Solve 4x - Y = 8 For Y

by Andrew McMorgan 23 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of mathematics, and more specifically, we're going to tackle a common algebra problem that can trip some people up: solving an equation for a specific variable. We've got a cracking question here: Which equation is equivalent to 4xβˆ’y=84x - y = 8 when solved for yy? This might seem a bit tricky at first glance, but trust me, once you break it down step-by-step, it's totally manageable. We'll go through the process together, explaining each move so you can confidently solve similar problems in the future. So, grab your notebooks, get comfy, and let's get this algebra party started! We'll explore the logic behind isolating 'y' and why the other options just don't cut the mustard. Stick around, and by the end of this, you'll be a pro at manipulating equations!

Understanding the Goal: Isolating 'y'

Alright, let's get down to business. When we're asked to solve an equation for a specific variable, in this case, y, our main goal is to get that variable all by itself on one side of the equals sign. Think of it like trying to get your favorite snack all to yourself – you want it isolated from everything else! The original equation we're working with is 4xβˆ’y=84x - y = 8. To achieve our goal, we need to perform a series of operations that will move all the other terms (the '4x4x' and the '88') to the other side of the equation, leaving 'yy' alone. It's crucial to remember the golden rule of algebra: whatever you do to one side of the equation, you MUST do to the other side to maintain the balance. If you add something to the left, you have to add it to the right. If you subtract from the left, you subtract from the right, and so on. This ensures that the equality remains true. We'll start by looking at the term with 'yy' in it, which is 'βˆ’y-y'. Our ultimate aim is to have '+y+y' on one side, or to have 'yy' isolated. Let's think about how we can achieve this. We can either add 'yy' to both sides of the equation, or we can work with 'βˆ’y-y' and then multiply the entire equation by -1 at the end. Both methods are valid, but sometimes one is a bit more straightforward than the other depending on the specific problem. For this particular equation, adding 'yy' to both sides often feels like the most direct route to getting a positive 'yy' term, which is usually what we aim for when solving for 'yy'. So, let's mentally prepare ourselves for that move, keeping in mind the balance we need to maintain throughout the entire process. The options provided (A, B, C, and D) are all potential outcomes, and our job is to figure out which one is the mathematically correct result of correctly manipulating the original equation. It’s a bit like a puzzle, and we’re finding the missing piece!

Step-by-Step Solution

So, we've got our equation: 4xβˆ’y=84x - y = 8. Our mission, should we choose to accept it, is to get 'yy' by itself. The first thing we want to do is get the 'yy' term away from the '4x4x' term. Since '4x4x' is being added to the 'β€²βˆ’y'-y' term (or more accurately, 'β€²βˆ’y'-y' is on the same side as '4x4x'), we need to move the '4x4x' to the other side. The best way to do this is to subtract '4x4x' from both sides of the equation. This is a fundamental algebraic technique for isolating terms. So, here's what happens:

4xβˆ’yextbfβˆ’4x=8extbfβˆ’4x4x - y extbf{ - 4x} = 8 extbf{ - 4x}

On the left side, the '+4x+4x' and the 'βˆ’4x-4x' cancel each other out, leaving us with just 'β€²βˆ’y'-y'. On the right side, we simply have '8βˆ’4x8 - 4x'. So, after this first step, our equation looks like this:

βˆ’y=8βˆ’4x-y = 8 - 4x

Now, we're super close! We have 'β€²βˆ’y'-y' isolated, but we actually want '+y+y'. To change 'β€²βˆ’y'-y' into '+y+y', we need to multiply the entire equation by βˆ’1-1. Remember, multiplying by βˆ’1-1 is the same as changing the sign of every term. So, we'll multiply both sides by βˆ’1-1:

(βˆ’1)βˆ—(βˆ’y)=(βˆ’1)βˆ—(8βˆ’4x)(-1) * (-y) = (-1) * (8 - 4x)

On the left side, 'βˆ’1-1 times βˆ’y-y' gives us '+y+y'. On the right side, we need to distribute the βˆ’1-1 to both terms inside the parentheses: 'βˆ’1-1 times 88' is 'βˆ’8-8', and 'βˆ’1-1 times βˆ’4x-4x' is '+4x+4x'. So, the right side becomes βˆ’8+4x-8 + 4x.

