Solve (2y-3)(y+4)-8y=10: A Friendly Guide

Hey there, math enthusiasts and curious minds! Ever looked at an equation like (2y - 3)(y + 4) - 8y = 10 and thought, "Whoa, where do I even begin?" Well, you're in luck because today, we're going to break down this beast of an equation step-by-step, making it super easy to understand. This isn't just about finding 'y'; it's about building your confidence in tackling algebraic expressions and showing you how powerful these tools can be. We're going to dive deep into solving this specific problem, but also give you the foundational knowledge that will help you conquer any similar challenge you face. So grab a comfy seat, maybe a snack, and let's unravel the mystery of this equation together. This guide is designed for you, whether you're just starting your algebra journey or need a quick refresher. We'll cover everything from expanding terms to isolating variables, all while keeping things light and friendly. Let's make algebra fun, shall we? Solving equations like (2y - 3)(y + 4) - 8y = 10 is a fundamental skill in algebra, opening doors to understanding more complex mathematical concepts. It’s not just about memorizing steps; it’s about understanding the logic behind each transformation. By the end of this article, you'll not only know how to solve this particular equation but also have a clearer picture of the algebraic principles involved. We'll explore topics such as the distributive property, combining like terms, and isolating variables, all crucial steps in arriving at the correct solution. Remember, everyone learns at their own pace, and it's totally okay to pause, re-read, or even grab a scratch pad and work through the examples alongside us. Our goal here is to demystify algebra and turn what might look like a daunting problem into a clear, manageable sequence of steps. Ready to become an algebra rockstar? Let's dive right into understanding our equation and then systematically dismantle it piece by piece to find that elusive 'y'.

Unpacking the Equation: What Are We Dealing With?

Alright, folks, before we jump into solving the equation (2y - 3)(y + 4) - 8y = 10, let's take a moment to really look at it. What exactly are we working with here? Understanding the components is half the battle, trust me. First off, we've got something called binomials – those expressions with two terms inside parentheses, like (2y - 3) and (y + 4). These two binomials are being multiplied together, which is often the first hurdle for many students. This multiplication step is key to simplifying the entire expression. Then, we have a subtraction of another term, -8y, which is a linear term because 'y' is raised to the power of one. Finally, everything on the left side of the equals sign (=) has to balance out with the number on the right side, which is 10. So, our mission, should we choose to accept it (and we do!), is to find the value of 'y' that makes this entire statement true. Understanding these parts is crucial for effective problem-solving. We're essentially dealing with a mix of multiplication and subtraction, all leading to a specific numerical outcome. Many beginners rush into solving without fully appreciating the structure, which can lead to mistakes down the line. Take a moment to mentally break down the equation: we have a product of two expressions, followed by a simple subtraction, and the whole thing equals a constant. This structure tells us a lot about the order of operations we'll need to follow. We need to handle the multiplication first, then the subtraction, and finally, we'll start isolating our variable. It’s a bit like assembling IKEA furniture; you need to follow the instructions step-by-step, understanding what each component does before moving on. By thoroughly unpacking the equation, we set ourselves up for success, ensuring we don't overlook any crucial details or make premature simplifications. This analytical approach is fundamental not just for this problem, but for mastering any algebraic challenge you encounter. So, now that we've properly introduced ourselves to the problem, let's get down to business and start simplifying this expression!

Step 1: Expand and Conquer the Parentheses (FOIL Method)

Okay, guys, the very first thing we need to tackle when solving an equation like (2y - 3)(y + 4) - 8y = 10 is that pesky multiplication of the two binomials: (2y - 3)(y + 4). This is where the FOIL method comes in super handy! FOIL is a mnemonic that helps us remember the steps for multiplying two binomials: First, Outer, Inner, Last. Let's break it down.

First Terms

First, we multiply the first terms of each binomial. In (2y - 3)(y + 4), the first terms are 2y and y. Multiplying them gives us 2y * y = 2y². This is our first piece of the expanded puzzle. Remember, when you multiply variables, you add their exponents. Here, y * y = y to the power of (1+1) = y². Don't forget that exponent! This step is crucial for transforming our equation into a quadratic form, meaning it will have a y² term.

