How To Find The LCD Of (x-5)/(x^2 - 7x + 12) + (x+3)/(x^2 + 2x - 15)

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In mathematics, particularly when dealing with algebraic fractions, finding the Least Common Denominator (LCD) is a crucial step in simplifying expressions and solving equations. The LCD allows us to combine fractions with different denominators, making them easier to work with. In this article, we will delve into the process of finding the LCD for the given expression: (xβˆ’5)x2βˆ’7x+12+(x+3)x2+2xβˆ’15{ \frac{(x-5)}{x^2 - 7x + 12} + \frac{(x+3)}{x^2 + 2x - 15} }. We will break down the steps, explain the underlying concepts, and provide a comprehensive guide to understanding and calculating the LCD in such scenarios.

Understanding the Least Common Denominator (LCD)

To effectively grasp the concept of the Least Common Denominator (LCD), it is essential to first understand what it represents and why it is so vital in mathematical operations involving fractions. The LCD is, in essence, the smallest multiple that is common to a set of denominators. When dealing with numerical fractions, this is a relatively straightforward concept. However, when algebraic expressions are involved, the process becomes a bit more intricate, requiring a solid understanding of factoring and algebraic manipulation.

The primary purpose of the LCD is to enable the addition and subtraction of fractions that have different denominators. Think of it like this: you can't directly add or subtract apples and oranges; you need a common unit. Similarly, you can't combine fractions with different denominators without first finding a common denominator. The LCD provides this common ground, allowing us to rewrite the fractions with the same denominator so that they can be combined. This process is fundamental in simplifying complex expressions and solving equations involving fractions. Without the LCD, many algebraic manipulations would be significantly more challenging, if not impossible. This concept is not just a theoretical tool; it is a practical necessity in various mathematical contexts, from basic algebra to advanced calculus. The LCD ensures that we can perform these operations accurately and efficiently. Thus, mastering the technique of finding the LCD is a cornerstone skill for anyone working with algebraic fractions.

Step-by-Step Guide to Finding the LCD

To find the LCD for the expression (xβˆ’5)x2βˆ’7x+12+(x+3)x2+2xβˆ’15{ \frac{(x-5)}{x^2 - 7x + 12} + \frac{(x+3)}{x^2 + 2x - 15} }, we will follow a systematic approach. This involves factoring the denominators, identifying common factors, and constructing the LCD from these factors. Let's break down each step:

1. Factor the Denominators

The first crucial step in finding the LCD is to factor each denominator completely. Factoring involves breaking down the quadratic expressions into their simplest factors. This process is essential because it allows us to identify the building blocks of each denominator, making it easier to find the common multiples. For the given expression, we have two denominators: xΒ² - 7x + 12 and xΒ² + 2x - 15. Let's factor each one individually.

For the first denominator, xΒ² - 7x + 12, we look for two numbers that multiply to 12 and add up to -7. These numbers are -3 and -4. Therefore, we can factor the quadratic expression as follows:

xΒ² - 7x + 12 = (x - 3)(x - 4)

Now, let's factor the second denominator, xΒ² + 2x - 15. We need to find two numbers that multiply to -15 and add up to 2. These numbers are 5 and -3. So, the factored form of the quadratic expression is:

xΒ² + 2x - 15 = (x + 5)(x - 3)

By factoring the denominators, we have transformed the original expressions into a more manageable form. This step is not just a preliminary task; it is the foundation upon which the LCD will be built. Accurate factoring is paramount to finding the correct LCD, which in turn is essential for correctly simplifying and solving the overall expression. With the denominators now factored, we can clearly see the individual factors that make up each expression, setting the stage for the next step in our process.

2. Identify Unique Factors

After factoring the denominators, the next step in finding the LCD is to identify the unique factors present in each denominator. This involves examining the factored forms and noting each distinct factor that appears. Understanding the unique factors is crucial because the LCD must include each of these factors to be a common multiple of both denominators. In our case, the factored denominators are:

  • (x - 3)(x - 4)
  • (x + 5)(x - 3)

Looking at these factored forms, we can identify the unique factors as (x - 3), (x - 4), and (x + 5). It's important to note that even though (x - 3) appears in both denominators, it is only counted once when identifying unique factors. The reason for this is that the LCD needs to include each factor to the highest power it appears in any of the denominators. Since (x - 3) appears only once in each denominator, we include it once in our list of unique factors.

Identifying these unique factors is a critical step because it ensures that the LCD will be a multiple of each denominator. If we were to miss a factor, the resulting expression would not be a common denominator, and we would not be able to combine the fractions correctly. This step is not merely a formality; it is a safeguard that ensures the mathematical integrity of our process. By carefully identifying each unique factor, we set the stage for constructing the LCD in the next step, ensuring that it will indeed be the least common multiple of the denominators.

3. Construct the LCD

Once the unique factors have been identified, the final step in determining the LCD is to construct it by multiplying these factors together. The LCD must include each unique factor to the highest power it appears in any of the denominators. This ensures that the LCD is a multiple of each denominator, allowing us to combine the fractions. In our case, the unique factors we identified are (x - 3), (x - 4), and (x + 5). Since each of these factors appears only once in the factored denominators, we simply multiply them together to form the LCD.

Therefore, the LCD is:

(x - 3)(x - 4)(x + 5)

This expression is the Least Common Denominator for the given fractions. It is the smallest expression that is divisible by both (x - 3)(x - 4) and (x + 5)(x - 3). By using this LCD, we can rewrite the original fractions with a common denominator, which is necessary for adding or subtracting them. The process of constructing the LCD is the culmination of the previous steps. It brings together the individual factors in a way that ensures the resulting expression meets the criteria of being a common multiple. This step is not just a matter of mathematical procedure; it is a synthesis of understanding the factors and their roles in creating a common denominator.

