💡First, I need to determine how many bounces it takes for the ball to rise less than 40 cm after being dropped from a height of 320 cm. Each bounce, the ball reaches 75% of the previous height.I'll start by setting up the inequality to represent the height after ( b ) bounces:[320 times (0.75)^b < 40]Next, I'll simplify this inequality by dividing both sides by 320:[(0.75)^b < 0.125]To solve for ( b ), I'll take the logarithm of both sides. Since the base of the exponential term is 0.75, I'll use a logarithm with the same base:[log_{0.75}(0.75)^b < log_{0.75}(0.125)]This simplifies to:[b > log_{0.75}(0.125)]Using a calculator, I find that:[log_{0.75}(0.125) approx 5.047]Since the number of bounces must be an integer, I'll round up to the next whole number:[b = 6]Therefore, the ball will first rise less than 40 cm after 6 bounces.

💡Okay, so I have this equation: |||x - a| - 1| - 1| = |b|. I need to figure out how many solutions this equation can have. The options are from 2 to 6, so I need to consider different scenarios based on the values of a and b.First, let me try to understand the structure of the equation. It's a nested absolute value equation, which can sometimes be tricky because absolute value functions create V-shaped graphs, and nesting them can lead to more complex shapes.Let me break it down step by step. Let's start from the innermost absolute value:1. |x - a|: This is a V-shaped graph shifted to the right by 'a' units. It has its vertex at (a, 0).2. | |x - a| - 1 |: This takes the previous graph and shifts it down by 1 unit, then takes the absolute value again. So, if |x - a| - 1 is negative, it reflects it above the x-axis. This will create another V-shape, but it might have a different vertex.3. | | |x - a| - 1 | - 1 |: This is another layer of absolute value. So, we take the previous result, subtract 1, and then take the absolute value again. This can potentially create more V-shapes or flatten parts of the graph.Finally, we set this equal to |b|, which is a constant. So, the number of solutions will depend on how many times the graph of the left-hand side intersects the horizontal line y = |b|.To visualize this, I think it would help to sketch the graph of the left-hand side function. Let me denote the function as f(x) = |||x - a| - 1| - 1|.Let's analyze f(x):- Start with |x - a|: This is a V-shape with vertex at (a, 0).- Then, | |x - a| - 1 |: This shifts the previous V down by 1 unit. If |x - a| is less than 1, the expression inside the absolute value becomes negative, so it flips to positive. This creates a W-shape or another V-shape depending on the shift.- Then, | | |x - a| - 1 | - 1 |: This shifts the previous result down by another 1 unit and takes the absolute value again. This can create additional V-shapes or flatten parts of the graph.I think the key is to find the critical points where the expression inside the absolute values changes sign, which will help in determining the number of intersections with y = |b|.Let's find the critical points:1. The innermost absolute value |x - a| changes sign at x = a.2. The next absolute value | |x - a| - 1 | changes sign when |x - a| - 1 = 0, which is at |x - a| = 1, so x = a ± 1.3. The outermost absolute value | | |x - a| - 1 | - 1 | changes sign when | |x - a| - 1 | - 1 = 0, which is when | |x - a| - 1 | = 1. This happens when |x - a| - 1 = ±1, so |x - a| = 0 or |x - a| = 2. Therefore, x = a, x = a ± 2.So, the critical points are x = a, a ± 1, a ± 2.These points divide the real line into intervals where the function f(x) behaves differently. Let's consider each interval:1. x < a - 22. a - 2 ≤ x < a - 13. a - 1 ≤ x < a4. a ≤ x < a + 15. a + 1 ≤ x < a + 26. x ≥ a + 2In each interval, the expression inside the absolute values will have consistent signs, so we can