Outer Terms

Next up, we multiply the outer terms. These are the terms on the very outside of our expression: 2y from the first binomial and +4 from the second. So, 2y * 4 = 8y. Simple enough, right? This contributes another linear term to our expanding expression. Keeping track of the signs here is also very important; since both are positive, our result is positive.

Inner Terms

Now, let's look at the inner terms. These are the two terms closest to each other: -3 from the first binomial and y from the second. Multiplying these, we get -3 * y = -3y. Notice the negative sign stays with the 3. This is a common place for small errors, so always be mindful of the signs, folks! This also adds a linear term, but this time, it's a negative one.

Last Terms

Finally, we multiply the last terms of each binomial. That's -3 from the first and +4 from the second. Multiplying these gives us -3 * 4 = -12. Just like with the inner terms, the negative sign is important here. So, our constant term from the expansion is -12.

Combining the Expanded Terms

Now that we've done all the FOIL steps, we put them all together: 2y² + 8y - 3y - 12. This is the expanded form of (2y - 3)(y + 4). Pretty neat, huh? But remember, we still have the -8y = 10 part of our original equation. So, the equation now looks like this: 2y² + 8y - 3y - 12 - 8y = 10. This entire expansion process is fundamental in algebra, as it allows us to convert products of binomials into a more manageable polynomial form. Without correctly applying the FOIL method or the distributive property, the rest of your calculations will be incorrect. This step truly sets the stage for the subsequent simplifications. It requires careful attention to detail, especially with the signs of each term as you multiply. Many students find it helpful to draw "rainbow" arcs connecting the terms they are multiplying to ensure they don't miss any pair. Once you have all four terms from the FOIL expansion, the next step will be to bring them together with any other terms from the original equation. We're making great progress towards solving for y!

Step 2: Simplify by Combining Like Terms

Alright, team, we've successfully expanded our binomials in (2y - 3)(y + 4) - 8y = 10 to get 2y² + 8y - 3y - 12 - 8y = 10. Now, the next crucial step is to simplify this expression by combining all the like terms. What are like terms, you ask? They are terms that have the exact same variable parts raised to the exact same powers. Think of it like sorting different types of fruit: you can add apples to apples, but you can't add apples to oranges directly. Here, we've got a y² term, several y terms (linear terms), and a constant term. Let's gather them up!

First, let's look for any other y² terms. In our current equation, 2y² + 8y - 3y - 12 - 8y = 10, we only have one y² term: 2y². So, that one stands alone for now. Easy peasy.

Next, let's identify all the y terms. We have +8y, -3y, and -8y. Now, we just combine their coefficients (the numbers in front of the variables). So, 8 - 3 - 8. Let's do that arithmetic: 8 - 3 = 5, and then 5 - 8 = -3. This means all our 'y' terms combine to -3y. See how that works? It’s about careful addition and subtraction, paying close attention to the positive and negative signs. Many errors happen right here if signs are overlooked. Always double-check your arithmetic, guys, it's super important!

Finally, we have our constant term. These are the numbers without any variables attached. In our expanded expression, we only have -12. So, that term also stands alone.

Now, let's put it all back together with the combined like terms. Our simplified equation on the left side becomes: 2y² - 3y - 12. Don't forget the right side of the equation! So, our entire equation now looks much cleaner: 2y² - 3y - 12 = 10. Isn't that a breath of fresh air? This simplification step is vital because it reduces the complexity of the equation, making it much easier to manage for the subsequent steps. If you skip this, or do it incorrectly, you'll be carrying extra, uncombined terms, leading to a much harder and likely incorrect solution. Properly combining like terms transforms a messy expression into a neat, standard quadratic form, which is the cornerstone for applying further algebraic techniques. This entire process relies on your understanding of signed numbers and basic arithmetic, reinforcing why foundational math skills are so crucial. Keep up the great work, we're making excellent progress in solving for y!

Step 3: Isolate the Variable – Getting 'y' All By Itself

Alright, rockstars, we've successfully brought our equation down to a much more manageable form: 2y² - 3y - 12 = 10. Now, our goal is to isolate the variable 'y'. However, because we have a y² term, this isn't going to be a simple linear equation where we just move numbers around and divide. This is a quadratic equation, which usually means we need to set it equal to zero and then use techniques like factoring, completing the square, or the quadratic formula. Let's get it into standard form for a quadratic equation, which is ax² + bx + c = 0.

