Solving Algebraic Equations: A Simple Guide

Hey guys! Ever stared at a math problem that looks like a secret code? Well, today we're going to crack one of those codes together, focusing on the left hand side of the equation. We'll break down how to figure out what's going on, making it super clear and, dare I say, even a little bit fun!

Understanding the Left Hand Side

So, what exactly is the left hand side of an equation, you ask? Think of an equation like a balanced seesaw. The equals sign (=) is the pivot point. Everything to the left of the equals sign is the left hand side (LHS), and everything to the right is the right hand side (RHS). Our mission, should we choose to accept it, is to determine the value of whatever is chilling on the LHS. In the problem you're looking at, we've got a few steps laid out for us. We start with n = 3.ar{1}. This little notation, the bar over the 1, means that the '1' repeats forever: $3.11111...$. It's a repeating decimal, and these can sometimes feel tricky, but we've got a neat trick up our sleeves.

The next line tells us 10n = 31.ar{1}. How did they get that? Well, when you multiply a number by 10, you move the decimal point one place to the right. So, $3.11111...$ becomes $31.11111...$, which is 31.ar{1}. Makes sense, right? Now, the really cool part is the third line: 10n - n = 31.ar{1} - 3.ar{1}. This is where the magic happens for finding the value of the LHS. They're subtracting the original equation ($n$) from the multiplied equation ($10n$). This is a common technique when dealing with repeating decimals because it often eliminates the repeating part, leaving us with a simple number. So, the left hand side of this specific equation is $10n - n$. Our job is to simplify this expression and find its numerical value. It's like peeling back layers of an onion to get to the core. We're aiming to isolate 'n' or find its value. This whole process is a cornerstone of algebra, helping us solve for unknowns and understand relationships between numbers. We're not just crunching numbers; we're learning a powerful way to think and solve problems. So, stick with me, and we'll get to the bottom of this!

Step-by-Step Simplification

Alright guys, let's get down to business and simplify that left hand side (LHS): $10n - n$. This is a pretty straightforward algebraic simplification. Think of 'n' as a variable, like an 'x' or a 'y'. If you have 10 apples and you take away 1 apple, how many apples do you have left? You have 9 apples, right? The same logic applies here. We have 10 'n's and we're subtracting 1 'n'. So, $10n - n$ simplifies to 9n. It's like combining like terms. When you have terms with the same variable, you can add or subtract their coefficients (the numbers in front of the variable). In this case, the coefficient of $10n$ is 10, and the coefficient of $n$ (which is the same as $1n$) is 1. So, $10 - 1 = 9$, and we keep the variable 'n', giving us $9n$. This is the simplified form of the left hand side of the equation 10n - n = 31.ar{1} - 3.ar{1}.

Now, let's look at the right hand side (RHS) of that same equation: 31.ar{1} - 3.ar{1}. Remember, ar{1} means the '1' repeats infinitely. So, we have $31.11111... - 3.11111...$. When we subtract these two numbers, something really cool happens. Let's line them up vertically for subtraction:

  31.11111...
-  3.11111...
-----------

Notice how all the digits after the decimal point are the same? When you subtract them, they all cancel each other out, leaving you with all zeros after the decimal. So, we are left with $31 - 3$. And what is $31 - 3$? It's 28! So, the entire equation 10n - n = 31.ar{1} - 3.ar{1} simplifies to $9n = 28$. This is a huge step because it gets rid of the repeating decimals and gives us a clear relationship between $9n$ and a simple number. The beauty of this method is its efficiency in transforming a problem involving repeating decimals into a standard linear equation.

Finding the Value of n

We've simplified the equation to $9n = 28$. The question asks for the value of the left hand side of the original equation before the subtraction step, which was $10n - n$. We found that $10n - n$ simplifies to $9n$. We also found that $9n$ is equal to 28. Therefore, the value of the left hand side, $10n - n$, is 28. We didn't even need to find the value of 'n' itself to answer this specific question! But, if we did want to find 'n', we could easily do it from $9n = 28$. To isolate 'n', we would divide both sides by 9:

$n = \frac{28}{9}$

Let's quickly check if this value of $n$ actually works. If $n = \frac{28}{9}$, then $n$ as a decimal is $3.11111...$, which is 3.ar{1}. This is exactly what we started with! So, our calculations are spot on. The process of multiplying by 10 and subtracting the original equation is a classic technique to convert repeating decimals into fractions. It works because multiplying by 10 shifts the decimal place, and when you subtract the original number, the repeating part aligns perfectly and cancels out. This leaves you with an integer or a terminating decimal on the right side, making it much easier to solve for the variable.

This method is super powerful for understanding and manipulating numbers, especially those pesky repeating decimals. It shows how algebraic manipulation can unlock the true value hidden within seemingly complex expressions. So, the next time you see a repeating decimal, remember this trick – it might just be the key to solving your problem. The elegance of this approach lies in its ability to transform an infinite series of digits into a simple fraction or integer, a fundamental concept in number theory and algebra. We've successfully navigated through the algebraic steps, simplified the expression, and arrived at a concrete numerical value. It's a testament to the power of systematic problem-solving in mathematics. Remember, the goal is often to simplify and reveal underlying structures, and this example perfectly illustrates that principle. We've confirmed our answer and even explored how to find the value of 'n' itself, reinforcing our understanding of the entire process. It's all about breaking down complex problems into manageable steps, and this problem provided a great opportunity to do just that. Keep practicing, and you'll become a math whiz in no time!

Conclusion: The Answer Revealed

So, to wrap things up, the question asked for the value of the left hand side of the equation. Looking at the provided steps, the equation we are focusing on is 10n - n = 31.ar{1} - 3.ar{1}. The left hand side (LHS) of this equation is $10n - n$. Through our algebraic simplification, we found that $10n - n$ is equal to $9n$. We also simplified the right hand side (RHS), 31.ar{1} - 3.ar{1}, and found it to be equal to 28. This means our equation boils down to $9n = 28$. Since the left hand side, $10n - n$, simplified to $9n$, and we know $9n$ equals 28, the value of the left hand side is simply 28.

It’s pretty neat how these steps work together, right? We started with a repeating decimal, used a bit of algebra to clear it out, and ended up with a straightforward value for the expression $10n - n$. This technique is a lifesaver for anyone working with repeating decimals and trying to express them in a more manageable form, like fractions. It highlights the beauty of algebra in simplifying complex numerical relationships. The key takeaway is that by performing the subtraction $10n - n$, we are directly evaluating the expression that represents the LHS of the equation presented. And as we've shown, this expression simplifies to 28. It's a direct answer derived from the algebraic steps provided, requiring no further manipulation to find 'n' itself. This is a common scenario in math problems where you might be asked to evaluate an intermediate expression rather than the final variable. So, confidently, we can say the value is 28. Keep applying these methods, and you'll be solving even tougher problems in no time! Practice makes perfect, and understanding these fundamental algebraic manipulations will serve you well in all your future math endeavors. It's all about building that mathematical intuition, one problem at a time.