Putting it all together, our final equation is:

y=βˆ’8+4xy = -8 + 4x

Now, let's reorder the terms on the right side to match a more standard form where the 'xx' term comes first. This is just a matter of convention and doesn't change the mathematical value. So, '4x4x' comes before 'βˆ’8-8'.

y=4xβˆ’8y = 4x - 8

And there you have it, guys! We've successfully solved the equation for 'yy' and found the equivalent equation. Now, let's compare this to the options given in the question to see which one matches our result. It's always a good idea to double-check your work, especially when multiple choice options are involved, to make sure you haven't made any silly mistakes along the way. This step-by-step breakdown should make it crystal clear how we arrived at our answer.

Evaluating the Options

Okay, mathletes, we've done the hard work and figured out that the equation equivalent to 4xβˆ’y=84x - y = 8 when solved for yy is y=4xβˆ’8y = 4x - 8. Now, let's look at the options provided and see which one matches our triumphant result. This is where we confirm our awesomeness!

  • A. y=4x+8y = 4x + 8: Does this match our result? Nope. We got '$ - 8β€²ontherightside,notβ€²' on the right side, not '+ 8
. This option likely comes from incorrectly adding '4x4x' to both sides instead of subtracting, or perhaps a sign error when moving terms.
  • B. y=βˆ’4x+8y = -4x + 8: This one's also a no-go. While it has a '+8+ 8', the 'xx' term is negative ('βˆ’4x-4x'). This could happen if someone forgot to multiply by βˆ’1-1 at the end, leaving us with 'β€²βˆ’y=8βˆ’4x'-y = 8 - 4x' and then incorrectly wrote the 'xx' term as positive.
  • C. y=8βˆ’4xy = 8 - 4x: This is almost right, but not quite the standard form we usually aim for. Remember, we got βˆ’y=8βˆ’4x-y = 8 - 4x after the first step. If we had stopped there and tried to match it, this might look appealing. However, when we multiplied by βˆ’1-1, we got y=βˆ’8+4xy = -8 + 4x, which is the same as y=4xβˆ’8y = 4x - 8. The order of terms matters for matching the exact option, and this option has the '88' term as positive instead of negative, and the '4x4x' term as negative instead of positive. This option is actually equivalent to βˆ’y=βˆ’8+4x-y = -8 + 4x if we were to rearrange it and make βˆ’y-y positive. Wait, let's re-evaluate. The option C is y=8βˆ’4xy = 8 - 4x. Our intermediate step was βˆ’y=8βˆ’4x-y = 8 - 4x. If we multiply both sides by -1, we get y=βˆ’(8βˆ’4x)y = -(8 - 4x), which is y=βˆ’8+4xy = -8 + 4x. So, option C is not our answer. It seems I might have confused myself for a second there, which is why double-checking is key, guys!
  • D. y=4xβˆ’8y = 4x - 8: Bingo! This matches our calculated result perfectly. We successfully isolated 'yy' and ended up with 'y=4xβˆ’8y = 4x - 8'. This is the correct equivalent equation.
  • So, after carefully working through the algebra and checking each option, we can confidently say that option D is the correct answer. It’s always satisfying when your calculated result lines up perfectly with one of the choices, right? It means you’ve nailed the process!

    Why the Other Options Are Incorrect

    Let's take a moment to really hammer home why the other options just don't work out. Understanding these common pitfalls can be super helpful for avoiding mistakes in the future. When you're solving equations, little sign errors can snowball into completely wrong answers, and that's exactly what happens with options A, B, and C.

    Option A, y=4x+8y = 4x + 8, suggests that '+8+8' should be on the right side. If we look back at our original equation, 4xβˆ’y=84x - y = 8, the '88' is positive. To get 'yy' by itself, we first subtracted '4x4x' from both sides, resulting in βˆ’y=8βˆ’4x-y = 8 - 4x. If we incorrectly decided to add '4x4x' to both sides instead of subtracting, we would have gotten 8xβˆ’y=8+4x8x - y = 8 + 4x, which is way off track. Or, perhaps someone might have gotten to βˆ’y=8βˆ’4x-y = 8 - 4x and then decided to add '88' to both sides to make it positive, which isn't a valid algebraic step to isolate 'yy'. The most common way people might arrive at y=4x+8y = 4x + 8 is if they incorrectly add '4x4x' to both sides and then somehow end up with a positive '88'. It really highlights the importance of performing the correct inverse operation. Subtracting '4x4x' is the inverse of adding '4x4x', and that's what we needed to do to move the '4x4x' term.