Moving Constants to One Side

First things first, we want all the terms on one side of the equation, with zero on the other. Currently, we have -12 on the left side and 10 on the right. To move the 10 from the right side to the left, we need to perform the inverse operation. Since it's a positive 10, we'll subtract 10 from both sides of the equation to maintain balance. Remember, whatever you do to one side, you must do to the other!

So, we have: 2y² - 3y - 12 - 10 = 10 - 10 2y² - 3y - 22 = 0

Voila! We now have our quadratic equation in standard form: 2y² - 3y - 22 = 0. Now, we're ready to actually solve for 'y'. Since this is a quadratic equation, there are a few common methods we can use: factoring, completing the square, or the quadratic formula. Factoring is often the quickest if it's possible. Let's check if we can factor this. We need two numbers that multiply to (2 * -22 = -44) and add to -3. After a little thought, it seems like -11 and 4 fit the bill (-11 * 4 = -44, and -11 + 4 = -7, which is not -3). Hmm, it looks like simple factoring might not work directly or easily. In such cases, the quadratic formula is our best friend! It always works, no matter how complex the numbers are.

Using the Quadratic Formula

The quadratic formula is a magical tool that gives us the solutions for 'x' (or 'y' in our case) in any quadratic equation ax² + bx + c = 0. The formula is:

y = [-b ± sqrt(b² - 4ac)] / 2a

In our equation, 2y² - 3y - 22 = 0:

• a = 2
• b = -3
• c = -22

Let's plug these values into the formula:

y = [-(-3) ± sqrt((-3)² - 4 * 2 * -22)] / (2 * 2) y = [3 ± sqrt(9 - (-176))] / 4 y = [3 ± sqrt(9 + 176)] / 4 y = [3 ± sqrt(185)] / 4

Since 185 is not a perfect square, our solutions for 'y' will be irrational (meaning they can't be expressed as simple fractions). We'll have two solutions:

y1 = (3 + sqrt(185)) / 4 y2 = (3 - sqrt(185)) / 4

These are our exact solutions! Sometimes, math gives us neat whole numbers, and other times, we get these more complex expressions, but they are just as valid. The process of isolating the variable often depends on the type of equation. For linear equations, it's simpler, but for quadratics, understanding how to apply the quadratic formula is an indispensable skill. It's the ultimate backup plan when factoring seems impossible. This entire step of transforming the equation and then correctly applying the quadratic formula demonstrates a deep understanding of algebraic manipulation. You've done a fantastic job getting this far, folks! We’ve successfully found the values of 'y' that satisfy our original equation, even if they look a bit intimidating. Don't worry, we're almost at the finish line!

Step 4: Verify Your Solution (The Ultimate Check!)

Alright, amazing mathematicians, we've done all the hard work and found our solutions for 'y': y1 = (3 + sqrt(185)) / 4 and y2 = (3 - sqrt(185)) / 4. Now, how do we know if we're right? The absolute best way to confirm your answer in algebra is to verify your solution by plugging it back into the original equation (2y - 3)(y + 4) - 8y = 10. This step is like being your own math detective, checking your work and ensuring everything balances out. While plugging in exact irrational numbers can be a bit tedious for verification (especially if you don't have a calculator handy that handles square roots precisely), the principle is what's important here. For tests or when dealing with simpler solutions, this is a non-negotiable step.

Let's consider how we would verify. If we had found a simpler, integer solution, say y=2, we would plug 2 into every 'y' in the original equation:

(2(2) - 3)(2 + 4) - 8(2) = 10 (4 - 3)(6) - 16 = 10 (1)(6) - 16 = 10 6 - 16 = 10 -10 = 10

Since -10 = 10 is false, we would know that y=2 is not a solution. This hypothetical example illustrates why verification is so crucial. It immediately flags errors. For our actual solutions involving sqrt(185), the verification process would involve calculating the decimal approximations or carefully substituting the exact forms back into the equation. For example, if we approximate sqrt(185) as about 13.60, then:

y1 approx (3 + 13.60) / 4 = 16.60 / 4 = 4.15 y2 approx (3 - 13.60) / 4 = -10.60 / 4 = -2.65

Then, we'd take one of these approximations, say y ≈ 4.15, and plug it back into (2y - 3)(y + 4) - 8y = 10:

