Equivalent Expression For (x^2+1)/3 Over (x+1)/3

by Alex Johnson 49 views

Understanding how to manipulate and simplify complex fractions is a crucial skill in mathematics. In this article, we'll break down the process of finding an equivalent expression for the given fraction x2+13x+13\frac{\frac{x^2+1}{3}}{\frac{x+1}{3}}. We will explore the underlying principles, step-by-step methods, and common pitfalls to avoid. By the end of this discussion, you'll not only be able to solve this specific problem but also gain a solid foundation for tackling similar challenges. Let's dive into the world of algebraic fractions!

Breaking Down the Complex Fraction

When dealing with complex fractions, the key is to remember that a fraction bar represents division. So, the given expression x2+13x+13\frac{\frac{x^2+1}{3}}{\frac{x+1}{3}} can be understood as dividing the fraction x2+13\frac{x^2+1}{3} by the fraction x+13\frac{x+1}{3}. This understanding is fundamental to simplifying the expression. Rewriting the complex fraction as a division problem is the first step toward finding an equivalent expression.

To accurately express the equivalent form, we need to remember the rule for dividing fractions. Dividing by a fraction is the same as multiplying by its reciprocal. This means we will invert the second fraction (the divisor) and then multiply it by the first fraction (the dividend). This transformation allows us to convert a complex division problem into a simpler multiplication problem, which is often easier to manage and solve. By carefully applying this rule, we can eliminate the complex fraction structure and move toward a simplified algebraic expression. This process not only helps in solving the immediate problem but also reinforces a core mathematical principle that applies across various algebraic manipulations.

The Division Rule

The division rule for fractions is a cornerstone of fraction manipulation. It states that to divide one fraction by another, you multiply the first fraction by the reciprocal of the second. In mathematical terms, if you have two fractions, ab{\frac{a}{b}} and cd{\frac{c}{d}}, then abΓ·cd=abβ‹…dc{\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}}. Here, dc{\frac{d}{c}} is the reciprocal of cd{\frac{c}{d}}. Applying this rule correctly is essential for simplifying complex fractions and avoiding common mistakes. The reciprocal is found by simply swapping the numerator and the denominator of the fraction. For instance, the reciprocal of 23{\frac{2}{3}} is 32{\frac{3}{2}}. Understanding and applying this rule is not just about solving problems; it’s about building a solid foundation in algebraic manipulation. It allows you to transform seemingly complex expressions into manageable forms, opening the door to further simplification and problem-solving. Remember, the key is to convert the division problem into a multiplication problem using the reciprocal, making the subsequent steps much clearer and easier to execute. This rule is not just a trick; it’s a fundamental principle rooted in the very definition of division as the inverse operation of multiplication.

Applying the Rule to Our Problem

Now, let’s apply the division rule to our specific problem: x2+13x+13\frac{\frac{x^2+1}{3}}{\frac{x+1}{3}}. Following the rule, we can rewrite this expression as the fraction x2+13\frac{x^2+1}{3} divided by the fraction x+13\frac{x+1}{3}. To perform this division, we multiply the first fraction by the reciprocal of the second fraction. The reciprocal of x+13\frac{x+1}{3} is 3x+1\frac{3}{x+1}. Therefore, our expression transforms into x2+13β‹…3x+1{\frac{x^2+1}{3} \cdot \frac{3}{x+1}}. This step is crucial because it converts the complex fraction into a simpler multiplication problem, making it easier to identify the correct equivalent expression from the given options. By applying the division rule, we've effectively navigated the initial complexity of the problem and set the stage for further simplification if needed. This process underscores the power of understanding and applying fundamental mathematical rules to untangle complex expressions and pave the way for solutions.

Evaluating the Options

Now that we've transformed the complex fraction into a multiplication problem, we can evaluate the given options to find the equivalent expression. Let's revisit the transformed expression: x2+13β‹…3x+1\frac{x^2+1}{3} \cdot \frac{3}{x+1}. This form is much easier to compare with the provided choices.

Analyzing Option A

Option A states: x2+13β‹…x+13\frac{x^2+1}{3} \cdot \frac{x+1}{3}. This option incorrectly multiplies the original fractions without inverting the second fraction. Comparing this to our transformed expression, x2+13β‹…3x+1\frac{x^2+1}{3} \cdot \frac{3}{x+1}, we can see that the second fraction is not the reciprocal. Therefore, Option A is not equivalent to the original expression. The key difference lies in the numerator of the second fraction. In our transformed expression, it is 3, while in Option A, it is x+1{x+1}. This discrepancy highlights the importance of correctly applying the division rule for fractions.

Analyzing Option B

Option B presents: x+13Γ·x2+13\frac{x+1}{3} \div \frac{x^2+1}{3}. This option represents the division in the reverse order. While it does involve division, it does not accurately reflect the original complex fraction, which has x2+13\frac{x^2+1}{3} as the numerator and x+13\frac{x+1}{3} as the denominator. Dividing in reverse order will yield a completely different result. To confirm this, we can rewrite Option B using the division rule: x+13β‹…3x2+1\frac{x+1}{3} \cdot \frac{3}{x^2+1}. This is clearly not the same as our transformed expression, x2+13β‹…3x+1\frac{x^2+1}{3} \cdot \frac{3}{x+1}. Therefore, Option B is not the correct equivalent expression. The critical point here is the order of operations in division, which directly affects the outcome.

Identifying the Correct Option: Option C

Option C is given as: x2+13Γ·x+13\frac{x^2+1}{3} \div \frac{x+1}{3}. This option directly reflects our initial understanding of the complex fraction as a division problem. It correctly represents the fraction x2+13\frac{x^2+1}{3} being divided by the fraction x+13\frac{x+1}{3}. Applying the division rule, we transform this into x2+13β‹…3x+1\frac{x^2+1}{3} \cdot \frac{3}{x+1}, which perfectly matches our transformed expression. Therefore, Option C is the correct equivalent expression. This option demonstrates a clear understanding of how complex fractions translate into division problems and the subsequent application of the division rule. The equivalence is evident when we rewrite the division as multiplication by the reciprocal.

Ruling Out Option D

Finally, let's consider Option D: x+13β‹…x2+13\frac{x+1}{3} \cdot \frac{x^2+1}{3}. This option involves multiplication, but it multiplies the original fractions without any inversion. It's essentially the product of the two original fractions in the complex fraction without applying the division rule. Comparing this to our transformed expression, x2+13β‹…3x+1\frac{x^2+1}{3} \cdot \frac{3}{x+1}, we see that neither fraction has been inverted to its reciprocal. Therefore, Option D is not the equivalent expression. This option fails to recognize the division inherent in the complex fraction and thus does not apply the necessary reciprocal transformation. The correct application of the division rule is what distinguishes the correct answer from this incorrect option.

Conclusion

In conclusion, by carefully applying the division rule for fractions, we've determined that Option C, x2+13Γ·x+13\frac{x^2+1}{3} \div \frac{x+1}{3}, is the correct equivalent expression for the given complex fraction. This exercise highlights the importance of understanding fundamental mathematical principles and their application in simplifying complex problems. Remember, when dealing with complex fractions, the key is to recognize the division and apply the rule of multiplying by the reciprocal. By mastering these concepts, you'll be well-equipped to tackle a wide range of algebraic challenges.

For further exploration of fraction manipulation and algebraic expressions, you can visit Khan Academy's Algebra Section.