remove the absolute values accordingly and find the expression for f(x).Let's compute f(x) in each interval:1. x < a - 2: - |x - a| = a - x - | |x - a| - 1 | = |a - x - 1| = a - x - 1 (since a - x - 1 > 0 because x < a - 2) - | | |x - a| - 1 | - 1 | = |a - x - 1 - 1| = |a - x - 2| = a - x - 2 (since a - x - 2 > 0 because x < a - 2) So, f(x) = a - x - 22. a - 2 ≤ x < a - 1: - |x - a| = a - x - | |x - a| - 1 | = |a - x - 1| = a - x - 1 (since a - x - 1 > 0 because x < a - 1) - | | |x - a| - 1 | - 1 | = |a - x - 1 - 1| = |a - x - 2| = a - x - 2 (since a - x - 2 ≥ 0 because x ≥ a - 2) So, f(x) = a - x - 2Wait, that seems the same as the previous interval. Maybe I made a mistake.Wait, in the interval a - 2 ≤ x < a - 1, |x - a| = a - x, which is greater than 1 because x < a - 1, so |x - a| - 1 = a - x - 1 > 0, so | |x - a| - 1 | = a - x - 1. Then, | | |x - a| - 1 | - 1 | = |a - x - 1 - 1| = |a - x - 2|. Since x ≥ a - 2, a - x - 2 ≤ 0, so |a - x - 2| = x - a + 2.Wait, that's different. So, f(x) = x - a + 2 in this interval.Similarly, let's correct that:1. x < a - 2: - |x - a| = a - x - | |x - a| - 1 | = a - x - 1 - | | |x - a| - 1 | - 1 | = |a - x - 2| = a - x - 2 (since a - x - 2 > 0) So, f(x) = a - x - 22. a - 2 ≤ x < a - 1: - |x - a| = a - x - | |x - a| - 1 | = a - x - 1 - | | |x - a| - 1 | - 1 | = |a - x - 2| = x - a + 2 (since a - x - 2 ≤ 0) So, f(x) = x - a + 23. a - 1 ≤ x < a: - |x - a| = a - x - | |x - a| - 1 | = |a - x - 1| = x - a + 1 (since a - x - 1 ≤ 0 because x ≥ a - 1) - | | |x - a| - 1 | - 1 | = |x - a + 1 - 1| = |x - a| = a - x So, f(x) = a - x4. a ≤ x < a + 1: - |x - a| = x - a - | |x - a| - 1 | = |x - a - 1| = a - x + 1 (since x - a - 1 ≤ 0 because x < a + 1) - | | |x - a| - 1 | - 1 | = |a - x + 1 - 1| = |a - x| = x - a So, f(x) = x - a5. a + 1 ≤ x < a + 2: - |x - a| = x - a - | |x - a| - 1 | = |x - a - 1| = x - a - 1 (since x - a - 1 ≥ 0 because x ≥ a + 1) - | | |x - a| - 1 | - 1 | = |x - a - 1 - 1| = |x - a - 2| = a - x + 2 (since x - a - 2 ≤ 0 because x < a + 2) So, f(x) = a - x + 26. x ≥ a + 2: - |x - a| = x - a - | |x - a| - 1 | = x - a - 1 - | | |x - a| - 1 | - 1 | = |x - a - 2| = x - a - 2 (since x - a - 2 ≥ 0) So, f(x) = x - a - 2Now, let's summarize the expressions for f(x) in each interval:1. x < a - 2: f(x) = a - x - 22. a - 2 ≤ x < a - 1: f(x) = x - a + 23. a - 1 ≤ x < a: f(x) = a - x4. a ≤ x < a + 1: f(x) = x - a5. a + 1 ≤ x < a + 2: f(x) = a - x + 26. x ≥ a + 2: f(x) = x - a - 2Now, let's analyze the shape of f(x):- For x < a - 2, f(x) is a line with slope -1.- For a - 2 ≤ x < a - 1, f(x) is a line with slope +1.- For a - 1 ≤ x < a, f(x) is a line with slope -1.- For a ≤ x < a + 1, f(x) is a line with slope +1.- For a + 1 ≤ x < a + 2, f(x) is a line with slope -1.- For x ≥ a + 2, f(x) is a line with slope +1.Plotting these, we can see that f(x) has a series of V-shapes, creating a sort of zigzag pattern. The critical points at x = a - 2, a - 1, a, a + 1, a + 2 are where the slope changes.Now, we set f(x) = |b| and want to find the number of solutions. The number of solutions depends on how many times the horizontal line y = |b| intersects the graph of f(x).Let's consider different cases for |b|:Case 1: |b| > 2In this case, the horizontal line y = |b| will intersect the graph of f(x) only in the outermost regions where f(x) is increasing or decreasing without bound. Specifically, for x < a - 2 and x ≥ a + 2, f(x) is linear with slopes -1 and +1, respectively. Since |b| > 2, the intersections will occur in these outer regions, giving two solutions.Case 2: |b| = 2Here, y = 2 will touch the graph of f(x) at the points where f(x) reaches its maximum in the middle regions. Specifically, at x = a - 2 and x = a + 2, f(x) = 0, but wait, no. Wait, f(x) at x = a - 2 is f(a - 2) = a - (a - 2) - 2 = 0. Similarly, at x = a + 2, f(x) = (a + 2) - a - 2 = 0. Wait, that's not right. Let me recalculate.Wait, f(a - 2) = a - (a - 2) - 2 = 0. Similarly, f(a + 2) = (a + 2) - a - 2 = 0. So, at x = a - 2 and x = a + 2, f(x) = 0. But we set f(x) = 2, which is greater than 0. So, the intersections will be in the regions where f(x) is increasing or decreasing beyond these points. Specifically, for x < a - 2, f(x) = a - x - 2. Setting this equal to 2: a - x - 2 = 2 => x = a - 4. Similarly, for x ≥ a + 2, f(x) = x - a - 2. Setting this equal to 2: x - a - 2 = 2 => x = a + 4. So, two solutions.Wait, but also, in the middle regions, f(x) reaches a maximum of 1 at x = a - 1 and x = a + 1. So, if |b| = 2, which is greater than 1, the intersections are only in the outer regions, giving two solutions.Case 3: 1 < |b| < 2In this case, the horizontal line y = |b| will intersect the graph of f(x) in four regions: two in the middle regions and two in the outer regions. Let's see:- In the region a - 2 ≤ x < a - 1, f(x) = x - a + 2. Setting this equal to |b|: x - a + 2 = |b| => x = a - 2 + |b|. Since 1 < |b| < 2, x will be between a - 1 and a - 2 + 2 = a. So, this is a valid solution in this interval.- In the region a - 1 ≤ x < a, f(x) = a - x. Setting this equal to |b|: a - x = |b| => x = a - |b|. Since 1 < |b| < 2, x will be between a - 2 and a - 1. So, this is another solution.Similarly, on the right side:- In the region a ≤ x < a + 1, f(x) = x - a. Setting this equal to |b|: x - a = |b| => x = a + |b|. Since 1 < |b| < 2, x will be between a + 1 and a + 2. So, this is another solution.- In the region a + 1 ≤ x < a + 2, f(x) = a - x + 2. Setting this equal to |b|: a - x + 2 = |b| => x = a + 2 - |b|. Since 1 < |b| < 2, x will be between a + 0 and a + 1. So, this is another solution.Additionally, in the outer regions:- For x < a - 2, f(x) = a - x - 2. Setting this equal to |b|: a - x - 2 = |b| => x = a - 2 - |b|. Since |b| > 1, x will be less than a - 3, which is valid in this interval.- For x ≥ a + 2, f(x) = x - a - 2. Setting this equal to |b|: x - a - 2 = |b| => x = a + 2 + |b|. Since |b| > 1, x will be greater than a + 3, which is valid in this interval.Wait, so that's actually six solutions? But that contradicts my earlier thought. Let me check.Wait, no. When |b| is between 1 and 2, the horizontal line y = |b| will intersect the graph in four points: two in the middle regions (a - 2 ≤ x < a - 1 and a + 1 ≤ x < a + 2) and two in the outer regions (x < a - 2 and x ≥ a + 2). So, that's four solutions.Wait, but earlier I thought of six, but that's incorrect because in the middle regions, each region contributes one solution, and the outer regions contribute one each, totaling four.Case 4: |b| = 1Here, y = 1 will intersect the graph of f(x) at points where f(x) = 1. Let's see:- In the region a - 2 ≤ x < a - 1, f(x) = x - a + 2. Setting this equal to 1: x - a + 2 = 1 => x = a - 1. But x = a - 1 is the boundary of this interval, so it's a solution.- In the region a - 1 ≤ x < a, f(x) = a - x. Setting this equal to 1: a - x = 1 => x = a - 1. Again, this is the boundary.Similarly, on the right side:- In the region a ≤ x < a + 1, f(x) = x - a. Setting this equal to 1: x - a = 1 => x = a + 1.- In the region a + 1 ≤ x < a + 2, f(x) = a - x + 2. Setting this equal to 1: a - x + 2 = 1 => x = a + 1.So, at |b| = 1, the horizontal line touches the graph at four points: x = a - 1, a - 1, a + 1, a + 1. But since these are boundaries, they are counted as two distinct points, each with multiplicity two. So, technically, it's two solutions, but each with multiplicity two, making it four solutions in total.Wait, but in reality, at x = a - 1 and x = a + 1, the function f(x) reaches its maximum of 1. So, setting f(x) = 1 will intersect exactly at these points, giving two solutions. However, considering