Solving (z+9)/(z^2-81) = 1 A Step-by-Step Guide

Hey guys! Today, we're diving into a fun little algebra problem where we need to solve for z in the equation (z+9)/(z^2-81) = 1. This might seem tricky at first glance, but don't worry, we'll break it down step by step and make sure you understand every single move. Math can be a bit intimidating sometimes, but it is really a puzzle to be solved, and you can totally nail this with the right approach. We'll not only find the solution but also discuss the important concepts behind it, such as factoring, restrictions, and how to check your answers. By the end of this guide, you'll be confidently solving similar equations and impressing your friends with your algebraic prowess!

Understanding the Equation

Before we start crunching numbers, let's take a good look at our equation: (z+9)/(z^2-81) = 1. The first thing we notice is that it's a rational equation, meaning it involves fractions with variables in the denominator. These types of equations require a bit of extra care because we need to avoid any values of z that would make the denominator equal to zero. Why? Because division by zero is a big no-no in the math world – it's undefined and leads to all sorts of problems. So, before we even think about solving, we need to identify any restrictions on z. The denominator here is z^2 - 81. We need to figure out what values of z would make this equal to zero. Setting z^2 - 81 = 0, we can factor it as (z-9)(z+9) = 0. This tells us that z cannot be 9 or -9 because either of these values would make the denominator zero. Keep these restrictions in mind as we move forward; they're super important for ensuring our final solution is valid.

The next thing we might notice is that the denominator, z^2 - 81, looks familiar. It's actually a difference of squares! Recognizing patterns like this is key in algebra. The difference of squares pattern is a^2 - b^2 = (a - b)(a + b). In our case, a is z and b is 9, so we can factor z^2 - 81 as (z - 9)(z + 9). This factorization is going to be crucial for simplifying the equation and making it easier to solve. Factoring is one of the fundamental skills in algebra. It allows us to break down complex expressions into simpler ones, which can then be manipulated more easily. Mastering factoring techniques will significantly improve your ability to tackle a wide range of algebraic problems, from solving equations to simplifying expressions and even working with more advanced concepts like calculus. So, understanding how to factor different types of expressions, such as the difference of squares, quadratic trinomials, and others, is a worthwhile investment of your time and effort.

Now that we've identified the restrictions on z (z ≠ 9 and z ≠ -9) and factored the denominator, we're ready to start solving the equation. The goal here is to isolate z on one side of the equation. We can do this by manipulating the equation using algebraic operations, making sure to perform the same operation on both sides to maintain the balance. We'll use our factored form of the denominator to simplify the equation and hopefully cancel out some terms. This is where things start to get interesting, and you'll see how all the pieces we've discussed so far come together to help us find the solution. Remember, the key to solving any equation is to be organized, methodical, and to double-check your work along the way. So, let's get started and see how we can crack this equation!

Step-by-Step Solution

Okay, let's get our hands dirty and walk through the step-by-step solution to the equation (z+9)/(z^2-81) = 1. Remember, we've already identified that z cannot be 9 or -9. Now, the first thing we want to do is get rid of the fraction. To do this, we can multiply both sides of the equation by the denominator, which is z^2 - 81. But remember, we already factored z^2 - 81 as (z - 9)(z + 9), so we can write this as:

(z+9)/( (z-9)(z+9) ) = 1

Multiply both sides by (z - 9)(z + 9):

(z+9) = (z-9)(z+9)

This step is crucial because it eliminates the fraction and transforms our rational equation into a more manageable linear equation. By multiplying both sides by the denominator, we're essentially "undoing" the division, which allows us to work with a simpler expression. However, it's important to remember that we need to keep track of our restrictions on z. Even though we've eliminated the fraction, the restrictions still apply, and we'll need to check our final answer against them.

Now, notice that we have a (z + 9) term on both sides of the equation. This is where things get interesting. It might be tempting to divide both sides by (z + 9), but we need to be careful! Dividing by a variable expression can sometimes lead to losing solutions. Why? Because if that expression could be zero, we're essentially dividing by zero, which is a big no-no. However, we already know that z cannot be -9, so it's safe to divide both sides by (z+9) now. Doing so gives us:

1 = z - 9

This simplification is a major step forward. We've gone from a complex rational equation to a simple linear equation that we can easily solve. This highlights the power of factoring and simplification in algebra. By recognizing the structure of the equation and applying the appropriate algebraic techniques, we've transformed a seemingly difficult problem into a straightforward one. This is a common theme in mathematics – complex problems can often be broken down into simpler steps with the right approach.

