Problem 19
Question
Find the \(L C D\) for each list of rational expressions. $$ \frac{2}{x+3}, \frac{5}{x-2} $$
Step-by-Step Solution
Verified Answer
The LCD is \((x+3)(x-2)\).
1Step 1: Identify the Denominators
First, observe the denominators of the provided rational expressions. For the expressions \(\frac{2}{x+3}\) and \(\frac{5}{x-2}\), the denominators are \(x+3\) and \(x-2\).
2Step 2: Factor the Denominators
Next, factor each denominator if possible. Both \(x+3\) and \(x-2\) are already in their simplest form as linear factors.
3Step 3: Determine the LCD
The Least Common Denominator (LCD) is the product of the distinct factors from all denominators. Here, the denominators are \(x+3\) and \(x-2\). Thus, the LCD is the product: \((x+3)(x-2)\).
4Step 4: Verify the LCD with Each Fraction
To verify, consider each rational expression: \(\frac{2}{x+3}\) and \(\frac{5}{x-2}\), and ensure the LCD \((x+3)(x-2)\) includes the necessary factors to convert each denominator into the LCD.
Key Concepts
Rational ExpressionsFactoring DenominatorsLinear Factors
Rational Expressions
Rational expressions can be a fascinating journey of numbers! They're similar to fractions but use polynomials in their numerators and denominators instead of simple integers. When dealing with rational expressions, you might encounter functions like \( \frac{2}{x+3} \) or \( \frac{5}{x-2} \). These show just how versatile mathematical expressions can be. This makes solving problems a bit like assembling a puzzle, where each piece must fit perfectly.
Key points to remember when dealing with rational expressions include:
Key points to remember when dealing with rational expressions include:
- They're undefined when their denominators are zero. For example, in \( \frac{2}{x+3} \), if \( x = -3 \), the expression becomes undefined.
- Also, simplifying rational expressions involves factoring and canceling common factors in numerators and denominators.
- The concepts of Least Common Denominator (LCD) often become necessary for performing operations such as addition or subtraction.
Factoring Denominators
Factoring the denominator is a crucial step when working with rational expressions. By transforming a polynomial denominator into its prime factors, we can simplify the expression or find the Least Common Denominator with other rational expressions.
In the given problem, the denominators \( x+3 \) and \( x-2 \) are already linear factors in their simplest form. Here’s why factoring is essential:
In the given problem, the denominators \( x+3 \) and \( x-2 \) are already linear factors in their simplest form. Here’s why factoring is essential:
- Factoring reduces complex fractions into simpler forms, which is handy for future operations like addition or subtraction. Instead of dealing with complicated terms, we target more manageable elements.
- It reveals common factors, if any, which can be canceled out to simplify the expression further.
Linear Factors
Understanding linear factors is essential when dealing with expressions like \( x+3 \) or \( x-2 \). Linear factors are polynomials of degree one; they appear as straightforward expressions that significantly affect how we handle rational expressions.
In our case, both provided denominators \( x+3 \) and \( x-2 \) are perfect examples. Why are linear factors important? Consider these reasons:
In our case, both provided denominators \( x+3 \) and \( x-2 \) are perfect examples. Why are linear factors important? Consider these reasons:
- Linear factors simplify calculations. You can quickly identify the needed factors for the least common denominator or simplify complicated expressions.
- They help identify solutions for expressions when set equal to zero. For instance, \( x+3=0 \) gives \( x=-3 \), a crucial point for understanding where an expression might be undefined.
- Since linear factors are straightforward, they form the building blocks for solving more complex polynomial equations.
Other exercises in this chapter
Problem 18
Solve the following. Twelve divided by the sum of \(x\) and 2 equals the quotient of 4 and the difference of \(x\) and 2 . Find \(x\).
View solution Problem 18
Perform each indicated operation. Simplify if possible. \(\frac{-9}{25 x^{2}-1}+\frac{7}{1-25 x^{2}}\)
View solution Problem 19
Find any numbers for which each rational expression is undefined. $$ \frac{3 x^{2}+9}{x^{2}-5 x-6} $$
View solution Problem 19
Simplify each complex fraction. $$ \frac{\frac{1}{5}-\frac{1}{x}}{\frac{7}{10}+\frac{1}{x^{2}}} $$
View solution