the behavior of the function, at these points, the function changes direction, so it's a single point of intersection, not two. So, actually, it's two solutions.But earlier, I thought it's four, but that's incorrect because the function only touches the line y = 1 at two points: x = a - 1 and x = a + 1. So, two solutions.Case 5: 0 < |b| < 1In this case, the horizontal line y = |b| will intersect the graph of f(x) in six regions:- In the region a - 2 ≤ x < a - 1, f(x) = x - a + 2. Setting this equal to |b|: x = a - 2 + |b|. Since |b| < 1, x will be between a - 2 and a - 1, which is valid.- In the region a - 1 ≤ x < a, f(x) = a - x. Setting this equal to |b|: x = a - |b|. Since |b| < 1, x will be between a - 1 and a, which is valid.- In the region a ≤ x < a + 1, f(x) = x - a. Setting this equal to |b|: x = a + |b|. Since |b| < 1, x will be between a and a + 1, which is valid.- In the region a + 1 ≤ x < a + 2, f(x) = a - x + 2. Setting this equal to |b|: x = a + 2 - |b|. Since |b| < 1, x will be between a + 1 and a + 2, which is valid.Additionally, in the outer regions:- For x < a - 2, f(x) = a - x - 2. Setting this equal to |b|: x = a - 2 - |b|. Since |b| > 0, x will be less than a - 2, which is valid.- For x ≥ a + 2, f(x) = x - a - 2. Setting this equal to |b|: x = a + 2 + |b|. Since |b| > 0, x will be greater than a + 2, which is valid.So, in total, six solutions.Case 6: |b| = 0Here, y = 0. The equation becomes |||x - a| - 1| - 1| = 0, which implies ||x - a| - 1| - 1 = 0 => ||x - a| - 1| = 1.This equation has solutions where |x - a| - 1 = ±1.So:1. |x - a| - 1 = 1 => |x - a| = 2 => x = a ± 2.2. |x - a| - 1 = -1 => |x - a| = 0 => x = a.So, three solutions: x = a - 2, x = a, x = a + 2.Wait, but earlier in the analysis, when |b| = 0, f(x) = 0 occurs at x = a - 2, a, a + 2. So, three solutions.But in the initial breakdown, I thought of six solutions when |b| < 1, but when |b| = 0, it's three solutions. So, the number of solutions varies depending on |b|.Putting it all together:- If |b| > 2: 2 solutions.- If |b| = 2: 2 solutions.- If 1 < |b| < 2: 4 solutions.- If |b| = 1: 2 solutions.- If 0 < |b| < 1: 6 solutions.- If |b| = 0: 3 solutions.Wait, that seems inconsistent. Let me double-check.Wait, when |b| = 2, f(x) = 2 intersects the graph at two points: x = a - 4 and x = a + 4, as I calculated earlier.When |b| = 1, f(x) = 1 intersects at x = a - 1 and x = a + 1, which are two points.When |b| = 0, f(x) = 0 intersects at x = a - 2, a, a + 2, which are three points.When 0 < |b| < 1, f(x) = |b| intersects at six points: x = a - 2 - |b|, a - 2 + |b|, a - |b|, a + |b|, a + 2 - |b|, a + 2 + |b|.When 1 < |b| < 2, f(x) = |b| intersects at four points: x = a - 2 + |b|, a - |b|, a + |b|, a + 2 - |b|.Wait, but earlier I thought for 1 < |b| < 2, it's four solutions, and for 0 < |b| < 1, it's six solutions.So, the maximum number of solutions is six, which occurs when 0 < |b| < 1.But the question is asking "how many solutions can the equation have?" So, it's asking for the maximum number of solutions possible, not the number of solutions for a specific |b|.Therefore, the equation can have up to six solutions when 0 < |b| < 1.But wait, let me confirm:When 0 < |b| < 1, we have six solutions:1. x = a - 2 - |b|2. x = a - 2 + |b|3. x = a - |b|4. x = a + |b|5. x = a + 2 - |b|6. x = a + 2 + |b|Yes, that's six distinct solutions.When |b| = 1, we have two solutions.When |b| = 0, we have three solutions.When |b| > 2, we have two solutions.When 1 < |b| < 2, we have four solutions.Therefore, the maximum number of solutions is six.However, looking back at the options, the choices are A) 2, B) 3, C) 4, D) 5, E) 6.So, the answer should be E) 6.But wait, in the initial analysis, I thought the answer was 4, but after a more detailed breakdown, it's 6.I think the confusion arises from whether the function f(x) can indeed have six intersections with y = |b| when |b| is between 0 and 1.Yes, because in each of the six intervals, the function f(x) is linear and can intersect the horizontal line y = |b| once in each interval, leading to six solutions.Therefore, the maximum number of solutions is six.

