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Magazine Chapter 1: Problem 53
Perform the addition or subtraction and simplify. $$\frac{1}{x+3}+\frac{1}{x^{2}-9}$$
Short Answer
Expert verified
\( \frac{x-2}{(x-3)(x+3)} \)
Step by step solution
01
Identify the Denominators
The expression involves two fractions: \( \frac{1}{x+3} \) and \( \frac{1}{x^2 - 9} \). The denominators are \( x + 3 \) and \( x^2 - 9 \) respectively.
02
Factor the Denominators
Recognize that \( x^2 - 9 \) is a difference of squares. Therefore, factor it as \( (x-3)(x+3) \). This means we have \( \frac{1}{x+3} + \frac{1}{(x-3)(x+3)} \).
03
Find the Least Common Denominator (LCD)
The least common denominator for the fractions is \((x-3)(x+3)\). This is because the second fraction already has this as its denominator, and the first fraction's denominator \( x+3 \) is part of it.
04
Rewrite Each Fraction with the Common Denominator
Adjust \( \frac{1}{x+3} \) to have the denominator \((x-3)(x+3)\). Multiply the numerator and denominator of the first fraction by \( x-3 \), resulting in \( \frac{x-3}{(x-3)(x+3)} \). The second fraction, \( \frac{1}{(x-3)(x+3)} \), already has the desired denominator.
05
Add the Fractions
Now the expression is \( \frac{x-3}{(x-3)(x+3)} + \frac{1}{(x-3)(x+3)} \). Combine the numerators over the common denominator: \( \frac{x-3+1}{(x-3)(x+3)} = \frac{x-2}{(x-3)(x+3)} \).
06
Simplify the Result
The resulting fraction \( \frac{x-2}{(x-3)(x+3)} \) cannot be simplified further, as there are no common factors between the numerator and the denominator.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference of Squares
The difference of squares is a neat and handy mathematical trick that helps us simplify expressions, including rational expressions like the one in this exercise. An expression that looks like \( a^2 - b^2 \) fits the formula for the difference of squares. This is because it can be factored into two binomials: \( (a-b)(a+b) \).
In our exercise, the expression \( x^2 - 9 \) is a classic example of the difference of squares:
In our exercise, the expression \( x^2 - 9 \) is a classic example of the difference of squares:
- \( x^2 \) is the square of \( x \).
- \( 9 \) is the square of \( 3 \), since \( 9 = 3^2 \).
Least Common Denominator
The least common denominator (LCD) is the smallest polynomial that can serve as the denominator for all fractions involved in solving an expression. Having a common denominator allows us to easily add or subtract rational expressions.
In this problem, we have two denominators: \( x+3 \) and \( (x-3)(x+3) \). The LCD for these fractions needs to encompass each distinct factor present in the denominators.
In this problem, we have two denominators: \( x+3 \) and \( (x-3)(x+3) \). The LCD for these fractions needs to encompass each distinct factor present in the denominators.
- The denominator \( x+3 \) is already a factor in \((x-3)(x+3)\).
- Thus, the least common denominator is \( (x-3)(x+3) \).
Factorization
Factorization is all about breaking down numbers or expressions into multiples that, when combined, produce the original number or expression. It's a critical step in simplifying complex algebraic expressions.
For the rational expression in the exercise, factoring helps to find the least common denominator. In this case, the expression \( x^2 - 9 \) had to be factored:
For the rational expression in the exercise, factoring helps to find the least common denominator. In this case, the expression \( x^2 - 9 \) had to be factored:
- We recognized it as a difference of squares: \( (x-3)(x+3) \).
