Calculating Expressions: Find The Value When X=2 And Y=3

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Calculating Expressions: Find the Value When x=2 and y=3

Hey math enthusiasts! Let's dive into a fun little problem today, where we'll figure out the value of an expression when given specific values for our variables. We'll be looking at the expression 2x0yβˆ’22x^0y^{-2} and figuring out its value when x=2x = 2 and y=3y = 3. Sounds easy, right? Well, it is! Just a couple of simple steps, and we'll have our answer. This kind of problem is fundamental in algebra, and understanding it sets the stage for more complex equations. So, let's break it down and make sure we all get it. Remember, practice makes perfect, so don't be shy about working through similar examples after we're done here. Let's make sure we have a solid understanding of how exponents work, because they're the real MVPs in this problem.

We need to remember our exponent rules, which are the backbone of this problem, and it will help us understand what we are dealing with. Specifically, anything raised to the power of zero equals one. Think of it like this: x0=1x^0 = 1 no matter what x is (as long as it's not zero itself). The other part we need is to understand what a negative exponent does. A negative exponent means we take the reciprocal of the base raised to the positive version of that exponent. In other words, yβˆ’2y^{-2} is the same as 1y2\frac{1}{y^2}. So, keep these two rules in mind, and you'll do great! We'll go step by step through the expression, applying these rules as we go. It's like a recipe; follow the steps, and you'll get the right result. Let's get started. Also, keep in mind that the variables are only placeholders. We can swap them out for the numbers we're given, and the math still works the same way. It's like a secret code, and we're about to crack it!

Step-by-Step Calculation

Alright, let's get down to business and break down the calculation step by step, guys. First, we're given the expression 2x0yβˆ’22x^0y^{-2}. We know that x=2x = 2 and y=3y = 3. Now, let's substitute these values into the expression, so we get 2βˆ—(2)0βˆ—(3)βˆ’22 * (2)^0 * (3)^{-2}. See? Easy peasy! Now, we will simplify this expression. Next, we look at (2)0(2)^0. Anything to the power of zero equals 1, so (2)0=1(2)^0 = 1. The expression becomes 2βˆ—1βˆ—(3)βˆ’22 * 1 * (3)^{-2}. Now, let's address that negative exponent. We know that (3)βˆ’2(3)^{-2} means 132\frac{1}{3^2}. So, let's calculate 323^2, which is 3βˆ—3=93 * 3 = 9. Thus, (3)βˆ’2=19(3)^{-2} = \frac{1}{9}. Our expression is now 2βˆ—1βˆ—192 * 1 * \frac{1}{9}. Finally, let's multiply everything together. We have 2βˆ—1=22 * 1 = 2, and then 2βˆ—19=292 * \frac{1}{9} = \frac{2}{9}. And there you have it, folks! The value of the expression 2x0yβˆ’22x^0y^{-2} when x=2x = 2 and y=3y = 3 is 29\frac{2}{9}. Pretty straightforward, right? We just applied the exponent rules and did a little bit of arithmetic. That's the beauty of math; even complex-looking expressions can be simplified with the right knowledge. Remember the steps: substitute the values, simplify the terms, and then calculate. Practice with different numbers, and you'll become a pro in no time.

Detailed Breakdown and Explanation

Let's break it down even further, just to make sure everyone is on the same page. We started with 2x0yβˆ’22x^0y^{-2}. First, we substituted x with 2 and y with 3: 2βˆ—(2)0βˆ—(3)βˆ’22 * (2)^0 * (3)^{-2}. Now, the exponent rules come into play. A number raised to the power of 0 is always 1, so (2)0=1(2)^0 = 1. This simplifies our expression to 2βˆ—1βˆ—(3)βˆ’22 * 1 * (3)^{-2}. Next, let's tackle the negative exponent. A negative exponent tells us to take the reciprocal of the base raised to the positive exponent. So, (3)βˆ’2=132(3)^{-2} = \frac{1}{3^2}. Then, we calculate 323^2, which is 3βˆ—3=93 * 3 = 9. So, (3)βˆ’2=19(3)^{-2} = \frac{1}{9}. Now, our expression is 2βˆ—1βˆ—192 * 1 * \frac{1}{9}. Finally, we multiply all the terms together. 2βˆ—1=22 * 1 = 2, and 2βˆ—19=292 * \frac{1}{9} = \frac{2}{9}. The answer is 29\frac{2}{9}. You can think of it like this: first, we eliminate the variable using the value given. Then, we simplify by applying the exponent rules. After that, we calculate the remaining numerical expression. This is a common process in algebra, and it's essential for solving more advanced problems. This approach, in which we plug in numbers for variables and simplify, is a cornerstone of algebra. Make sure you practice these types of problems so you feel comfortable with them. If you can handle these, you can handle almost anything they throw at you! Also, keep in mind that the variables are only placeholders. We can swap them out for the numbers we're given, and the math still works the same way. It's like a secret code, and we're about to crack it!

Understanding Exponents and Variables

Now, let's dive a bit deeper into the concepts of exponents and variables, as these are fundamental to understanding the problem. Variables, like x and y, are simply symbols that represent unknown values or values that can change. In our case, we know the values of x and y, which makes our job easier. Exponents, on the other hand, tell us how many times to multiply a number by itself. For example, 323^2 means 3 multiplied by itself twice (3βˆ—33 * 3). The rules of exponents, like the zero power rule and the negative exponent rule, are crucial. The zero power rule states that any non-zero number raised to the power of 0 is equal to 1. The negative exponent rule tells us that a base raised to a negative exponent is equal to the reciprocal of the base raised to the positive exponent. We can think of this as moving the base to the opposite side of the fraction bar and changing the sign of the exponent. So, aβˆ’n=1ana^{-n} = \frac{1}{a^n}. Understanding these concepts will allow us to handle algebraic equations like a pro. With a solid grasp of exponents and variables, you can begin to work on more complex algebraic equations. This foundation will serve you well as you go further in math.

Conclusion: The Final Answer

So, after all the calculations, we've arrived at our final answer: the value of the expression 2x0yβˆ’22x^0y^{-2} when x=2x = 2 and y=3y = 3 is 29\frac{2}{9}. This result corresponds to option A in the multiple-choice question. Congratulations, guys, on making it through this problem! Remember, practice is key. Try some similar problems with different numbers and expressions to solidify your understanding. The more you practice, the more comfortable you'll become with algebraic expressions. Keep up the great work, and happy calculating!

Summary of Key Points

Here’s a quick recap of the key points we covered:

  • Substitution: We replaced the variables x and y with their given values.
  • Zero Exponent Rule: Any number raised to the power of zero equals one.
  • Negative Exponent Rule: A negative exponent means to take the reciprocal.
  • Calculation: We performed the arithmetic operations to arrive at the final answer.
  • Final Answer: The value of the expression is 29\frac{2}{9}.

By following these steps, you can confidently solve similar problems. Math is all about understanding the concepts and applying them in a systematic way. With consistent practice, you'll become more confident in tackling various types of algebraic expressions. Keep practicing, and you'll surely ace those math tests and problems! Remember, math is like a muscle; the more you use it, the stronger it gets. So, keep flexing your math muscles, and you'll do great. Cheers to mastering mathematical expressions!