Unraveling Equations: Solutions And Types Explained

Hey Plastik Magazine readers! Let's dive into the world of equations, where we'll solve a tricky one and then figure out what kind of equation we're dealing with. It's like a math detective game, and trust me, it's more fun than it sounds! We'll break down the equation, find its solution, and then classify it as an identity, a conditional equation, or an inconsistent equation. Ready to sharpen those math skills? Let's get started!

Solving the Equation: A Step-by-Step Guide

Alright, guys, let's tackle this equation head-on. Our equation is: rac{4}{x-3}+rac{3}{x+9}=rac{20}{(x+9)(x-3)}. The goal here is to find the value(s) of 'x' that make this equation true. First things first, we need to get rid of those pesky fractions. To do that, we're going to multiply both sides of the equation by the common denominator, which is $(x-3)(x+9)$. Remember, this is a crucial step to simplify the equation and make it easier to solve. Always keep an eye out for potential values of x that would make the denominator zero, as these are values that x cannot equal (we'll come back to this later, trust me, it is very important!).

So, multiplying both sides by $(x-3)(x+9)$, we get: $4(x+9) + 3(x-3) = 20$. See how those fractions just magically disappeared? That's the power of finding a common denominator! Now, we can simplify this further by distributing the numbers outside the parentheses: $4x + 36 + 3x - 9 = 20$. Next, combine like terms. This means adding the 'x' terms together (4x + 3x = 7x) and the constant terms together (36 - 9 = 27). This leaves us with: $7x + 27 = 20$. Now, we need to isolate 'x'. Subtract 27 from both sides: $7x = 20 - 27$, which simplifies to $7x = -7$. Finally, divide both sides by 7 to solve for x: $x = -1$. Boom! We've found a solution. Now, before we get too excited, let's make sure our solution is valid and doesn't contradict any of our initial conditions. Always remember to check your solution by plugging it back into the original equation to ensure it works. This is like a double-check to make sure we didn't make any mistakes along the way. Additionally, we need to consider if this solution is consistent with the domain of the original equation. This is a very important step and the solution may not be valid, because the original equation contains fractions. The original equation is undefined if the denominator is zero.

Checking for Extraneous Solutions and Domain Considerations

Okay, guys, we found a solution: $x = -1$. But before we celebrate, let's make sure our solution is legit. In the original equation rac{4}{x-3}+rac{3}{x+9}=rac{20}{(x+9)(x-3)}, we had fractions. And remember what happens with fractions? They get a little grumpy when the denominator is zero. So, we need to check if our solution, $x = -1$, causes any of the denominators to become zero. If it does, our solution is no good! That solution is not valid. The denominators in the original equation are $(x-3)$ and $(x+9)$ and $(x+9)(x-3)$. Let's check them out.

• If $x = -1$, then $x - 3 = -1 - 3 = -4$. Not zero! Good.
• If $x = -1$, then $x + 9 = -1 + 9 = 8$. Not zero! Fantastic.
• If $x = -1$, then $(x + 9)(x - 3) = (8)(-4) = -32$. Still not zero! Excellent.

So, our solution, $x = -1$, doesn't make any of the denominators zero. Therefore, it's a valid solution, no problem. It satisfies the domain of the equation. Also, in the original equation, we have denominators $(x - 3)$ and $(x + 9)$, which means $x$ cannot be 3 or -9. Our solution is $x = -1$, which is not 3 or -9. Now, let's move on to the next part of our math adventure: classifying the equation!

Classifying the Equation: Identity, Conditional, or Inconsistent

Alright, now that we've solved the equation and confirmed our solution, it's time to classify it. Is it an identity, a conditional equation, or an inconsistent equation? This classification tells us something about the nature of the equation and its solutions.

• Identity: An identity is an equation that is true for all values of the variable (except for those that make the denominator zero, if there are any fractions). In other words, no matter what number you plug in for 'x' (as long as it doesn't break the equation, like making a denominator zero), the equation will always be true. Think of it like a mathematical fact that always holds true.
• Conditional Equation: A conditional equation is true for some values of the variable, but not all. Our equation is conditional because it is true only when x = -1. This is the most common type of equation you'll encounter. It has specific solutions that make the equation true.
• Inconsistent Equation: An inconsistent equation is an equation that has no solutions. No matter what value you plug in for the variable, the equation is never true. This usually happens when you end up with a contradiction, such as 2 = 5, after simplifying the equation.

In our case, the equation is true only when x = -1. Therefore, our equation is a conditional equation. The solution set for this equation is {-1}. Pretty cool, right? We solved it, and now we understand its nature. Let's recap what we've done.

Recap and Conclusion

So, to recap, we started with the equation rac{4}{x-3}+rac{3}{x+9}=rac{20}{(x+9)(x-3)}. We solved it step-by-step, simplified it, and found that x = -1. Then, we checked our solution to make sure it didn't cause any denominators to become zero, ensuring our solution was valid. Finally, we classified the equation as a conditional equation because it is true only for the specific value of x = -1. We've successfully navigated the world of equations, solved for 'x', and categorized the equation type! Congratulations, everyone! You've successfully conquered an equation and learned about identities, conditional equations, and inconsistent equations. Keep practicing, and you'll become equation-solving pros in no time. Thanks for reading, and see you next time, Plastik Magazine readers! Keep those math skills sharp, and always remember to double-check your work.