Solve 2x+5y=-13 & 3x-4y=-8: Easy Steps

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Solve 2x+5y=-13 & 3x-4y=-8: Easy Steps\n\n## Unlocking the Secrets of Systems of Linear Equations\n\nHey there, math enthusiasts! Ever found yourself staring at a couple of equations, wondering how on earth to find the values that make *both* of them true? Well, you've just encountered a **system of linear equations**, and trust me, they're super common in the real world, not just in textbooks! Today, we're diving deep into _solving two-variable equations_ like the ones you've probably seen: `2x + 5y = -13` and `3x - 4y = -8`. These aren't just random numbers; they represent situations where two different conditions or relationships must hold true simultaneously. Think of it like a puzzle where you need to find the perfect `x` and `y` values that satisfy *every* rule given. We'll explore powerful strategies to tackle these _simultaneous equations_ head-on, giving you the confidence to conquer similar _mathematical problems_. Our goal is to make these _equation basics_ crystal clear, transforming what might seem daunting into a straightforward, almost fun, process. Whether you're a student looking to ace your next algebra test or just someone curious about the practical side of math, understanding how to solve these systems is an incredibly valuable skill. We're going to break down the process step-by-step, using a friendly, conversational tone so you can truly grasp the logic and magic behind it all. So, grab your favorite beverage, maybe a snack, and let's get ready to become _systems of linear equations_ pros! We'll show you exactly how to approach these types of challenges, ensuring you walk away with a solid understanding and the tools to solve any system thrown your way. This isn't just about getting the right answer; it's about understanding *why* the answer is right and building a strong foundation in algebraic _equation basics_. Prepare to demystify these mathematical puzzles with us!\n\n## Why Mastering These Equations is Super Important (Real-World Value!)\n\nNow, you might be thinking, "Why do I even need to learn how to solve these _systems of linear equations_? Are they just for math class?" Oh, but my friends, nothing could be further from the truth! _Real-world applications of math_ are everywhere, and solving these kinds of _practical problem-solving_ scenarios is a fundamental skill that goes far beyond the classroom. Imagine you're running a small business. You might have one equation representing your revenue based on the number of products sold (`x`) and services offered (`y`), and another equation showing your costs. To find your break-even point, or to maximize profits, you'd be solving a system of equations! Or consider a chemist mixing solutions; they need to balance quantities and concentrations, leading directly to _mathematics in daily life_. Engineers use them to design bridges and circuits, economists model market trends, and even everyday budgeting can involve balancing income and expenses from different sources. Learning to manipulate these equations sharpens your _critical thinking skills_ and enhances your ability to approach complex problems logically. It teaches you to break down big challenges into smaller, manageable parts, a skill that's invaluable in *any* field. From determining the best investment strategies in finance to optimizing delivery routes in logistics, the principles of _solving two-variable equations_ are at play. It's not just about finding `x` and `y`; it's about developing the analytical prowess to make informed _decision making_ in various contexts. So, when we tackle `2x + 5y = -13` and `3x - 4y = -8` today, remember that you're not just doing math for math's sake; you're building a powerful toolkit for understanding and influencing the world around you. This foundational knowledge will serve you well, opening doors to advanced studies and practical applications that you might not even anticipate right now. It’s truly about empowering yourself with effective problem-solving strategies for life!