(2 * 4.15 - 3)(4.15 + 4) - 8 * 4.15 (8.30 - 3)(8.15) - 33.20 (5.30)(8.15) - 33.20 43.195 - 33.20 9.995

This 9.995 is very close to 10, indicating our approximation is quite good, and thus our exact solution is correct. The slight difference is due to rounding sqrt(185). Verification is your personal quality control step. It’s where you solidify your understanding and ensure that all your hard work hasn't gone astray. While it might seem like an extra step, it saves you from handing in incorrect answers and reinforces the principles of algebraic equality. So, whenever you solve an equation, especially a complex one, make sure to take that extra minute to verify your solution. It's a hallmark of a meticulous and confident problem-solver!

Common Pitfalls and How to Avoid Them

Alright, folks, you've conquered a pretty complex equation: (2y - 3)(y + 4) - 8y = 10! But here's the deal: even the pros make mistakes sometimes, especially in algebra. So, let's chat about some common pitfalls when solving equations like this, and more importantly, how you can avoid them to keep your math journey smooth sailing. Being aware of these traps is half the battle, trust me!

One of the biggest blunders we often see is with the distributive property, specifically when expanding the binomials using FOIL. People sometimes forget to multiply every term in the first binomial by every term in the second. Forgetting an "inner" or "outer" term, or mixing up the signs, is super common. For example, in (2y - 3)(y + 4), if you forget that (-3) is negative, you might end up with +3y instead of -3y, or +12 instead of -12. Always double-check your signs! A neat trick is to verbally say "first, outer, inner, last" as you write each term down, ensuring all four multiplications are accounted for, paying extra attention to the signs involved. Also, remember that a negative times a negative equals a positive, and a negative times a positive equals a negative. Simple rules, but easy to overlook in the heat of the moment.

Another frequent mistake happens during the combining like terms step. Students might incorrectly combine terms that aren't truly "alike." For instance, trying to add 2y² and -3y would be a no-go because they have different powers of 'y'. Only terms with the exact same variable and exponent can be combined. Also, miscalculating the coefficients when combining can happen, especially with negative numbers. For example, 8y - 3y - 8y could easily be mistaken as +13y if you're not careful with the subtractions. Take your time, draw lines under like terms to group them visually, and perform the arithmetic carefully. It sounds basic, but precision here is paramount.

Then there's the classic sign error when moving terms across the equals sign. Remember, when you move a term from one side of the equation to the other, you must change its sign. If it was +10 on the right, it becomes -10 when you bring it to the left. Forgetting to do this, or incorrectly changing the sign, will throw your entire solution off. This is where the principle of inverse operations comes into play; addition undoes subtraction, multiplication undoes division. Apply the opposite operation to both sides to cancel terms out correctly.

Finally, for quadratic equations like ours, misapplying the quadratic formula is a common pitfall. People might mix up 'b' and 'c', forget the +/- part, or make arithmetic errors under the square root (the discriminant). Always write down the formula first, then clearly identify your 'a', 'b', and 'c' values, including their signs, before plugging them in. Then, proceed step-by-step with the arithmetic. Don't try to do too many steps in your head. Write it out! By being mindful of these common algebraic pitfalls, you're not just solving equations; you're building a robust foundation for all future math challenges. Practice makes perfect, and recognizing these traps will make you a much more efficient and accurate problem-solver!

Why Bother with Equations Like This? Real-World Connections

Okay, guys, after all that hard work solving the equation (2y - 3)(y + 4) - 8y = 10, you might be thinking, "This is great and all, but when am I ever going to use this in real life?" That's a totally fair question! The truth is, while you might not directly see an equation like (2y - 3)(y + 4) - 8y = 10 popping up on your grocery list, the principles and skills you've just honed are absolutely foundational to countless real-world applications. Algebra isn't just a subject in school; it's a language used to describe the world around us, from the smallest atoms to the vastness of space. Understanding how to manipulate and solve equations, especially quadratic ones like the one we just tackled, is a superpower in many fields.

Think about engineering. Whether it's designing a bridge, a circuit board, or the aerodynamics of a new car, engineers constantly use algebraic equations to model forces, calculate stresses, and predict performance. The path of a projectile, for example, is often described by a quadratic equation. If you want to know how far a ball will travel or how high a rocket will go, you're essentially solving for 'y' in a more complex version of what we just did. Architects use these principles to ensure buildings are stable and aesthetically pleasing, calculating areas, volumes, and structural loads that often involve quadratic relationships. The ability to solve for an unknown variable in such scenarios is critical for safety and efficiency.

