Solve $x-\frac{7}{8}=-5$: Easy Steps To Find X!

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Solve $x-\frac{7}{8}=-5$: Easy Steps to Find X!

Unlocking the Mystery of Algebraic Equations

Hey there, math enthusiasts and curious minds! Ever looked at an equation like xβˆ’78=βˆ’5x-\frac{7}{8}=-5 and thought, "Whoa, where do I even begin?" Well, you're in the right place, because today we're going to demystify this type of problem and show you exactly how to solve it. It might look a little intimidating with that fraction, but trust me, by the end of this, you'll feel like an algebraic superstar. We're going to break down how to solve xβˆ’78=βˆ’5x-\frac{7}{8}=-5 for xx into simple, easy-to-follow steps. This isn't just about getting the right answer; it's about understanding the logic behind it, which is way more powerful.

Think of an equation as a perfectly balanced scale. Whatever you do to one side, you absolutely must do to the other side to keep it balanced. Our main goal when we're trying to find the value of x is to get x all by itself on one side of that scale. This process, often called isolating the variable, is a fundamental skill in algebra and something you'll use over and over again. Whether you're dealing with fractions, decimals, or just plain old whole numbers, the core principles remain the same. We'll be focusing specifically on what number should be added to both sides in our example, demonstrating a key technique in solving linear equations. This concept is super important for anyone diving into mathematics, from high school students to adults brushing up on their skills. We're going to make sure you grasp not just the "how" but also the "why," so you can confidently tackle similar problems in the future. So, grab a notepad, maybe a snack, and let's jump into the awesome world of algebraic problem-solving! We're talking about making equations easy and understandable, getting rid of that common fear of fractions, and empowering you to solve for x with confidence. This foundational knowledge is crucial for future math success, building blocks for more complex topics like systems of equations, inequalities, and even calculus. Our focus on balancing equations with additive inverses is a cornerstone principle that, once mastered, opens up a world of possibilities. So get ready to conquer this equation and many more like it!

Breaking Down the Equation: xβˆ’78=βˆ’5x - \frac{7}{8} = -5

Alright, guys, let's zoom in on our specific challenge: the equation xβˆ’78=βˆ’5x-\frac{7}{8}=-5. Before we even think about what number should be added to both sides, let's really understand what this equation is telling us. At its heart, it's saying: "There's some unknown number, x, and if you subtract the fraction 78\frac{7}{8} from it, you'll end up with -5." Our mission, should we choose to accept it (and we definitely should!), is to uncover the true value of x. This process of finding x is what algebra is all about.

The key term here is "inverse operation." In simple terms, to undo something, you do the opposite. If someone adds 5 to your number, you subtract 5 to get back to where you started. If they multiply by 2, you divide by 2. In our equation, we see that 78\frac{7}{8} is being subtracted from x. To isolate x, we need to undo that subtraction. And what's the opposite, or inverse operation, of subtraction? You guessed it: addition! This is where what number should be added to both sides comes into play. We need to add the same amount that was subtracted, but as a positive value, to effectively "cancel out" the subtraction and leave x alone.

Understanding fractions is also super important here. Don't let them scare you off! A fraction like 78\frac{7}{8} is just a part of a whole. In this case, it's 7 out of 8 equal parts. When we see xβˆ’78x - \frac{7}{8}, it just means x minus that specific portion. The equals sign, "=", is your best friend. It signifies balance. Whatever is on the left side must be equivalent to what's on the right side. Our goal is to manipulate this equation, using legitimate algebraic moves, to reveal x. By focusing on the inverse operation, we're not just guessing; we're applying a fundamental mathematical principle. We're setting ourselves up for success in solving for x by systematically removing anything that's "attached" to x on its side of the equation. This careful breakdown ensures we're on the right track before we even make our first move. So, understanding the terms, recognizing the operation, and remembering the inverse is the solid foundation for finding the number to add to solve xβˆ’78=βˆ’5x-\frac{7}{8}=-5. Get ready to apply this knowledge and see x reveal itself!