\n\n## Getting to Know Our Challenge: The Specific System We're Tackling Today\n\nAlright, guys, let's zoom in on the specific _linear equation system_ we're going to conquer together. We have two distinct equations, and our mission, should we choose to accept it, is to find the *unique* pair of values for `x` and `y` that makes *both* statements true simultaneously. Our equations are:\n\nEquation 1: `2x + 5y = -13`\nEquation 2: `3x - 4y = -8`\n\nWhen we talk about _finding x and y_ for a _two equations two variables_ setup, we're looking for that special point `(x, y)` on a coordinate plane where the lines represented by these equations intersect. Graphically, that intersection is the solution! But today, we're focusing on the algebraic methods, which are often more precise and always reliable. Think of this as an _algebra challenge_ that requires a methodical approach. It's not about guessing; it's about applying tried-and-true techniques to arrive at the correct answer. The beauty of these systems is that once you understand the core methods, you can apply them to an endless variety of problems. This particular _problem setup_ involves integers, which makes the calculations relatively straightforward, but the principles remain the same even with fractions or decimals. Notice that the coefficients (the numbers in front of `x` and `y`) are different in each equation, and the constant terms on the right side are also unique. This means we'll need a strategy that allows us to either combine or substitute parts of the equations to isolate one variable at a time. This methodical process ensures that we're always working towards simplifying the problem until we can easily identify the values of `x` and `y`. Understanding this initial setup is key before we jump into the fun part of solving. It lays the groundwork for choosing the most efficient method and executing it flawlessly. So, let's keep these equations in mind as we explore the steps to unravel their secrets! We’re about to transform this pair of equations from a mystery into a clear, concise solution, proving that this _linear equation system_ is totally solvable with the right tools.\n\n## The Elimination Method: Your Go-To Strategy for These Equations\n\nWhen it comes to _solving by elimination_, this method is often a superstar, especially when you have equations like ours where the variables aren't easily isolated or coefficients have opposite signs, making _variable removal_ a breeze. The core idea behind the _elimination method_ is pretty clever: we want to manipulate our equations (by multiplying them by carefully chosen numbers) so that when we add or subtract them, one of the variables completely disappears. This transforms our _system solution_ from two equations with two unknowns into a single equation with just one unknown, which is way easier to solve! This _algebraic technique_ is incredibly powerful because it streamlines the process significantly. It's all about strategic scaling of the equations to create matching or opposite coefficients. Let's walk through this process with our specific challenge, `2x + 5y = -13` and `3x - 4y = -8`, making sure you understand every crucial step. We'll aim to eliminate one variable first, then solve for the other, and finally, find the value of the variable we initially eliminated. It's a systematic approach that guarantees a precise answer if followed correctly. This method shines when the coefficients are already multiples of each other or can be easily made so, allowing for a clean _variable removal_. Keep in mind, the goal is always to reduce complexity, and the elimination method excels at this by simplifying your equation setup into a solvable form. This is definitely one of your most valuable tools for any _system solution_.\n\n### Step 1: Prepping Our Equations for the Big Show\n\nOur first move in the elimination game is all about _aligning coefficients_. We need to make the coefficients of either `x` or `y` identical (or opposite) so they'll cancel out when we add or subtract the equations. Looking at our system: `2x + 5y = -13` and `3x - 4y = -8`. Let's aim to eliminate `y` because the `y` terms already have opposite signs (`+5y` and `-4y`), which makes _preparing for elimination_ a bit smoother as we'll just add the equations. To make the `y` coefficients match, we can find the least common multiple of 5 and 4, which is 20. So, we'll _multiplying equations_ strategically: Multiply the first equation by 4 and the second equation by 5. This will give us `20y` in the first and `-20y` in the second. Don't forget to multiply *every term* in the equation, including the constant on the right side!\n\nEquation 1 (multiplied by 4): `4 * (2x + 5y) = 4 * (-13)` which becomes `8x + 20y = -52`\nEquation 2 (multiplied by 5): `5 * (3x - 4y) = 5 * (-8)` which becomes `15x - 20y = -40`\n\nSee? Now we have our _matching variables_ all set up for elimination. This careful preparation is crucial for a clean and error-free next step.