💡First, I need to simplify each square root separately. For the first square root, √(9⋅4), I can multiply 9 and 4 to get 36, and then take the square root of 36, which is 6.Next, for the second square root, √(3³⋅4³), I recognize that both 3 and 4 are raised to the power of 3. I can rewrite this as (3⋅4)³, which simplifies to 12³. Taking the square root of 12³ gives me 12 raised to the power of 3/2.To simplify 12³/², I can break it down into 12 times the square root of 12. The square root of 12 can be further simplified to 2√3, so multiplying this by 12 gives me 24√3.Finally, I multiply the two simplified results together: 6 times 24√3 equals 144√3. Therefore, the simplified expression is 144√3.

💡First, I need to determine the total number of students taking at least one of the subjects: music, art, or science. This can be done using the principle of inclusion-exclusion.I'll start by adding the number of students taking each individual subject:- Music: 80 students- Art: 60 students- Science: 50 studentsTotal = 80 + 60 + 50 = 190 studentsNext, I need to account for the students who are taking more than one subject, as they have been counted multiple times. I'll subtract the number of students taking two subjects at a time:- Both Music and Art: 30 students- Both Music and Science: 25 students- Both Art and Science: 20 studentsTotal to subtract = 30 + 25 + 20 = 75 studentsAdjusted total after subtracting overlaps = 190 - 75 = 115 studentsHowever, students taking all three subjects have been subtracted too many times. I'll add them back once:- All three subjects: 15 studentsFinal total of students taking at least one subject = 115 + 15 = 130 studentsFinally, to find the number of students taking neither music, nor art, nor science, I'll subtract the number of students taking at least one subject from the total number of students:- Total students: 800- Students taking at least one subject: 130Number of students taking neither = 800 - 130 = 670 students

💡Okay, so I need to figure out how much money Porter will earn in a month if he works overtime every week. Let's see, he earns 8 per day and works 5 days a week normally. His manager is asking him to work an extra day, and for that extra day, he'll get an extra 50% on top of his daily rate. Hmm, okay, so I need to calculate his total earnings for a month, including this overtime.First, let's break down his regular earnings. If he earns 8 per day and works 5 days a week, then his weekly earnings would be 8 times 5, which is 40. Now, assuming there are 4 weeks in a month, his monthly earnings without any overtime would be 40 times 4, which is 160.But he's also working an extra day each week. On that extra day, he gets 50% more than his regular daily rate. So, what's 50% of 8? That would be 8 times 0.5, which is 4. So, his overtime pay for that extra day is 8 plus 4, totaling 12 per overtime day.Since he's working overtime every week, and there are 4 weeks in a month, his total overtime pay for the month would be 12 times 4, which is 48.Now, to find his total earnings for the month, I need to add his regular earnings and his overtime earnings together. So, 160 plus 48 equals 208.Wait, let me double-check that. His regular pay is 40 a week, times 4 weeks is 160. His overtime pay is 12 a week, times 4 weeks is 48. Adding them together gives 208. Yeah, that seems right.I think I've got it. So, if Porter works overtime every week, he'll earn a total of 208 in a month.

