divide both sides by 3:a² = 4/3Take square root of both sides:a = √(4/3) = (2)/√3 = (2√3)/3Since a > 1, and √3 ≈ 1.732, so 2√3 ≈ 3.464, divided by 3 is approximately 1.154, which is greater than 1. So, it's valid.Looking back at the options:A: 2√3 / 3B: √2C: √3D: √6So, the answer is option A: 2√3 / 3.Wait, but let me double-check my steps to make sure I didn't make a mistake.First, for C₂:a = 2, b = 1, c = √(4 - 1) = √3, e₂ = √3 / 2. That seems correct.For C₁:e₁ = √(a² - 1)/a.Given e₂ = √3 e₁, so √3 / 2 = √3 * (√(a² - 1)/a).Divide both sides by √3: 1/2 = √(a² - 1)/a.Then, √(a² - 1) = a/2.Square both sides: a² - 1 = a² / 4.Multiply by 4: 4a² - 4 = a².Subtract a²: 3a² - 4 = 0.So, 3a² = 4, a² = 4/3, a = 2√3 / 3. Yes, that seems correct.I think my steps are correct. So, the answer is A.**Final Answer**boxed{A}

answer：Okay, so I need to find the intersection of two lines given by the equations (3x - 2y = 8) and (5x + 3y = 27). Hmm, intersection means the point where both lines meet, right? So, basically, I need to find the values of (x) and (y) that satisfy both equations simultaneously. Alright, how do I approach this? I remember from class that there are a couple of methods to solve systems of equations like this: substitution and elimination. I think elimination might be easier here because the coefficients of (y) are different, but maybe I can manipulate the equations to eliminate one variable. Let me try that.First, let me write down the equations again to have them clear:1. (3x - 2y = 8)2. (5x + 3y = 27)I need to eliminate one of the variables. Let's see, if I can make the coefficients of (y) the same or opposites, I can add or subtract the equations to eliminate (y). The coefficients of (y) are (-2) and (3). If I multiply the first equation by 3 and the second equation by 2, then the coefficients of (y) will be (-6) and (6), which are opposites. That should work.So, multiplying the first equation by 3:(3 times (3x - 2y) = 3 times 8)Which gives:(9x - 6y = 24) ...(3)And multiplying the second equation by 2:(2 times (5x + 3y) = 2 times 27)Which gives:(10x + 6y = 54) ...(4)Now, I can add equations (3) and (4) together to eliminate (y):(9x - 6y + 10x + 6y = 24 + 54)Simplifying the left side:(9x + 10x - 6y + 6y = 19x + 0 = 19x)And the right side:(24 + 54 = 78)So, (19x = 78)Now, solving for (x):(x = frac{78}{19})Hmm, that's a fraction. Okay, so (x) is (frac{78}{19}). Now, I need to find (y). I can substitute this value of (x) back into one of the original equations. Let me choose the first equation because it looks simpler:(3x - 2y = 8)Substituting (x = frac{78}{19}):(3 times frac{78}{19} - 2y = 8)Calculating (3 times frac{78}{19}):(frac{234}{19} - 2y = 8)Now, I need to solve for (y). Let me subtract (frac{234}{19}) from both sides:(-2y = 8 - frac{234}{19})But wait, 8 is a whole number. I should convert it to a fraction with denominator 19 to subtract:(8 = frac{152}{19})So,(-2y = frac{152}{19} - frac{234}{19})Subtracting the numerators:(-2y = frac{152 - 234}{19} = frac{-82}{19})Now, dividing both sides by -2 to solve for (y):(y = frac{-82}{19} div (-2) = frac{-82}{19} times frac{1}{-2} = frac{82}{38})Simplifying (frac{82}{38}), both numerator and denominator are divisible by 2:(frac{82 ÷ 2}{38 ÷ 2} = frac{41}{19})So, (y = frac{41}{19})Alright, so the solution is (x = frac{78}{19}) and (y = frac{41}{19}). Let me just double-check to make sure I didn't make any mistakes.Plugging (x = frac{78}{19}) and (y = frac{41}{19}) into the first equation:(3x - 2y = 8)(3 times frac{78}{19} - 2 times frac{41}{19} = frac{234}{19} - frac{82}{19} = frac{152}{19} = 8)That works out.Now, checking the second equation:(5x + 3y = 27)(5 times frac{78}{19} + 3 times frac{41}{19} = frac{390}{19} + frac{123}{19} = frac{513}{19} = 27)Yes, that also works. So, both equations are satisfied with these values of (x) and (y). Therefore, the intersection point is (left(frac{78}{19}, frac{41}{19}right)).I think that's it. I followed the elimination method, multiplied the equations appropriately, solved for (x), then substituted back to find (y), and checked both solutions in the original equations to make sure they work. Everything seems consistent.