\n\n### Step 2: Making a Variable Disappear (The Magic of Elimination)\n\nNow for the really cool part! Since we've prepped our equations so beautifully, we have `+20y` in the first modified equation and `-20y` in the second. These are perfect for _eliminating a variable_ by _adding equations_ together. When you add `20y` and `-20y`, they simply cancel each other out, disappearing entirely! This is the core magic of the _elimination method_, giving us a much simpler, _single variable equation_. Let's add the two new equations we created:\n\n`(8x + 20y) + (15x - 20y) = -52 + (-40)`\n\nCombine the `x` terms: `8x + 15x = 23x`\nCombine the `y` terms: `20y - 20y = 0` (Woohoo, `y` is gone!)\nCombine the constant terms: `-52 - 40 = -92`\n\nSo, our new, much simpler equation is: `23x = -92`. How awesome is that? We've successfully transformed a system of two equations with two unknowns into a single, solvable equation with just one unknown. This _simplifying equations_ step is what makes elimination so powerful and often preferred by many math wizards.\n\n### Step 3: Finding the Value of Our First Mystery Variable (Solving for _x_)\n\nWith our newly _simplified equation_, `23x = -92`, we're just one tiny step away from _solving for x_! This is pure _basic algebra_ at its finest. To _isolate the variable_ `x`, all we need to do is divide both sides of the equation by its coefficient, which is 23. This will leave `x` all by itself, revealing its value.\n\n`23x = -92`\n`x = -92 / 23`\n`x = -4`\n\nAnd there you have it! Our _first solution_ for `x` is **-4**. See? That wasn't so bad, right? The trickiest part was getting rid of one variable, and once that's done, solving for the remaining one is typically very straightforward. This step confirms the power of the elimination method in breaking down complex systems into manageable parts. Now that we have one piece of the puzzle, we're ready to find the other, bringing us closer to the complete solution for our system of linear equations.\n\n### Step 4: Bringing Back the Second Variable (Solving for _y_)\n\nFantastic! We've successfully found `x = -4`. Now it's time for _finding y_ and completing our solution. To do this, we'll take the value of `x` we just found and _substituting back_ it into *one of the original equations*. It doesn't matter which one you choose, as both should yield the same `y` value. Let's pick the first original equation: `2x + 5y = -13` because it looks a bit friendlier with smaller numbers.\n\nSubstitute `x = -4` into `2x + 5y = -13`:\n`2(-4) + 5y = -13`\n`-8 + 5y = -13`\n\nNow, we just need to solve this simple linear equation for `y`. Add 8 to both sides to start _isolating y_:\n`5y = -13 + 8`\n`5y = -5`\n\nFinally, divide both sides by 5:\n`y = -5 / 5`\n`y = -1`\n\nBoom! We've got it! The _second variable_, `y`, is **-1**. So, our complete solution for this _system of equations_ is `(x, y) = (-4, -1)`. This step is crucial for _completing the solution_ and ensures that both `x` and `y` are correctly identified. Always remember to substitute back into one of the *original* equations to maintain accuracy and prevent any errors from compounded manipulations.\n\n### Step 5: Double-Checking Our Awesome Work (Verification!)\n\nAlright, guys, you've done the hard work, but we're not done just yet! The absolute *most important* step in _solving systems of linear equations_ is to _verify solution_. This means taking our proposed solution `(x, y) = (-4, -1)` and plugging it back into *both* of the original equations to make sure they hold true. This _checking answers_ step is your ultimate safeguard against arithmetic errors or missteps during the elimination process. It's like having a built-in _equation proof_ that confirms your _accuracy check_. If both equations are satisfied, you can be 100% confident in your answer!\n\nLet's check Equation 1: `2x + 5y = -13`\nSubstitute `x = -4` and `y = -1`:\n`2(-4) + 5(-1) = -8 + (-5) = -8 - 5 = -13`\n`-13 = -13` (This one works! ✅)\n\nNow, let's check Equation 2: `3x - 4y = -8`\nSubstitute `x = -4` and `y = -1`:\n`3(-4) - 4(-1) = -12 - (-4) = -12 + 4 = -8`\n`-8 = -8` (This one works too! ✅)\n\nSince both equations are satisfied by `x = -4` and `y = -1`, we know that our solution `(-4, -1)` is absolutely correct! This final step provides immense peace of mind and is a hallmark of truly mastering _systems of linear equations_. Never skip it – it's a quick way to catch potential errors and ensure you get full credit for your hard work!