The Crucial Step: What Number Do We Add?

Alright, party people, this is where the magic happens! We've established that our goal is to isolate x in the equation xβˆ’78=βˆ’5x-\frac{7}{8}=-5. We also know that to undo the subtraction of 78\frac{7}{8} from x, we need to perform the inverse operation, which is addition. So, what number should be added to both sides of this equation to achieve our goal? The answer is the same number that is being subtracted, but as a positive value. In this specific case, we need to add 78\frac{7}{8} to both sides. Why 78\frac{7}{8}? Because βˆ’78+78-\frac{7}{8} + \frac{7}{8} equals zero, effectively canceling out the fraction on the left side and leaving x all by its lonesome self. This is the cornerstone of solving xβˆ’78=βˆ’5x-\frac{7}{8}=-5 for xx.

Let's write it out and see what happens:

xβˆ’78=βˆ’5x - \frac{7}{8} = -5

Now, we add 78\frac{7}{8} to the left side and, to keep our equation perfectly balanced, we must also add 78\frac{7}{8} to the right side:

xβˆ’78+78=βˆ’5+78x - \frac{7}{8} + \frac{7}{8} = -5 + \frac{7}{8}

On the left side, as planned, βˆ’78+78-\frac{7}{8} + \frac{7}{8} simplifies to 00. So we're left with:

x+0=βˆ’5+78x + 0 = -5 + \frac{7}{8}

Which further simplifies to:

x=βˆ’5+78x = -5 + \frac{7}{8}

Now, our task boils down to solving the arithmetic problem on the right side: βˆ’5+78-5 + \frac{7}{8}. This is where knowing how to work with integers and fractions comes in handy. To add or subtract a whole number and a fraction, it's easiest to express the whole number as a fraction with a common denominator. In this case, our denominator is 8. So, we can rewrite βˆ’5-5 as a fraction with a denominator of 8. Since 5=5Γ—88=4085 = \frac{5 \times 8}{8} = \frac{40}{8}, then βˆ’5=βˆ’408-5 = -\frac{40}{8}.

So the equation becomes:

x=βˆ’408+78x = -\frac{40}{8} + \frac{7}{8}

Now that they have the same denominator, we can simply add the numerators:

x=βˆ’40+78x = \frac{-40 + 7}{8}

x=βˆ’338x = \frac{-33}{8}

And there you have it! The value of x is βˆ’338-\frac{33}{8}. This fraction can also be expressed as a mixed number: βˆ’418-4 \frac{1}{8}.

This entire process, from identifying the inverse operation to performing the arithmetic, is crucial for mastering equation solving. We've successfully determined the number to add to solve xβˆ’78=βˆ’5x-\frac{7}{8}=-5 and meticulously applied it. The elegance lies in its simplicity: whatever is being done to x, do the exact opposite to both sides of the equation. This ensures that the equality holds true while allowing us to zero in on our unknown variable. Remember, patience and careful calculation, especially with fractions and negative numbers, are your best friends here. Don't rush, and always double-check your arithmetic! This method is a game-changer for finding x and truly understanding the mechanics of algebraic manipulation. We're not just moving numbers; we're strategically balancing the equation to reveal the hidden truth of x.

Step-by-Step Guide to Solving xβˆ’78=βˆ’5x - \frac{7}{8} = -5

Let's condense that whole process into a clear, actionable guide, perfect for solving xβˆ’78=βˆ’5x-\frac{7}{8}=-5 for xx. This is your go-to checklist for tackling similar problems:

  1. Identify the operation: Look at the side with x. What operation is being applied to x? In xβˆ’78=βˆ’5x-\frac{7}{8}=-5, we see 78\frac{7}{8} is being subtracted from x.
  2. Determine the inverse operation: To undo subtraction, you add.
  3. Apply the inverse to both sides: This is the golden rule! Since we are subtracting 78\frac{7}{8}, we must add 78\frac{7}{8} to both sides of the equation. xβˆ’78+78=βˆ’5+78x - \frac{7}{8} + \frac{7}{8} = -5 + \frac{7}{8}
  4. Simplify both sides:
    • On the left: βˆ’78+78-\frac{7}{8} + \frac{7}{8} cancels out, leaving just x.
    • On the right: You need to perform the addition βˆ’5+78-5 + \frac{7}{8}.
      • Convert βˆ’5-5 to a fraction with a denominator of 8: βˆ’5=βˆ’408-5 = -\frac{40}{8}.
      • Add the fractions: βˆ’408+78=βˆ’40+78=βˆ’338-\frac{40}{8} + \frac{7}{8} = \frac{-40+7}{8} = -\frac{33}{8}.
  5. State your final answer: x=βˆ’338x = -\frac{33}{8}.
  6. (Optional but recommended) Verify your answer: Plug the value of x back into the original equation to ensure both sides are equal. βˆ’338βˆ’78=βˆ’5-\frac{33}{8} - \frac{7}{8} = -5 βˆ’33βˆ’78=βˆ’5\frac{-33 - 7}{8} = -5 βˆ’408=βˆ’5\frac{-40}{8} = -5 βˆ’5=βˆ’5-5 = -5 It works! This verification step is super important for confirming your solution and building confidence in your algebraic skills.

Following these steps will make finding the value of x straightforward and accurate every single time. This systematic approach ensures that you don't miss any critical details and confidently arrive at the correct solution for x.

Why Balance Matters: The Golden Rule of Algebra

Guys, we've talked about it a few times, but it's worth dedicating a whole section to the absolute core principle behind solving equations: The Golden Rule of Algebra. This rule is not just a suggestion; it's the fundamental law that governs all equation solving, and it's what allows us to reliably find the value of x. What is it? Simply put: Whatever you do to one side of an equation, you MUST do the exact same thing to the other side. This principle is crucial when we consider what number should be added to both sides or any other operation.

Imagine an old-school balance scale, the kind with two pans. When you first write down an equation like xβˆ’78=βˆ’5x-\frac{7}{8}=-5, it's like both sides of that scale are perfectly level, holding equal weight. The expression xβˆ’78x-\frac{7}{8} has the exact same value as βˆ’5-5. If you were to, say, add a weight (or in our case, add a number) to just one side of that scale, what would happen? It would tip! The balance would be broken. The equality would be destroyed, and your equation would no longer be true. That's why, when we decided to add 78\frac{7}{8} to the left side to get x by itself, we absolutely had to add 78\frac{7}{8} to the right side as well. This maintains the equilibrium, ensuring that the new equation is just a different form of the original, but still holds the same truth about x. This constant rebalancing is key to correctly solving xβˆ’78=βˆ’5x-\frac{7}{8}=-5 for xx.

This Golden Rule applies to every single operation you perform on an equation: addition, subtraction, multiplication, division, even more complex operations like taking square roots or logarithms. If you multiply the left side by 2, you must multiply the right side by 2. If you divide the left side by 3, you must divide the right side by 3. This isn't just arbitrary; it's what ensures that the solution for x you ultimately find is valid for the original equation. Without this rule, algebra would be chaos, and we'd never be able to consistently find the value of x in a way that makes sense. So, as you continue your journey in mathematics, always keep that mental image of the balanced scale. It will guide you through countless algebraic challenges and empower you to confidently solve for x in any equation. This rule is truly the bedrock of algebraic manipulation, making sense of every step we take to find the number to add to solve xβˆ’78=βˆ’5x-\frac{7}{8}=-5 and beyond. Mastering this foundational concept is what separates mere memorization from true mathematical understanding, equipping you with the lifelong skill of equation balancing.

Beyond This Problem: Applying Your New Skills

Awesome work, everyone! You've just mastered solving xβˆ’78=βˆ’5x-\frac{7}{8}=-5 for xx, a fantastic accomplishment. But guess what? The techniques you learned today, especially the concept of using inverse operations and the Golden Rule of Algebra, are not just for this one specific equation. They are universally applicable, foundational skills that will empower you to tackle a vast array of algebraic problems. The method of determining what number should be added to both sides to isolate a variable extends far beyond simple addition and subtraction with fractions.