💡Okay, so I have this problem where I need to find the points where at least two of the three given lines intersect. The lines are given by the equations:1. (2y - 3x = 4)2. (3x + y = 5)3. (6x - 4y = 8)Hmm, I remember that to find the intersection of two lines, I need to solve their equations simultaneously. That means finding the values of (x) and (y) that satisfy both equations at the same time. First, maybe I should write each equation in slope-intercept form ((y = mx + b)) so I can easily see if any lines are parallel or if any are the same line. That might help me figure out if they intersect or not.Starting with the first equation: (2y - 3x = 4). Let me solve for (y).Add (3x) to both sides: (2y = 3x + 4).Now divide both sides by 2: (y = frac{3}{2}x + 2).Okay, so the first line has a slope of (frac{3}{2}) and a y-intercept at 2.Moving on to the second equation: (3x + y = 5). Let's solve for (y).Subtract (3x) from both sides: (y = -3x + 5).This line has a slope of (-3) and a y-intercept at 5.Now the third equation: (6x - 4y = 8). Let's solve for (y).First, subtract (6x) from both sides: (-4y = -6x + 8).Now divide both sides by (-4): (y = frac{6}{4}x - 2).Simplify (frac{6}{4}) to (frac{3}{2}): (y = frac{3}{2}x - 2).So, the third line also has a slope of (frac{3}{2}) but a y-intercept at (-2).Wait a minute, the first and third lines both have the same slope, (frac{3}{2}). That means they are parallel. If two lines are parallel, they never intersect unless they are the same line. But looking at their y-intercepts, the first line has a y-intercept of 2, and the third line has a y-intercept of (-2). Since these are different, the lines are distinct and parallel, so they don't intersect. That means there's no intersection point between the first and third lines.Alright, so now I need to check the intersections between the first and second lines, and between the second and third lines.Starting with the first and second lines:First line: (y = frac{3}{2}x + 2)Second line: (y = -3x + 5)To find their intersection, set the two equations equal to each other:(frac{3}{2}x + 2 = -3x + 5)Let me solve for (x). First, I'll get all the (x) terms on one side and constants on the other.Add (3x) to both sides:(frac{3}{2}x + 3x + 2 = 5)Combine like terms:(frac{3}{2}x + frac{6}{2}x + 2 = 5)Which is:(frac{9}{2}x + 2 = 5)Now, subtract 2 from both sides:(frac{9}{2}x = 3)To solve for (x), multiply both sides by (frac{2}{9}):(x = 3 times frac{2}{9} = frac{6}{9} = frac{2}{3})Wait, that doesn't seem right. Let me double-check my steps.Starting from:(frac{3}{2}x + 2 = -3x + 5)Subtract (frac{3}{2}x) from both sides:(2 = -3x - frac{3}{2}x + 5)Combine the (x) terms:(-3x - frac{3}{2}x = -frac{6}{2}x - frac{3}{2}x = -frac{9}{2}x)So, (2 = -frac{9}{2}x + 5)Subtract 5 from both sides:(2 - 5 = -frac{9}{2}x)Which is:(-3 = -frac{9}{2}x)Multiply both sides by (-1):(3 = frac{9}{2}x)Now, multiply both sides by (frac{2}{9}):(x = 3 times frac{2}{9} = frac{6}{9} = frac{2}{3})Okay, so (x = frac{2}{3}). Now, plug this back into one of the original equations to find (y). I'll use the second equation because it looks simpler: (y = -3x + 5).Substitute (x = frac{2}{3}):(y = -3 times frac{2}{3} + 5 = -2 + 