\n\n## The Substitution Method: Another Cool Tool in Your Math Toolbox\n\nWhile the elimination method is often super efficient, especially for our given system, it's always great to have more than one trick up your sleeve! The _substitution method_ is another fantastic way to solve _systems of linear equations_, and it's particularly handy when one of your variables already has a coefficient of 1 or -1, making _variable isolation_ very straightforward. With substitution, the main idea is to solve one of the equations for one variable in terms of the other, and then literally *substitute* that expression into the second equation. This transforms the second equation into a single-variable equation, which, as we've seen, is much easier to solve. This _alternative technique_ might seem a bit more involved if you have tricky coefficients, but for systems where a variable is already almost by itself, it can be a real time-saver. Let's see how the _solving by substitution_ process works with our equations: `2x + 5y = -13` and `3x - 4y = -8`. We’ll meticulously go through each step, demonstrating how this method also leads us to the same correct solution, `(-4, -1)`. Understanding both methods gives you flexibility and a deeper grasp of _algebraic problem-solving_. This _system solution_ approach, like elimination, eventually boils down to isolating one variable at a time, just through a different pathway. Having a solid command of both techniques ensures you’re well-equipped for any equation system that comes your way, solidifying your mathematical toolkit.\n\n### Step 1: Isolating a Variable – Setting Up for Substitution\n\nThe first move in the _substitution method_ is to _express one variable_ in terms of the other. This means picking one of your equations and _rearrange equation_ to solve for either `x` or `y`. Ideally, you want to choose an equation and a variable that will result in the simplest expression, avoiding fractions if possible. Looking at our system: `2x + 5y = -13` and `3x - 4y = -8`. None of the variables have a coefficient of 1 or -1, so we'll end up with fractions, but that's okay – we'll handle it! Let's choose the first equation, `2x + 5y = -13`, and solve for `x`.\n\n`2x = -13 - 5y`\n`x = (-13 - 5y) / 2`\n\nNow we have an expression for `x` that we can use in our _pre-substitution step_. This expression tells us exactly what `x` is equal to in terms of `y`. This is the essential groundwork for the next stage of the substitution method, getting us ready to _isolate a variable_ in a way that allows for direct replacement into the other equation. This strategy helps us convert a two-variable problem into a simpler one-variable problem.\n\n### Step 2: Plugging It In – The Heart of Substitution\n\nThis is where the _substitution method_ gets its name! Now that we have `x = (-13 - 5y) / 2`, we're going to take this entire expression and _substituting expression_ it into the *other* original equation, which is `3x - 4y = -8`. Wherever you see `x` in that second equation, you replace it with `(-13 - 5y) / 2`. This step is crucial because it transforms our two-variable problem into a single-variable problem, which we know how to solve!\n\n`3 * ((-13 - 5y) / 2) - 4y = -8`\n\nSee? Now we have just `y` in the equation! This _one-variable equation_ is the goal of this step. It might look a little messy with the fraction, but don't fret – we're going to clean it up in the next step. This _direct replacement_ is the cornerstone of the substitution technique, effectively eliminating one variable from our immediate focus to simplify the solving process. It's a powerful transformation that sets us up to easily find the value of `y`.\n\n### Step 3: Solving for Your First Variable\n\nAlright, time to roll up our sleeves and tackle that equation with fractions we just created: `3 * ((-13 - 5y) / 2) - 4y = -8`. Our goal here is to _solve for y_ using straightforward _algebraic manipulation_. First, let's simplify the multiplication and clear the fraction. Multiply 3 into the numerator:\n\n`(-39 - 15y) / 2 - 4y = -8`\n\nTo get rid of the denominator (2), multiply *every term* in the entire equation by 2. This is a common and effective trick for _simplifying expressions_ with fractions.\n\n`2 * [(-39 - 15y) / 2] - 2 * (4y) = 2 * (-8)`\n`-39 - 15y - 8y = -16`\n\nNow, combine the `y` terms:\n`-39 - 23y = -16`\n\nNext, add 39 to both sides to start isolating the `y` term:\n`-23y = -16 + 39`\n`-23y = 23`\n\nFinally, divide by -23 to find `y`:\n`y = 23 / -23`\n`y = -1`\n\nAnd there it is! Our first variable, `y`, is **-1**. This step demonstrates that even with fractions, systematic _algebraic manipulation_ leads us to the correct value, setting the stage for finding `x` in the next part of this _substitution method_ journey.