Integer & Inverse Problems: Solve It Step-by-Step!

Hey guys! Let's break down these awesome math problems together. We've got some integer puzzles involving multiplicative inverses and even a cool pool-filling scenario. Don't worry, we'll take it one step at a time, making sure everyone understands the logic and math behind each solution. So, grab your calculators (or your brainpower!) and let’s dive in!

a) Finding an Integer with a Specific Inverse Difference

Okay, so the first problem asks us to find an integer where the difference between the number and its multiplicative inverse is 8/3. What does that even mean? Let's break it down. Remember, the multiplicative inverse of a number 'x' is simply 1/x. So, we need to find an integer, let's call it 'x', such that: x - (1/x) = 8/3.

Now, how do we solve this? This looks like an equation, and we love equations, right? The best way to tackle this is to get rid of the fractions. We can do this by multiplying both sides of the equation by 3x. Why 3x? Because that's the least common multiple of the denominators we have (3 and x). This will clear out those pesky fractions and make our lives easier. Trust me, it's a pro move in algebra!

Let's do it: 3x * [x - (1/x)] = 3x * (8/3). Distributing the 3x on the left side gives us 3x² - 3 = 8x. Aha! Now we have a quadratic equation! See how things are starting to fall into place? We can rearrange this equation to the standard quadratic form: 3x² - 8x - 3 = 0.

Now, we've got a classic quadratic equation. Time to put on our solving hats! There are a couple of ways we can solve this: factoring or using the quadratic formula. Factoring is often the quicker route if we can spot the factors easily. Let's see... Can we find two numbers that multiply to (3 * -3 = -9) and add up to -8? Yes! Those numbers are -9 and 1. So, we can rewrite the middle term (-8x) as -9x + x. This allows us to factor by grouping.

Our equation becomes: 3x² - 9x + x - 3 = 0. Now we factor by grouping: 3x(x - 3) + 1(x - 3) = 0. Notice the (x - 3) term is common? We can factor it out: (3x + 1)(x - 3) = 0. Amazing! We've factored the quadratic! This means that either (3x + 1) = 0 or (x - 3) = 0. Solving these two little equations will give us our possible values for x.

If 3x + 1 = 0, then 3x = -1, and x = -1/3. But wait! The problem specifically asked for an integer solution. -1/3 is not an integer. So, this solution doesn't work for us. Let's look at the other possibility.

If x - 3 = 0, then x = 3. Bingo! 3 is an integer! Let's just double-check if this solution works in our original equation: 3 - (1/3) = 9/3 - 1/3 = 8/3. It works perfectly! So, the integer we were looking for is 3. See? We solved it!

b) Finding an Integer with a Specific Inverse Sum

Alright, let's move on to the second problem. This time, we need to find an integer such that the sum of the number and its inverse is 26/5. Sounds similar, right? We’ll use the same awesome techniques we learned in the first one. Let's call our integer 'y' this time, just to mix things up. So, the equation we need to solve is: y + (1/y) = 26/5.

Just like before, our first step is to eliminate the fractions. The least common multiple of our denominators (y and 5) is 5y. So, let's multiply both sides of the equation by 5y: 5y * [y + (1/y)] = 5y * (26/5). Distributing the 5y on the left side, we get 5y² + 5 = 26y.

Time to rearrange this into the standard quadratic form: 5y² - 26y + 5 = 0. Now we have another quadratic equation to conquer! Are you feeling like a quadratic-solving pro yet? Let's try factoring again. We need two numbers that multiply to (5 * 5 = 25) and add up to -26. Hmm... -25 and -1 seem like good candidates!

So, we rewrite the middle term (-26y) as -25y - y, giving us: 5y² - 25y - y + 5 = 0. Now we factor by grouping: 5y(y - 5) - 1(y - 5) = 0. And we factor out the common (y - 5) term: (5y - 1)(y - 5) = 0. Excellent! We've factored another quadratic!

This gives us two possibilities: either (5y - 1) = 0 or (y - 5) = 0. If 5y - 1 = 0, then 5y = 1, and y = 1/5. But remember, we need an integer solution, and 1/5 is not an integer. So, let’s move on to the second case.

If y - 5 = 0, then y = 5. Aha! 5 is an integer! Let's check if it satisfies the original equation: 5 + (1/5) = 25/5 + 1/5 = 26/5. Perfect! It works. So, the integer we were searching for is 5. Another problem solved! You guys are crushing it!

c) The Pool-Filling Problem: Teamwork Makes the Dream Work!

Okay, this one is a bit different, but don't worry, we've got this! This is a classic work-rate problem, and we can totally handle it. The problem states that two friends, A and B, can fill a pool together in 2 hours. Friend A can fill the pool alone in 3 hours. The question is: how long would it take friend B to fill the pool alone?

The key to these types of problems is to think about the rate at which each person works. The rate is simply the fraction of the job completed in one unit of time (in this case, one hour). So, if friend A can fill the entire pool in 3 hours, then friend A's rate is 1/3 of the pool per hour.

Similarly, if friends A and B can fill the pool together in 2 hours, their combined rate is 1/2 of the pool per hour. Let's call friend B's rate 'r'. We want to find how many hours it would take B to fill the pool alone, which is 1/r (the inverse of B's rate).

The fundamental idea here is that the combined rate of A and B is equal to the sum of their individual rates. This makes perfect sense, right? If they're working together, their efforts add up! So, we can write the equation: (A's rate) + (B's rate) = (Combined rate), or (1/3) + r = (1/2).

Now we have a simple equation to solve for r. Let's subtract 1/3 from both sides: r = (1/2) - (1/3). To subtract these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6. So, we rewrite the fractions: r = (3/6) - (2/6) = 1/6.

So, friend B's rate is 1/6 of the pool per hour. This means it would take friend B 6 hours to fill the pool alone. We did it! This is a super common type of problem, and now you guys know how to solve it!

Key Takeaways: For word problems, the most important step is to identify what it says, and translate these english sentences to math sentences. It helps simplify the problem and can be easily solved.

Wrapping Up

Alright guys, we tackled some awesome math problems today! We solved for integers with specific inverse relationships and even figured out how long it takes to fill a pool with teamwork. The key takeaways are: break down the problem into smaller steps, identify the underlying concepts (like multiplicative inverses or work rates), and translate the words into mathematical equations. Once you've got the equation, you're well on your way to solving the problem!

Keep practicing, keep challenging yourselves, and most importantly, have fun with math! You've got this!