In finance and economics, algebra is indispensable. Financial models, interest calculations, investment growth, and even pricing strategies often involve solving equations. When you calculate compound interest or project future earnings, you're frequently dealing with variables and their relationships, which can often lead to quadratic or exponential equations. Businesses use these tools to optimize profits, analyze market trends, and make informed decisions. Economists use algebraic models to understand supply and demand, predict economic growth, and analyze policy impacts. Understanding how different factors relate and influence outcomes is at the heart of these fields, all powered by algebra.

Physics and chemistry are practically built on algebraic foundations. Every formula you encounter, from calculating velocity and acceleration to understanding chemical reactions and energy transfer, is an algebraic equation. When scientists are trying to find an unknown quantity, whether it's the mass of a substance or the time it takes for an object to fall, they are engaged in the same problem-solving process we just used. Think about designing a roller coaster; the loops and drops require precise calculations of potential and kinetic energy, often leading to quadratic equations to ensure riders have a thrilling yet safe experience.

Even in computer science and data analysis, algebra plays a huge role. Algorithms that power everything from search engines to artificial intelligence rely heavily on mathematical logic and algebraic operations. When you're optimizing code or analyzing large datasets, you're often working with systems of equations, matrices, and functions where finding unknown values is essential. Being comfortable with algebraic manipulation allows you to understand the underlying mechanics of these complex systems.

So, while (2y - 3)(y + 4) - 8y = 10 might look abstract, the skills you developed to solve it – systematic thinking, breaking down complex problems, careful calculation, and verifying your results – are universally valuable. These are the superpowers that enable innovation, discovery, and problem-solving across nearly every scientific, technological, and economic endeavor. You're not just learning math; you're learning how to think critically and solve real-world puzzles! Keep that enthusiasm going, because algebra is truly a gateway to understanding and shaping our world.

Wrapping Up: Your Journey to Algebraic Mastery!

Wow, what a journey we've had together, folks! From first encountering the seemingly complex equation (2y - 3)(y + 4) - 8y = 10 to systematically breaking it down and finding its solutions, you've demonstrated an amazing ability to tackle algebraic challenges head-on. Give yourselves a pat on the back! We navigated through several key algebraic concepts, including the distributive property (using the FOIL method), combining like terms, transforming the equation into its standard quadratic form, and finally, applying the quadratic formula to find our exact answers. We even touched upon the critical step of verifying your solutions and discussed some common pitfalls to help you avoid future headaches. This entire process wasn't just about getting two numbers; it was about building a robust framework for algebraic problem-solving that you can apply to countless other equations.

Remember, mastering algebra isn't a one-time event; it's a continuous journey of practice and understanding. Each equation you solve, each concept you grasp, adds another tool to your mathematical toolkit. Don't get discouraged if a problem seems tough at first. Break it down, identify the components, and apply the rules step-by-step, just like we did today. Persistence is key! The more you practice, the more intuitive these steps will become, and soon you'll be eyeing equations like (2y - 3)(y + 4) - 8y = 10 with confidence, knowing exactly how to approach them.

We also explored why these skills matter beyond the classroom, connecting algebra to real-world applications in engineering, finance, physics, and computer science. Hopefully, this gave you a deeper appreciation for the power and relevance of mathematics in our daily lives and various professional fields. The ability to think analytically, logically, and systematically—skills that are heavily honed through algebra—are invaluable in any career path you choose.

So, what's next for your algebraic adventure? Keep practicing! Look for similar problems, try variations, and don't be afraid to challenge yourself with new types of equations. Resources are abundant online, in textbooks, and through tutors. The more you engage with these concepts, the stronger your understanding will become. And always, always remember to check your work! That verification step is your ultimate safety net and a sign of true algebraic mastery.

Thank you for joining us on this deep dive into solving (2y - 3)(y + 4) - 8y = 10. We hope this guide was clear, helpful, and maybe even a little bit fun! Keep learning, keep exploring, and keep rocking those equations. You've got this, future math whizzes! Your journey to algebraic mastery is well underway. See you next time for more mathematical adventures!