Chapter 10: Problem 42
Simplify \(\frac{x+1}{2 x-4}+\frac{x-1}{2 x+4}-\frac{2 x}{x^2-4}\). Write your answer in lowest terms. A. \(\frac{x^2-x+1}{2(x-1)(x+1)}\) B. \(\frac{2 x^2-4 x+4}{2(x-2)(x+2)}\) C. \(\frac{x^2-2 x+2}{(x-2)(x+2)}\) D. \(\frac{x^2-x+1}{(x-1)(x+1)}\)
Short Answer
Expert verified
C. \(\frac{x^2-2x+2}{(x-2)(x+2)}\)
Step by step solution
01
Identify a Common Denominator
Identify a single common denominator for all the fractions. The denominators in this problem are \(2x-4\), \(2x+4\), and \(x^2-4\). Notice that these can be written as \(2(x-2)\), \(2(x+2)\), and \((x-2)(x+2)\) respectively.
02
Simplify Each Denominator
Write each denominator in factored form: \[\frac{x+1}{2(x-2)} + \frac{x-1}{2(x+2)} - \frac{2x}{(x-2)(x+2)}.\]
03
Express Each Fraction with Common Denominator
Make equivalent fractions with a common denominator \[2(x-2)(x+2)\]. \[\frac{(x+1)(x+2)}{2(x-2)(x+2)} + \frac{(x-1)(x-2)}{2(x-2)(x+2)} - \frac{2x}{(x-2)(x+2)}.\]
04
Distribute Numerators
Distribute in the numerators. \[\frac{x^2 + 3x + 2}{2(x-2)(x+2)} + \frac{x^2 - 3x + 2}{2(x-2)(x+2)} - \frac{2x}{(x-2)(x+2)}.\]
05
Combine Like Terms
Combine the like terms in the numerators: \[\frac{(x^2 + 3x + 2) + (x^2 - 3x + 2) - 2x}{2(x-2)(x+2)}.\]
06
Simplify the Numerator
Combine all terms in the numerator: \[\frac{x^2 + x^2 + 3x - 3x + 2 + 2 - 2x}{2(x-2)(x+2)}\]\[\frac{2x^2 - 2x + 4}{2(x-2)(x+2)}.\]
07
Factor and Finalize
Factor the numerator if possible, and write the simplified fraction:Factor out the common factor of 2 in the numerator:\[\frac{2(x^2 - x + 2)}{2(x-2)(x+2)} = \frac{x^2 - x + 2}{(x-2)(x+2)}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and arithmetic operations. They can be very simple, like \(x + 2\), or more complex, such as \(2x^2 - 3x + 4\). Understanding how to manipulate and simplify algebraic expressions is crucial in GED math problems.
In our problem, we deal with several algebraic expressions within the fractions. The goal is to simplify the overall expression by first rewriting it in forms that make the arithmetic operations easier.
Algebraic expressions often require us to apply the distributive property, which states that \(a(b + c) = ab + ac\), to expand or factor terms. Itβs essential to keep track of all variables, coefficients, and signs through each step to avoid mistakes.
In our problem, we deal with several algebraic expressions within the fractions. The goal is to simplify the overall expression by first rewriting it in forms that make the arithmetic operations easier.
Algebraic expressions often require us to apply the distributive property, which states that \(a(b + c) = ab + ac\), to expand or factor terms. Itβs essential to keep track of all variables, coefficients, and signs through each step to avoid mistakes.
Fractions
Understanding fractions is fundamental in algebra, especially when simplifying complex expressions. A fraction consists of a numerator and a denominator. In algebra, these can be polynomial expressions. When working with algebraic fractions, the key is to find a common denominator to combine them easily.
In our example, the common denominator is found by factoring each denominator and identifying the least common multiple (LCM). The LCM of \(2(x-2)\) and \(2(x+2)\) is \(2(x-2)(x+2)\).
Once a common denominator is established, each fraction is expressed in terms of this denominator, allowing for straightforward addition or subtraction. Then we distribute and combine like terms to further simplify the expression.
In our example, the common denominator is found by factoring each denominator and identifying the least common multiple (LCM). The LCM of \(2(x-2)\) and \(2(x+2)\) is \(2(x-2)(x+2)\).
Once a common denominator is established, each fraction is expressed in terms of this denominator, allowing for straightforward addition or subtraction. Then we distribute and combine like terms to further simplify the expression.
Factoring
Factoring is a technique used to break down polynomials into products of simpler polynomials, making them easier to work with. For instance, the polynomial \(x^2 - 4\) can be factored into \( (x-2)(x+2) \), because \(x^2-4=(x+2)(x-2)\).
In our problem, we began by rewriting the denominators in their factored forms. This helps identify the common denominator. Factoring alone can also simplify an expression through operations on the numerators and denominators.
At the final step, we factor out any common terms in the numerator to ensure that the expression is in its simplest form. This involves looking at the numerator \(2(x^2 - x + 2)\) and removing the common factor of 2, leading us to the solution \frac{x^2 - x + 2}{(x-2)(x+2)}\
. This form is much easier to understand and compare to the multiple choice answers. In our problem, we began by rewriting the denominators in their factored forms. This helps identify the common denominator. Factoring alone can also simplify an expression through operations on the numerators and denominators.
At the final step, we factor out any common terms in the numerator to ensure that the expression is in its simplest form. This involves looking at the numerator \(2(x^2 - x + 2)\) and removing the common factor of 2, leading us to the solution \frac{x^2 - x + 2}{(x-2)(x+2)}\