Simplifying Algebraic Expressions: A Step-by-Step Guide

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Simplifying Algebraic Expressions: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving into the exciting world of algebraic expressions. We'll be tackling a problem that involves simplifying a complex expression with fractions and division. Our goal? To find an equivalent expression from the given options. So, let's roll up our sleeves and get started! The question is: Which expression is equivalent to 2a+110aβˆ’5Γ·10a4a2βˆ’1\frac{2 a+1}{10 a-5} \div \frac{10 a}{4 a^2-1}?

Understanding the Problem

Before we jump into the solution, let's break down the problem. We're given a fraction divided by another fraction. Remember, dividing by a fraction is the same as multiplying by its reciprocal. So, our first step will be to flip the second fraction and change the division sign to multiplication. This is a crucial concept. We must understand it.

  • Original Expression: 2a+110aβˆ’5Γ·10a4a2βˆ’1\frac{2 a+1}{10 a-5} \div \frac{10 a}{4 a^2-1}
  • Reciprocal: To find the reciprocal of 10a4a2βˆ’1\frac{10 a}{4 a^2-1}, we simply swap the numerator and the denominator, resulting in 4a2βˆ’110a\frac{4 a^2-1}{10 a}.
  • Rewritten Expression: 2a+110aβˆ’5Γ—4a2βˆ’110a\frac{2 a+1}{10 a-5} \times \frac{4 a^2-1}{10 a}

Now, we have a multiplication problem. But before we start multiplying, let's see if we can simplify each fraction further. This is where factoring comes into play. Factoring makes the problem easier and more manageable. Always look for simplification opportunities before diving into complex multiplication. This approach can save time and reduce the chances of errors. It's like finding shortcuts that make the journey smoother and more efficient.

Step-by-Step Solution

Alright, let's take this step by step. We'll break down the expressions and factor where possible. We'll analyze each part to see how we can simplify it. This systematic approach is key to solving these kinds of problems.

Factor the Denominators

  • Factor out common terms: In the denominator of the first fraction (10aβˆ’510a - 5), we can factor out a common factor of 5, which gives us 5(2aβˆ’1)5(2a - 1).
  • Factor the difference of squares: The numerator of the second fraction (4a2βˆ’14a^2 - 1) is a difference of squares. This can be factored into (2a+1)(2aβˆ’1)(2a + 1)(2a - 1).

Now, our expression looks like this: 2a+15(2aβˆ’1)Γ—(2a+1)(2aβˆ’1)10a\frac{2 a+1}{5(2 a-1)} \times \frac{(2 a+1)(2 a-1)}{10 a}.

Simplify the Expression

Now that we've factored, we can start simplifying. We're looking for common factors in the numerators and denominators that can cancel each other out. This process is essential for simplifying algebraic fractions. We want to reduce the expression to its simplest form.

  • Cancel common factors: Notice that we have a (2aβˆ’1)(2a - 1) in both the numerator and the denominator. We can cancel these out. Also, we can see a common factor of (2a+1)(2a + 1).

  • Simplified expression: After canceling out the common factors, we get: (2a+1)5Γ—(2aβˆ’1)10a\frac{(2 a+1)}{5} \times \frac{(2 a-1)}{10 a}.

  • Multiply the fractions: Now, we multiply the remaining fractions. This yields (2a+1)(2aβˆ’1)50a\frac{(2a+1)(2a-1)}{50a}.

Comparing with the Options

  • Simplify the numerator: The numerator, (2a+1)(2aβˆ’1)(2a + 1)(2a - 1), can be expanded back to 4a2βˆ’14a^2 - 1.

  • Our Simplified Expression: 4a2βˆ’150a\frac{4 a^2-1}{50 a}

If we compare this final simplified form with the given options, we find that none of them match exactly. However, there might be an error in the given options or in our simplification process. But based on our meticulous step-by-step approach, we've arrived at the most simplified form. It's crucial to ensure that each step is correct and to double-check for any missed opportunities for simplification or any calculation errors.

Answering the Question

So, after careful simplification, we found an expression that isn't exactly the same as any of the options. This might suggest a typo in the options provided. However, we've successfully simplified the original expression step-by-step. Remember, in algebra, accuracy and a systematic approach are key. Always double-check your work and make sure each step is logical. If you find any discrepancies, it's always a good idea to revisit your calculations.

Tips for Solving Similar Problems

Here are some handy tips to help you conquer similar algebraic expression problems:

  • Always look for factoring opportunities. Factoring is your best friend when simplifying fractions. It allows you to find common factors that can be canceled out. Always start by identifying common factors, differences of squares, or other factoring patterns.

  • Remember the rules of exponents. When multiplying or dividing terms with exponents, remember the rules. This helps in simplifying expressions accurately.

  • Double-check your work. After each step, take a moment to review your work. Small mistakes can lead to big problems. Ensure that you have not missed any opportunities for simplification.

  • Practice makes perfect. The more you practice, the more comfortable you will become with these types of problems. Work through various examples to build your skills and confidence.

By following these steps and tips, you'll be well-equipped to tackle any algebraic expression problem that comes your way. Keep practicing, stay curious, and you'll become a master of algebra in no time!

Conclusion

Today, we’ve taken a deep dive into simplifying algebraic expressions. We’ve learned how to factor, cancel common terms, and ultimately simplify complex fractions. Remember, practice is key! The more problems you solve, the more confident you'll become. Keep up the great work, and happy simplifying!