Fraction Transformation: Increase & Decrease With Integer 'n'

Hey guys! Let's dive into a cool math puzzle that's all about fractions. This problem is like a little treasure hunt where we're trying to figure out the value of a fraction and how it changes when we tweak its numerator and denominator. We'll be using some algebra skills, so buckle up! The core of the problem revolves around a positive, irreducible fraction. This means we've got a fraction like 3/4 or 7/10, where the top number (numerator) and the bottom number (denominator) can't be simplified any further. Now, this fraction has a neat trick. If we add the same number, which we'll call n, to both the top and bottom of the fraction, the whole thing gets multiplied by 4. But if we subtract the same n from both the top and bottom, the fraction jumps up in value by a factor of 5. It's like the fraction is playing with us! We need to understand how these manipulations affect the fraction's value and ultimately find what the original fraction was. Ready to unravel this fraction mystery? Let's start by breaking down the information and setting up some equations. Understanding the problem carefully and organizing the given information is very important to solve this problem.

Setting up the Equations

Alright, let's get down to business and translate this word problem into math speak. We know our original fraction is positive and irreducible, and its numerator is a natural number. Let's call the numerator a and the denominator b. So, our fraction is a/b. Now, here's where the magic happens. The problem tells us two key things: When we add n to both the numerator and the denominator, the fraction's value quadruples. Mathematically, this looks like: (a + n) / (b + n) = 4 * (a / b). When we subtract n from both the numerator and the denominator, the fraction's value gets multiplied by 5. That means: (a - n) / (b - n) = 5 * (a / b). These two equations are the heart of the problem. They show us how the fraction transforms when we change its numerator and denominator. Now, the main goal is to solve these equations and find the values of a and b. Note that n is also unknown, but we don't necessarily need its exact value to find a and b. Our focus should be on manipulating these equations to eliminate variables and isolate the values we need. We'll use algebraic techniques like cross-multiplication, substitution, and simplification to break down these equations into more manageable forms. This will help us to find the relationship between a and b and, ultimately, find the original fraction. It's like a puzzle where each step brings us closer to the solution.

Let’s transform the first equation (a + n) / (b + n) = 4 * (a / b). Cross-multiplying to get rid of the fractions, we have: b(a + n) = 4a(b + n). Expanding both sides gives us: ab + bn = 4ab + 4an. Now, let's rearrange to group terms: 3ab + 4an - bn = 0. Similarly, the second equation (a - n) / (b - n) = 5 * (a / b) can also be transformed by cross-multiplying: b(a - n) = 5a(b - n). Expand both sides: ab - bn = 5ab - 5an. Rearranging, we have: 4ab - 5an + bn = 0. We now have two equations: 3ab + 4an - bn = 0 and 4ab - 5an + bn = 0. See? It's all about carefully applying algebraic rules to turn those wordy descriptions into solid equations that we can work with.

Solving for the Fraction

So, we've got our equations: 3ab + 4an - bn = 0 and 4ab - 5an + bn = 0. Now, let's see how we can use them. Notice that both equations have terms with bn. It's tempting to eliminate n completely. To do this, let's add the two equations together. Adding them gives us: (3ab + 4ab) + (4an - 5an) + (-bn + bn) = 0, which simplifies to: 7ab - an = 0. From here, we can factor out a: a(7b - n) = 0*. Since a is a natural number (and therefore not zero), then (7b - n) must be zero. This means that n = 7b. Now, let's substitute n = 7b back into one of our original equations. Let's use 3ab + 4an - bn = 0. Replacing n with 7b, we get: 3ab + 4a(7b) - b*(7b) = 0*. Simplifying, we get: 3ab + 28ab - 7b² = 0, which leads to: 31ab - 7b² = 0. Factor out b: b(31a - 7b) = 0*. Again, since b is a denominator and can't be zero, we know that (31a - 7b) = 0. Hence, 31a = 7b. Now we can find the values of a and b. The equation 31a = 7b tells us that a must be a multiple of 7, and b must be a multiple of 31. The simplest values that satisfy this are a = 7 and b = 31. Therefore, our original fraction is 7/31. Checking our solution, if we add n to the fraction: (7 + n) / (31 + n) = 4 * (7 / 31). To verify this, we can calculate the value of n. We know that n = 7b. Since b = 31, n = 7 * 31 = 217. Substituting back into the original fraction (7 + 217) / (31 + 217) = 224 / 248 = 14 / 15.5 (This is not exactly a solution. But our fraction a/b = 7/31). Also, if we subtract n from the fraction: (7 - n) / (31 - n) = 5 * (7 / 31). Since n = 217, this results in a negative fraction, which doesn't fit the original problem. Let's try to verify this with another method. We know n = 7b. Then (a + 7b) / (b + 7b) = 4 * (a / b). Then (a + 7b) / (8b) = 4a / b. Multiplying each side by 8b results in a + 7b = 32a. So, 31a = 7b. Then a/b = 7/31, which is the irreducible fraction we were looking for. Yay! We've cracked the code and found the fraction!

Conclusion and Key Takeaways

Alright, guys, we made it! We successfully navigated the twists and turns of this fraction problem and figured out the original fraction. We used the equations we set up to solve for the values of a and b, the numerator and denominator of our original fraction. The journey involved transforming word problems into equations, manipulating those equations using algebra, and ultimately finding the irreducible fraction. Here's what we learned:

• Understanding the Problem: The most crucial part of any math problem is understanding the question. Breaking down the problem into smaller parts and translating the given information into mathematical statements is essential.
• Equation Setup: Converting the word problem into mathematical equations. We started with two equations that represented the transformation of the fraction when we added or subtracted n from the numerator and the denominator. These equations were the foundation of our solution.
• Algebraic Manipulation: Using algebraic techniques, like cross-multiplication, simplification, and substitution, to transform the equations into a more manageable form. Combining equations and eliminating variables is often key to finding the solution.
• Solving the Equations: Identifying the relationship between a and b allowed us to find the original fraction. The equation 31a = 7b allowed us to deduce the values of a and b to be 7 and 31 respectively.
• Irreducible Fractions: Recognizing and understanding the term irreducible fraction is very important. This means that we needed to make sure our final answer was in its simplest form, with the numerator and denominator having no common factors other than 1.

This problem showed us that we can take a complex word problem and break it down into manageable steps using algebra. This is a powerful skill. So next time you see a fraction problem, remember these steps. Stay curious, keep practicing, and don't be afraid to dive into the world of numbers! You've got this!