\n\n### Step 4: Finding the Other Piece of the Puzzle\n\nExcellent work! We've found `y = -1` using the substitution method. Now, just like with elimination, we need to _find the other piece of the puzzle_ by calculating `x`. To do this, we'll take our value for `y` and substitute it back into one of the previous equations where `x` was already isolated. Remember, we had `x = (-13 - 5y) / 2` from Step 1? That's the perfect spot for _back-substitution_ to find our _final variable_ efficiently.\n\nSubstitute `y = -1` into `x = (-13 - 5y) / 2`:\n`x = (-13 - 5(-1)) / 2`\n`x = (-13 + 5) / 2`\n`x = -8 / 2`\n`x = -4`\n\nVoilà! We have `x = -4`. So, once again, our solution is `(x, y) = (-4, -1)`. Notice how both the elimination and substitution methods led us to the exact same correct answer. This consistency is a beautiful thing in mathematics and confirms the reliability of these algebraic techniques. This final step of using _back-substitution_ is critical to completely solving the system and determining both variable values, bringing our _substitution method_ application to a successful close.\n\n### Step 5: Quick Check! (Always a Good Idea)\n\nJust as we did with the elimination method, it's absolutely essential to perform a _quick check_ of our solution `(x, y) = (-4, -1)` with the substitution method. This step helps us _confirm solution_ and makes sure that all our hard work wasn't undone by a small arithmetic slip. It's about _validating answers_ and building confidence in our mathematical skills. Let's plug `x = -4` and `y = -1` back into *both* original equations to _ensure correctness_. \n\nOriginal Equation 1: `2x + 5y = -13`\nSubstitute: `2(-4) + 5(-1) = -8 - 5 = -13`\n`-13 = -13` (This checks out!)\n\nOriginal Equation 2: `3x - 4y = -8`\nSubstitute: `3(-4) - 4(-1) = -12 + 4 = -8`\n`-8 = -8` (Perfect, this one holds true too!)\n\nSince both equations are satisfied, we've successfully verified our solution. This final verification is a non-negotiable step that gives you the peace of mind knowing your answer is correct, and it's a great habit to develop for all your math endeavors! It's how we truly _confirm solution_ and show mastery of the material.\n\n## Elimination vs. Substitution: Which One is Your Best Friend?\n\nAlright, awesome job sticking with me through both the elimination and substitution methods! Now that you've seen both in action with the same problem, you might be asking yourself, "Which one should I use? Which is the _best method for systems_?" Honestly, guys, there's no single "best" method that fits *every* scenario. It often comes down to personal preference and the specific structure of the equations you're working with. However, there are some general guidelines for _choosing a solving strategy_ that can make your life a lot easier and improve your _equation efficiency_. The **elimination method** tends to be your go-to friend when variables in both equations have coefficients that are easy to make opposites (like our `+5y` and `-4y` becoming `+20y` and `-20y`) or when they are already the same or opposites. It's fantastic for neatly canceling out a variable with minimal fuss. If you see variables lined up with coefficients that are multiples of each other, or if they have opposite signs, think elimination first! On the other hand, the **substitution method** shines brightest when one of your equations *already* has a variable isolated (e.g., `x = 2y + 7`) or when a variable has a coefficient of 1 or -1. In such cases, solving for that variable is super quick, and you can jump straight into substituting. Trying to isolate a variable with fractions might make substitution a bit more prone to arithmetic errors if you're not super careful, as we saw in our example, but it's totally manageable. So, when deciding on the _best approach_, take a quick glance at your system. If an `x` or `y` is just begging to be isolated, go for substitution. If coefficients look like they could easily cancel out, elimination is probably your winner. Developing this intuition through practice is key. Ultimately, mastering both techniques gives you incredible flexibility and confidence to tackle any _equation system_ that comes your way, making you a versatile problem-solver. Each _method comparison_ reveals that while both lead to the same result, their pathways can differ significantly in terms of effort and speed depending on the initial problem setup. So, consider your options and pick the tool that makes the most sense for the job at hand!