    Option B, y=βˆ’4x+8y = -4x + 8, is another common mistake. This option looks like someone might have correctly subtracted '4x4x' from both sides to get βˆ’y=8βˆ’4x-y = 8 - 4x, but then forgot to multiply the entire equation by βˆ’1-1 to make 'yy' positive. If you leave it as βˆ’y=8βˆ’4x-y = 8 - 4x, it's not solved for 'yy'. If you then try to force it into a 'y=...y = ...' format without the proper multiplication by βˆ’1-1, you might end up with something like this, perhaps by just rearranging the terms of βˆ’y=8βˆ’4x-y = 8 - 4x into βˆ’y=βˆ’4x+8-y = -4x + 8 and then somehow flipping the sign of '4x4x' but not '88'. This is a classic sign error that happens when you're rushing through the final steps. Remember, the goal is '+y+y', not 'β€²βˆ’y'-y'.

    Option C, y=8βˆ’4xy = 8 - 4x, is so close it's almost sneaky! As we saw in our step-by-step, we reached an intermediate stage of βˆ’y=8βˆ’4x-y = 8 - 4x. Now, if we were to multiply both sides by βˆ’1-1, we get y=βˆ’(8βˆ’4x)y = -(8 - 4x), which expands to y=βˆ’8+4xy = -8 + 4x. If we reorder this, we get y=4xβˆ’8y = 4x - 8. Option C is y=8βˆ’4xy = 8 - 4x. These are not the same! The signs on the '88' and the '4x4x' are flipped. It suggests that maybe someone correctly isolated βˆ’y-y as βˆ’y=8βˆ’4x-y = 8 - 4x and then stopped, or perhaps made a sign error when they were supposed to multiply by βˆ’1-1. For example, if they thought βˆ’y=8βˆ’4x-y = 8 - 4x meant y=8βˆ’4xy = 8 - 4x, they'd be wrong because the βˆ’y-y requires a full sign flip of everything on the other side. The '88' should become negative, and the 'β€²βˆ’4x'-4x' should become positive. So, option C demonstrates a misunderstanding of how to properly handle the negative sign on the 'yy' variable.

    Ultimately, correctly solving for 'yy' requires careful attention to each step: applying inverse operations, maintaining balance, and correctly handling negative signs. Option D, y=4xβˆ’8y = 4x - 8, is the only one that results from applying these rules accurately to the original equation 4xβˆ’y=84x - y = 8. So, keep those minus signs in check, folks!

    Conclusion: Mastering Equation Manipulation

    So there you have it, amazing readers of Plastik Magazine! We've journeyed through the algebraic landscape, starting with the equation 4xβˆ’y=84x - y = 8 and successfully transformed it into its equivalent form when solved for 'yy'. The key takeaway here is the systematic approach: identify the variable you need to isolate, use inverse operations to move other terms to the opposite side while maintaining the equation's balance, and be extra vigilant with signs, especially negative ones. We saw how options A, B, and C represent common mistakes, often stemming from sign errors or incorrect application of inverse operations. But by diligently following the steps – subtracting '4x4x' from both sides to get βˆ’y=8βˆ’4x-y = 8 - 4x, and then multiplying the entire equation by βˆ’1-1 to finally arrive at y=4xβˆ’8y = 4x - 8 – we confirmed that option D is the only mathematically sound answer.

    Mastering this skill of solving for a variable is fundamental in algebra and opens the door to understanding more complex mathematical concepts. It's like learning to ride a bike; once you get the hang of it, you can go anywhere! Practice is your best friend here. Try solving other equations for different variables. Rearrange formulas, work through textbook examples, and don't be afraid to make mistakes – they are simply learning opportunities in disguise. Each time you work through a problem, you become a little more confident and a little more skilled. Remember that friendly tone we talked about? Keep that positive attitude! Math can be challenging, but it's also incredibly rewarding when you crack a tough problem. So, keep practicing, keep questioning, and keep exploring the fascinating world of numbers with us here at Plastik Magazine. Until next time, happy solving!