Now, it's just a matter of isolating z. To do this, we can add 9 to both sides of the equation:

1 + 9 = z

10 = z

So, we've found a potential solution: z = 10. But remember, we had restrictions on z. We need to check if this solution is valid. Does z = 10 violate our restrictions that z cannot be 9 or -9? Nope! 10 is perfectly fine. So, z = 10 is indeed a valid solution to our equation. This final check is absolutely crucial. It's easy to get caught up in the algebraic manipulations and forget about the initial restrictions. But failing to check your solution can lead to incorrect answers. So, always, always, always check your solutions against any restrictions!

Checking the Solution

Alright, we've got a potential solution: z = 10. But in math, it's always a good idea to check our work. Think of it like proofreading a paper before you turn it in. We want to make sure our answer actually works in the original equation. To check, we'll substitute z = 10 back into the original equation:

(z+9)/(z^2-81) = 1

(10+9)/(10^2-81) = 1

(19)/(100-81) = 1

(19)/(19) = 1

1 = 1

It checks out! This confirms that z = 10 is indeed the correct solution to our equation. Checking our solution is not just a formality; it's a crucial step in the problem-solving process. It gives us confidence that our answer is correct and helps us catch any errors we might have made along the way. In this case, the check was straightforward, and it clearly showed that z = 10 satisfies the original equation. But in more complex problems, the check might involve more intricate calculations. Regardless of the complexity, the principle remains the same: always check your solution!

Checking your solution is also a valuable learning opportunity. If your solution doesn't check out, it means there's an error somewhere in your work. Instead of just giving up, this is a chance to go back and review your steps. Where did you make a mistake? Did you forget to distribute a negative sign? Did you make an arithmetic error? By identifying and correcting your mistakes, you're not only solving the problem correctly but also strengthening your understanding of the underlying concepts. So, think of checking your solution as an integral part of the problem-solving process, not just an optional step.

Furthermore, checking solutions is particularly important when dealing with rational equations, like the one we solved today. As we discussed earlier, rational equations often have restrictions on the variable because certain values would make the denominator zero. These restrictions can sometimes lead to extraneous solutions, which are solutions that we obtain algebraically but don't actually satisfy the original equation. By checking our solution, we can identify and discard any extraneous solutions, ensuring that we only accept valid answers. So, in the context of rational equations, checking solutions is not just a good habit; it's an essential step.

Conclusion: The Value of Thoroughness

So, there you have it! We successfully solved for z in the equation (z+9)/(z^2-81) = 1 and found that z = 10. We walked through the entire process step by step, from understanding the equation and identifying restrictions to factoring, simplifying, and checking our solution. This problem highlights the importance of several key concepts in algebra, including factoring, working with rational expressions, and, most importantly, being thorough in our work. By carefully considering each step and checking our answer, we can confidently tackle even seemingly complex algebraic problems.

Algebra, like any branch of mathematics, is built on a foundation of logical reasoning and careful execution. There are often multiple approaches to solving a problem, but the key is to choose a method that makes sense to you and to follow it systematically. This means breaking down the problem into smaller, manageable steps, paying attention to details, and not being afraid to try different strategies. And, as we've emphasized throughout this guide, it also means checking your work every step of the way.

The value of thoroughness cannot be overstated in mathematics. It's not enough to just get the right answer; you need to understand why the answer is correct and how you arrived at it. This understanding comes from carefully reviewing your steps, checking your solution, and reflecting on the process. By being thorough, you not only increase your chances of getting the right answer but also deepen your understanding of the underlying mathematical concepts. This deeper understanding will serve you well as you continue to study mathematics and encounter more challenging problems.

Finally, remember that practice makes perfect! The more you work through algebraic problems, the more comfortable you'll become with the various techniques and strategies. Don't be afraid to make mistakes – they're a natural part of the learning process. The important thing is to learn from your mistakes and to keep practicing. So, go out there and tackle some more algebraic equations! You've got this!