Applying the LCD

Having found the LCD, the next step is to apply it to the original expression. This involves rewriting each fraction with the LCD as its denominator. To do this, we multiply both the numerator and the denominator of each fraction by the factors that are missing from its original denominator, compared to the LCD. This process ensures that the value of the fraction remains unchanged while allowing us to combine the fractions under a common denominator.

For the given expression, (xβˆ’5)x2βˆ’7x+12+(x+3)x2+2xβˆ’15{ \frac{(x-5)}{x^2 - 7x + 12} + \frac{(x+3)}{x^2 + 2x - 15} }, we found the LCD to be (x - 3)(x - 4)(x + 5). Now, let’s rewrite each fraction with this LCD.

The first fraction is (xβˆ’5)x2βˆ’7x+12{ \frac{(x-5)}{x^2 - 7x + 12} }, which we factored as (xβˆ’5)(xβˆ’3)(xβˆ’4){ \frac{(x-5)}{(x - 3)(x - 4)} }. To get the LCD, we need to multiply both the numerator and the denominator by the missing factor, which is (x + 5):

(xβˆ’5)(xβˆ’3)(xβˆ’4)Γ—(x+5)(x+5)=(xβˆ’5)(x+5)(xβˆ’3)(xβˆ’4)(x+5){ \frac{(x-5)}{(x - 3)(x - 4)} \times \frac{(x + 5)}{(x + 5)} = \frac{(x-5)(x + 5)}{(x - 3)(x - 4)(x + 5)} }

Similarly, for the second fraction, (x+3)x2+2xβˆ’15{ \frac{(x+3)}{x^2 + 2x - 15} }, which we factored as (x+3)(x+5)(xβˆ’3){ \frac{(x+3)}{(x + 5)(x - 3)} }, we need to multiply both the numerator and the denominator by the missing factor, which is (x - 4):

(x+3)(x+5)(xβˆ’3)Γ—(xβˆ’4)(xβˆ’4)=(x+3)(xβˆ’4)(xβˆ’3)(xβˆ’4)(x+5){ \frac{(x+3)}{(x + 5)(x - 3)} \times \frac{(x - 4)}{(x - 4)} = \frac{(x+3)(x - 4)}{(x - 3)(x - 4)(x + 5)} }

Now that both fractions have the same denominator, the LCD, we can combine them. This process of applying the LCD is a critical step in simplifying the expression. It transforms the fractions into a form where they can be easily added or subtracted, paving the way for further simplification and potentially solving for the variable. The careful application of the LCD ensures that the fractions are manipulated correctly, maintaining their values while making them compatible for combination. This skill is essential in various algebraic manipulations and is a cornerstone of working with rational expressions.

Simplifying the Expression

With the fractions now sharing a common denominator, we can proceed to simplify the expression by combining the numerators. This involves adding or subtracting the numerators while keeping the LCD as the denominator. After combining the numerators, we can further simplify the expression by expanding any products and combining like terms. This step is crucial for reducing the expression to its simplest form, making it easier to analyze and use in further calculations.

In our case, we have rewritten the original expression as:

(xβˆ’5)(x+5)(xβˆ’3)(xβˆ’4)(x+5)+(x+3)(xβˆ’4)(xβˆ’3)(xβˆ’4)(x+5){ \frac{(x-5)(x + 5)}{(x - 3)(x - 4)(x + 5)} + \frac{(x+3)(x - 4)}{(x - 3)(x - 4)(x + 5)} }

Now, let’s combine the numerators:

(xβˆ’5)(x+5)+(x+3)(xβˆ’4)(xβˆ’3)(xβˆ’4)(x+5){ \frac{(x-5)(x + 5) + (x+3)(x - 4)}{(x - 3)(x - 4)(x + 5)} }

Next, we expand the products in the numerator:

(x2βˆ’25)+(x2βˆ’xβˆ’12)(xβˆ’3)(xβˆ’4)(x+5){ \frac{(x^2 - 25) + (x^2 - x - 12)}{(x - 3)(x - 4)(x + 5)} }

Now, combine like terms in the numerator:

2x2βˆ’xβˆ’37(xβˆ’3)(xβˆ’4)(x+5){ \frac{2x^2 - x - 37}{(x - 3)(x - 4)(x + 5)} }

At this point, we have simplified the expression by combining the fractions and reducing the numerator to its simplest form. The simplified expression provides a clearer picture of the relationship between the variables and constants involved. This step of simplification is not just a cosmetic improvement; it often reveals underlying structures and patterns that were not immediately apparent in the original expression. Moreover, a simplified expression is generally easier to work with in subsequent steps, such as solving equations or evaluating the expression for specific values of the variable. Thus, simplification is a vital part of the mathematical process, enhancing both our understanding and our ability to manipulate algebraic expressions effectively.

Conclusion

In conclusion, finding the Least Common Denominator (LCD) is a fundamental skill in algebra, particularly when dealing with rational expressions. By factoring the denominators, identifying unique factors, and constructing the LCD, we can effectively combine fractions and simplify complex expressions. In the given example, the LCD of (xβˆ’5)x2βˆ’7x+12+(x+3)x2+2xβˆ’15{ \frac{(x-5)}{x^2 - 7x + 12} + \frac{(x+3)}{x^2 + 2x - 15} } is (x - 3)(x - 4)(x + 5). This process not only allows us to perform arithmetic operations on fractions but also provides a deeper understanding of algebraic structures and relationships. Mastering the concept of the LCD is essential for success in higher-level mathematics and its applications.

Final Answer:

The final answer is (x - 3)(x - 4)(x + 5). Option 2 is the correct answer.