Consider other scenarios. What if you had an equation like x+12=20x + 12 = 20? To find the value of x, you'd see that 12 is being added to x. The inverse operation is subtraction, so you'd subtract 12 from both sides. Or, how about 3x=153x = 15? Here, x is being multiplied by 3. The inverse operation is division, so you'd divide both sides by 3. Even if you encounter something like x4=7\frac{x}{4} = 7, you'd recognize that x is being divided by 4, and the inverse is multiplication, so you'd multiply both sides by 4. Notice a pattern here? It's always about identifying what's happening to x and then doing the exact opposite to both sides to maintain that balance and reveal x. This systematic approach is the power behind solving for x in various contexts.

Even when equations become more complex, involving multiple steps, decimals, or negative numbers, the underlying principles remain constant. You might need to combine like terms first, or deal with parentheses, but eventually, you'll always boil it down to operations that require you to add, subtract, multiply, or divide to isolate x. By practicing problems like solving xβˆ’78=βˆ’5x-\frac{7}{8}=-5, you're building a muscle memory for these algebraic manipulations. Don't stop here! Look for other equations, maybe with different fractions, or even mixed numbers and decimals. The more you practice finding the value of x using these inverse operations and the Golden Rule, the more confident and proficient you'll become. These skills are critical for success in higher-level mathematics, science, engineering, and even everyday problem-solving scenarios. So, keep that brain engaged, keep those equations balanced, and keep solving for x like the pro you're becoming! You've learned a super important tool for your math toolkit today, which is far more than just finding the number to add to solve xβˆ’78=βˆ’5x-\frac{7}{8}=-5; it's about building a robust understanding of algebra.

Wrapping It Up: Your Journey to Equation Mastery

Phew! We've covered a lot of ground today, guys, and you've done an amazing job diving deep into the world of algebraic equations. Starting with what might have seemed like a tricky fraction problem, xβˆ’78=βˆ’5x-\frac{7}{8}=-5, we've systematically broken it down, understood the core principles, and successfully arrived at the solution. You now know exactly what number should be added to both sides to solve for x in this particular equation, and more importantly, you understand why those steps are taken. The journey to finding the value of x isn't just about memorizing a formula; it's about understanding the logic of inverse operations and the non-negotiable principle of balancing an equation.

We started by identifying the term being subtracted from x, which was 78\frac{7}{8}. Our strategy was clear: to undo this subtraction, we needed to add 78\frac{7}{8}. And because of the Golden Rule of Algebra, we added 78\frac{7}{8} to both sides of the equation. This simple yet powerful move cancelled out the fraction on the left, leaving x alone, and transformed the right side into an arithmetic problem involving a whole number and a fraction. By converting the whole number to an equivalent fraction, we smoothly added them together, ultimately revealing that x = -338\frac{33}{8}. This step-by-step process for solving xβˆ’78=βˆ’5x-\frac{7}{8}=-5 for xx is a blueprint for tackling countless other linear equations.

The key takeaway here, beyond just the specific answer, is the fundamental understanding of algebraic manipulation. You've learned how to approach an equation, identify the operations at play, apply their inverses, and ensure the equation remains balanced throughout the entire process. This isn't just some abstract math concept; it's a practical skill that empowers you to decode mathematical puzzles and solve real-world problems. So, as you continue your mathematical adventure, remember these lessons. Don't be afraid of fractions or negative numbers. Approach each equation with confidence, break it down, apply the inverse operations to both sides, and you'll consistently be able to find the value of x. Keep practicing, keep questioning, and keep that scale balanced! You're well on your way to becoming an equation master, and that's super cool! You've gone from wondering what number to add to solve xβˆ’78=βˆ’5x-\frac{7}{8}=-5 to confidently executing the solution, a truly valuable step in your mathematical development. Keep up the fantastic work!