\n\n## Pro Tips for Conquering Any System of Equations!\n\nOkay, my fellow math adventurers, you've now got two powerful techniques for _solving systems of linear equations_ under your belt: elimination and substitution! But to truly become a master, it's not just about knowing the steps; it's about refining your approach and avoiding common traps. Here are some invaluable _math study tips_ to help you conquer *any* system of equations with confidence and precision. First off, **stay organized!** Write down each step clearly, especially when _algebraic precision_ is key in multiplying equations or substituting expressions. A messy workspace often leads to messy mistakes. Use separate lines for each step, and double-check your arithmetic as you go. Many errors in these problems are simple calculation errors, not conceptual misunderstandings. Secondly, **be meticulous with your signs!** A negative sign can completely derail your solution if misplaced or forgotten. Always pay close attention to positive and negative numbers when adding, subtracting, or multiplying. This is crucial for _avoiding common errors_. Thirdly, **verify your answer, always!** I cannot stress this enough. Taking those extra two minutes to plug your `x` and `y` values back into *both* original equations is the ultimate safety net. It immediately highlights any errors and lets you fix them before it costs you points on a test or leads to wrong conclusions in a real-world scenario. This _solving strategy_ isn't just a recommendation; it's a golden rule. Fourth, **don't be afraid of fractions!** As we saw with substitution, sometimes you'll end up with fractions. They're just numbers, and you can handle them! Multiply by the common denominator to clear them if they get too cumbersome. Finally, and perhaps most importantly, **practice, practice, practice!** Like learning to ride a bike or play an instrument, mastering _practice math problems_ requires repetition. The more systems you solve, the quicker you'll recognize the best method for each problem, and the more fluent you'll become in your calculations. Don't just do the assigned homework; seek out extra problems online or in textbooks. The goal is to make these problem-solving steps feel intuitive. By applying these _algebraic precision_ tips, you'll not only solve equations effectively but also develop strong analytical habits that will serve you well in all your academic and professional pursuits. These are truly _math tips_ that will elevate your equation-solving game from good to absolutely fantastic!\n\n## Wrapping It Up: You're a Systems of Equations Superstar!\n\nWow, guys, you've made it! By now, you should feel like a true superstar at _solving systems of linear equations_! We've journeyed through the intricacies of our specific problem, `2x + 5y = -13` and `3x - 4y = -8`, and successfully found its solution `(x, y) = (-4, -1)` using *both* the powerful elimination method and the versatile substitution method. This isn't just about getting the right answer; it's about building genuine _math confidence_ and developing robust _equation solving expertise_. Remember, these aren't just abstract numbers on a page; they're the language through which we can understand and model countless real-world scenarios, from economics to engineering. You've not only learned *how* to solve these equations but also *why* these methods work and *when* to choose one over the other. We covered key strategies like preparing equations, making variables disappear, back-substitution, and most importantly, verifying your answers to ensure flawless accuracy. The journey to _mastering linear equations_ is one of consistent effort and mindful practice. Don't stop here! Keep challenging yourself with different types of systems. Explore problems with fractions, decimals, or even those that lead to no solution or infinite solutions. Each new problem is an opportunity to strengthen your understanding and sharpen your skills. With the foundational knowledge and practical tips you've gained today, you're incredibly well-equipped for _continued learning_ and tackling more complex mathematical adventures. So, pat yourself on the back, because you've truly elevated your algebraic game. Go forth and solve, because the world needs your awesome _equation solving expertise_! You've got this! Keep practicing, keep exploring, and keep rocking the world of mathematics!