Solving Inequalities: Find The Sum Of Integer Solutions

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Hey guys! Today, we're diving into the exciting world of inequalities and integer solutions. We've got a fun problem on our hands: finding the sum of all integer solutions for a system of inequalities within a specific interval. Sounds like a challenge? Absolutely! But don't worry, we'll break it down step by step. So, grab your thinking caps, and let's get started!

Understanding the Problem

First, let's make sure we understand exactly what we're trying to do. We have a system of inequalities, which basically means we have two or more inequalities that need to be true at the same time. In our case, we've got two inequalities involving x, and we need to figure out what values of x satisfy both of them. But here's the catch: we're only interested in integer solutions. That means we only want whole numbers (no fractions or decimals) that make both inequalities true. And to make things even more specific, we're looking for these integer solutions within the interval [-10; 6]. This means our solutions must be greater than or equal to -10 and less than or equal to 6. So, to recap, our mission is to:

  1. Solve the system of inequalities.
  2. Identify the integer solutions.
  3. Make sure these solutions fall within the interval [-10; 6].
  4. Finally, calculate the sum of these integer solutions.

Sounds like a plan? Let's dive into the nitty-gritty of solving these inequalities!

Breaking Down the Inequalities

Now, let's take a closer look at the inequalities we're dealing with. We've got two of them:

  1. x² ≥ 25
  2. x² - 6x < 0

These might look a little intimidating at first, but don't sweat it! We'll tackle them one by one. Let's start with the first inequality, x² ≥ 25. This one involves a square, so we need to think about what numbers, when squared, are greater than or equal to 25. You might already be thinking of 5, since 5² = 25. But remember, negative numbers can also give us positive squares. For example, (-5)² = 25 as well. So, we need to consider both positive and negative possibilities. In fact, this inequality tells us that x must be either greater than or equal to 5, or less than or equal to -5. Makes sense? Now, let's move on to the second inequality, x² - 6x < 0. This one looks a little different, but we can still handle it. The key here is to factor the expression on the left side. We can factor out an x, which gives us x(x - 6) < 0. Now, we have a product of two terms that's less than zero. This means one of the terms must be positive, and the other must be negative. Think about it: a positive times a positive is positive, a negative times a negative is also positive, but a positive times a negative (or vice versa) is negative. So, how do we figure out when x and (x - 6) have opposite signs? We can analyze this by considering different intervals for x. We'll see how to do this in the next section.

Solving the System of Inequalities

Okay, we've broken down the individual inequalities, now it's time to put them together and solve the system. Remember, we need to find the values of x that satisfy both inequalities at the same time. Let's recap what we found:

  1. x² ≥ 25 implies x ≤ -5 or x ≥ 5
  2. x² - 6x < 0 implies x(x - 6) < 0

To solve the second inequality, x(x - 6) < 0, we need to figure out when the product of x and (x - 6) is negative. This happens when x and (x - 6) have opposite signs. Let's consider the critical points where either x or (x - 6) is equal to zero. These points are x = 0 and x = 6. These points divide the number line into three intervals: x < 0, 0 < x < 6, and x > 6. Now, we can test a value from each interval to see if it satisfies the inequality x(x - 6) < 0:

  • If x < 0 (e.g., x = -1), then x is negative and (x - 6) is also negative, so their product is positive. This interval doesn't work.
  • If 0 < x < 6 (e.g., x = 1), then x is positive and (x - 6) is negative, so their product is negative. This interval works!
  • If x > 6 (e.g., x = 7), then x is positive and (x - 6) is also positive, so their product is positive. This interval doesn't work.

So, the solution to x² - 6x < 0 is 0 < x < 6. Now we need to combine this with the solution to x² ≥ 25, which was x ≤ -5 or x ≥ 5. To satisfy both inequalities, we need to find the overlap between these solutions. Let's visualize this on a number line. We have x ≤ -5 or x ≥ 5 from the first inequality, and 0 < x < 6 from the second inequality. The overlap is the interval 5 ≤ x < 6. This means the values of x that satisfy both inequalities are those that are greater than or equal to 5, but strictly less than 6. Now, let's think about the integers in this range.

Identifying Integer Solutions and Their Sum

Alright, we're in the home stretch! We've solved the system of inequalities, and we know that the solutions must satisfy 5 ≤ x < 6. But remember, we're only interested in integer solutions, and we also have the restriction that our solutions must be within the interval [-10; 6]. So, let's put it all together. The integers that satisfy 5 ≤ x < 6 are simply x = 5. That's the only integer that falls within this range. Now, we need to check if this solution is within our given interval of [-10; 6]. Since 5 is indeed between -10 and 6, it's a valid solution. Great! So, we have only one integer solution: x = 5. Now, the final step: find the sum of all integer solutions. Since we only have one solution, the sum is just the solution itself. Therefore, the sum of all integer solutions is 5. And there you have it! We've successfully navigated through the inequalities, identified the integer solutions, and calculated their sum. High five!

Conclusion

Woohoo! We made it! We've successfully solved a system of inequalities, identified the integer solutions within a given interval, and calculated their sum. Remember, the key to tackling these problems is to break them down into smaller, manageable steps. First, we understood the problem, then we analyzed each inequality individually, then we combined the solutions, and finally, we identified the integer solutions and calculated their sum. It's like solving a puzzle, one piece at a time. And now, you've added another valuable tool to your problem-solving arsenal. So, the next time you encounter a system of inequalities, don't panic! Just remember the steps we've covered, and you'll be well on your way to finding the solutions. Keep practicing, keep exploring, and most importantly, keep having fun with math! You guys are awesome!