5 = 3)So, the intersection point of the first and second lines is (left(frac{2}{3}, 3right)).Wait, that doesn't match what I thought earlier. Let me check again. Maybe I made a mistake in substituting.Wait, no, if (x = frac{2}{3}), then (y = -3 times frac{2}{3} + 5 = -2 + 5 = 3). Yeah, that's correct.But earlier, I thought the intersection point was (2,5). Hmm, maybe I confused the lines. Let me check the equations again.First line: (2y - 3x = 4)Second line: (3x + y = 5)If I plug (x = frac{2}{3}) and (y = 3) into the first equation:(2(3) - 3(frac{2}{3}) = 6 - 2 = 4), which matches.And into the second equation:(3(frac{2}{3}) + 3 = 2 + 3 = 5), which also matches.So, the intersection point is indeed (left(frac{2}{3}, 3right)). I must have made a mistake earlier when I thought it was (2,5). Maybe I confused the equations.Now, moving on to the second and third lines.Second line: (y = -3x + 5)Third line: (y = frac{3}{2}x - 2)Set them equal to find the intersection:(-3x + 5 = frac{3}{2}x - 2)Let me solve for (x). First, get all (x) terms on one side.Add (3x) to both sides:(5 = frac{3}{2}x + 3x - 2)Combine like terms:(frac{3}{2}x + 3x = frac{3}{2}x + frac{6}{2}x = frac{9}{2}x)So, (5 = frac{9}{2}x - 2)Add 2 to both sides:(7 = frac{9}{2}x)Multiply both sides by (frac{2}{9}):(x = 7 times frac{2}{9} = frac{14}{9})Now, plug (x = frac{14}{9}) back into one of the equations to find (y). I'll use the second equation again: (y = -3x + 5).Substitute (x = frac{14}{9}):(y = -3 times frac{14}{9} + 5 = -frac{42}{9} + 5 = -frac{14}{3} + 5)Convert 5 to thirds: (5 = frac{15}{3})So, (y = -frac{14}{3} + frac{15}{3} = frac{1}{3})Therefore, the intersection point of the second and third lines is (left(frac{14}{9}, frac{1}{3}right)).Wait, let me verify this by plugging into the third equation as well.Third line: (6x - 4y = 8)Substitute (x = frac{14}{9}) and (y = frac{1}{3}):(6 times frac{14}{9} - 4 times frac{1}{3} = frac{84}{9} - frac{4}{3} = frac{28}{3} - frac{4}{3} = frac{24}{3} = 8), which matches.And into the second equation: (3x + y = 5)(3 times frac{14}{9} + frac{1}{3} = frac{42}{9} + frac{1}{3} = frac{14}{3} + frac{1}{3} = frac{15}{3} = 5), which also matches.So, both intersection points are correct.Wait, but earlier I thought the first and second lines intersect at (2,5), but now I'm getting (left(frac{2}{3}, 3right)). I must have made a mistake in my initial calculation. Let me check that again.First line: (2y - 3x = 4)Second line: (3x + y = 5)Let me solve them again using substitution.From the second equation: (y = 5 - 3x)Substitute into the first equation:(2(5 - 3x) - 3x = 4)Expand:(10 - 6x - 3x = 4)Combine like terms:(10 - 9x = 4)Subtract 10 from both sides:(-9x = -6)Divide by -9:(x = frac{6}{9} = frac{2}{3})Then, (y = 5 - 3 times frac{2}{3} = 5 - 2 = 3)So, yes, the intersection is indeed (left(frac{2}{3}, 3right)). I must have confused the equations earlier when I thought it was (2,5). That was a mistake.So, in summary, the first and second lines intersect at (left(frac{2}{3}, 3right)), and the second and third lines intersect at (left(frac{14}{9}, frac{1}{3}right)). The first and third lines are parallel and do not intersect.Therefore, there are two intersection points.