Hey guys! Ever get stumped by those word problems in math? You know, the ones that sound like riddles? Well, you're not alone! Let's break down a couple of those tricky problems today and turn them into something we can totally handle. We'll go step-by-step, making sure everything makes sense along the way. So grab your thinking caps, and let's dive into these number mysteries!
Unraveling the Mystery of Two Numbers
Problem (iii) One number is 3 more than the other, and the larger number is equal to their product. What are the numbers?
Okay, let's tackle this problem piece by piece. Our main goal here is to find two numbers that fit the clues we're given. The first clue is that “one number is 3 more than the other.” This tells us we're dealing with two numbers that are related, but not the same. To make things easier, we can use algebra. Let's call the smaller number "x." Since the other number is 3 more than the smaller one, we can call it "x + 3." This is where translating words into math symbols starts to make things clearer, guys! We've got our two numbers represented algebraically, which is a huge step.
Now for the second clue: “the larger number is equal to their product.” Remember, the larger number is "x + 3." The word "product" means we need to multiply the two numbers together. So, the product of our numbers is "x * (x + 3)." The clue tells us the larger number (x + 3) is equal to this product. In math terms, that's an equation: x + 3 = x * (x + 3). This is the key equation we need to solve. It looks a little intimidating, but don't worry, we've got this!
To solve it, we need to do some algebra magic. First, let's expand the right side of the equation: x + 3 = x^2 + 3x. Now, we've got a quadratic equation – an equation where the highest power of x is 2. To solve these, we usually want to get everything on one side and set it equal to zero. So, let's subtract "x" and "3" from both sides: 0 = x^2 + 2x - 3. See? It's becoming more manageable already! Next up is factoring. Factoring is like reverse multiplication – we're trying to find two expressions that multiply together to give us our quadratic. In this case, we're looking for two numbers that multiply to -3 and add to 2. Those numbers are 3 and -1. So, we can factor our equation as: 0 = (x + 3)(x - 1). Now we're in the home stretch! For the product of two things to be zero, at least one of them has to be zero. So, either x + 3 = 0 or x - 1 = 0. Solving these gives us two possible values for x: x = -3 or x = 1. These are our potential smaller numbers.
But wait, we're not done yet! We need to find the other number in each case. Remember, the other number is x + 3. If x = -3, then the other number is -3 + 3 = 0. So, one possible pair of numbers is -3 and 0. Let's check if they work. Is 0 equal to the product of -3 and 0? Yes, because -3 * 0 = 0. And is 0 three more than -3? Yes! This pair works. Now let’s check the other value. If x = 1, then the other number is 1 + 3 = 4. So, our second possible pair is 1 and 4. Is 4 equal to the product of 1 and 4? Yes, because 1 * 4 = 4. And is 4 three more than 1? Yes! This pair works too. So, we've found two sets of numbers that satisfy the conditions: -3 and 0, and 1 and 4. That's the power of breaking down a problem and tackling it step-by-step! We took a confusing word problem and turned it into a solvable equation, then we found our solutions. Nice work, guys!
Cracking the Code of Consecutive Even Integers
Problem (iv) There are three consecutive even integers such that twice their sum equals the product of the first and the second numbers. What are the numbers?
Alright, let's jump into another number puzzle! This time, we're dealing with consecutive even integers, which adds a fun little twist. The key here is understanding what “consecutive even integers” means. Think of it like this: they're even numbers that follow each other in order, like 2, 4, 6, or 10, 12, 14. The gap between each number is always 2. This pattern is crucial for setting up our problem. To represent these numbers algebraically, we can start with our first integer as "x." Since the next even integer is 2 more than the previous one, the second integer will be "x + 2," and the third will be "x + 4." We've now got our three mystery numbers represented in a way we can work with. That's the first hurdle cleared!
Now let's look at the heart of the problem: “twice their sum equals the product of the first and the second numbers.” Sounds complicated, right? Let's break it down. First, we need to find “their sum.” That means adding our three integers together: x + (x + 2) + (x + 4). This simplifies to 3x + 6. Next, we need “twice their sum,” so we multiply this by 2: 2 * (3x + 6), which equals 6x + 12. That's the left side of our equation. On the other side, we have “the product of the first and the second numbers.” That's x * (x + 2), which expands to x^2 + 2x. Now we can put it all together! The problem tells us that “twice their sum equals the product,” so we have the equation: 6x + 12 = x^2 + 2x. See how we've turned a sentence into an equation? That's the magic of algebra!
Now it's solving time! We've got a quadratic equation again, so let's get everything on one side and set it to zero. Subtract 6x and 12 from both sides: 0 = x^2 - 4x - 12. Time for some factoring! We need two numbers that multiply to -12 and add up to -4. Those numbers are -6 and 2. So, we can factor our equation as: 0 = (x - 6)(x + 2). Remember, for the product to be zero, at least one of the factors has to be zero. So, either x - 6 = 0 or x + 2 = 0. This gives us two possible values for x: x = 6 or x = -2. These are our potential first even integers. But we're not done until we find the other two integers in each set.
If x = 6, our first even integer is 6. The next two are x + 2 = 8 and x + 4 = 10. So, one possible set of consecutive even integers is 6, 8, and 10. Let’s check if this set works. Twice their sum is 2 * (6 + 8 + 10) = 2 * 24 = 48. The product of the first and second is 6 * 8 = 48. It checks out! Now let's check our other value. If x = -2, our first even integer is -2. The next two are x + 2 = 0 and x + 4 = 2. So, our second possible set is -2, 0, and 2. Let's see if this set works. Twice their sum is 2 * (-2 + 0 + 2) = 2 * 0 = 0. The product of the first and second is -2 * 0 = 0. This set works too! We've successfully found two sets of consecutive even integers that fit the problem: 6, 8, 10 and -2, 0, 2. How awesome is that? We've conquered another tricky math problem by breaking it down, translating the words into equations, and solving those equations step-by-step. You guys are math superstars!
Key Strategies for Tackling Word Problems
So, what did we learn today? More than just the answers to these specific problems. We learned a strategy for approaching those intimidating word problems. Here’s the secret sauce, guys:
- Read Carefully: Don't just skim the problem. Read it slowly and make sure you understand exactly what it's asking. Underlining key information can be super helpful.
- Translate into Algebra: This is where the magic happens. Turn the words into mathematical expressions and equations. Identify the unknowns (what you're trying to find) and represent them with variables like "x" or "y."
- Break it Down: Complex problems can feel overwhelming. Divide the problem into smaller, more manageable steps. Focus on one part at a time.
- Solve the Equation: Once you have an equation, use your algebra skills to solve for the unknown variables. Remember the order of operations and all those cool techniques you've learned!
- Check Your Answer: This is crucial! Does your answer make sense in the context of the problem? Plug your answer back into the original problem and make sure it fits all the conditions. There is no shame to double check! You might even find you have more than 1 answer as we found out in the problems above!
By following these strategies, you can turn those scary word problems into exciting puzzles to be solved. Math can be fun, guys, especially when you crack the code! Keep practicing, and you'll become a word problem master in no time!