Problem 51
Question
Simplify the expression \(\frac{x}{x-1}-\frac{1}{2 x+1}\) (A) \(\frac{x-1}{(x-1)(2 x+1)}\) (B) \(-\frac{x}{x-1}\) (C) \(\frac{2 x^{2}+1}{(x-1)(2 x+1)}\) (D) \(\frac{2 x^{2}-1}{(x-1)(2 x+1)}\)
Step-by-Step Solution
Verified Answer
The simplified form of the given expression is \(\frac{2x^2+1}{(x-1)(2x+1)}\), which corresponds to option (C).
1Step 1: Find the Least Common Denominator (LCD)
The LCD is the least common multiple of the denominators. Here, the denominators are \(x-1\) and \(2x+1\). In this case, because the two denominators have no common factor other than 1, the LCD is simply the product of the two denominators, which is \((x-1)(2x+1)\).
2Step 2: Convert each fraction to have the LCD
Make each fraction equivalent to a fraction with the LCD. For the first fraction, multiply the numerator and the denominator by \(2x+1\), and for the second fraction, multiply the numerator and the denominator by \(x-1\). This gives us \(\frac{x(2x+1)}{(x-1)(2x+1)}\) and \(\frac{(x-1)}{(x-1)(2x+1)}\).
3Step 3: Subtract second fraction from the first
Now that the two fractions have the same denominator, they can be combined. This gives us \(\frac{x(2x+1) - (x-1)}{(x-1)(2x+1)}\).
4Step 4: Simplify the numerator
Expand the numerator and collect like terms to simplify it, yielding \(\frac{2x^2+x-x+1}{(x-1)(2x+1)}\), which simplifies to \(\frac{2x^2+1}{(x-1)(2x+1)}\).
Key Concepts
Least Common DenominatorEquivalent FractionsAlgebraic Fractions
Least Common Denominator
Understanding the Least Common Denominator (LCD) is essential when working with fractions, especially when trying to add, subtract, or compare them. The LCD is essentially the smallest number that can be evenly divided by all the denominators in question. It provides a common ground for combining fractions.
When simplifying algebraic expressions that involve fractions, like \(\frac{x}{x-1}-\frac{1}{2 x+1}\), finding the LCD is your first step. This involves looking at both denominators and calculating the least common multiple of these algebraic expressions. If they share no common factors or are not multiples of each other, the LCD will be their product.
In our exercise, the denominators \(x-1\) and \(2x+1\) are unique prime polynomials, meaning their LCD is \(x-1)(2x+1)\). This common denominator enables us to perform operations across the fractions by ensuring they're on the same 'playing field'. Missing this step can lead to errors and frustration, so grasping the concept of the LCD can make algebraic adventures much smoother!
When simplifying algebraic expressions that involve fractions, like \(\frac{x}{x-1}-\frac{1}{2 x+1}\), finding the LCD is your first step. This involves looking at both denominators and calculating the least common multiple of these algebraic expressions. If they share no common factors or are not multiples of each other, the LCD will be their product.
In our exercise, the denominators \(x-1\) and \(2x+1\) are unique prime polynomials, meaning their LCD is \(x-1)(2x+1)\). This common denominator enables us to perform operations across the fractions by ensuring they're on the same 'playing field'. Missing this step can lead to errors and frustration, so grasping the concept of the LCD can make algebraic adventures much smoother!
Equivalent Fractions
Moving on to equivalent fractions is like finding a different outfit that looks exactly the same—it's technically different but essentially identical in value. To work with fractions that have different denominators, we must create equivalent fractions that have the same denominator. This process does not change the value of the fraction; it merely expresses it in a form that is compatible with another.
Look at the terms \(\frac{x}{x-1}\) and \(\frac{1}{2 x+1}\). To combine them, we change both fractions to have their LCD, which we've identified as \(x-1)(2x+1)\). You achieve this by multiplying the numerator and denominator of each fraction by the factor that it's missing from the LCD. It's like balancing the scales to make sure each side is even.
The clever trick here is ensuring you perform the same operation on the top (numerator) and bottom (denominator) of each fraction. This use of equivalent fractions is a fundamental skill in algebra that simplifies complex expressions and helps solve equations with greater ease.
Look at the terms \(\frac{x}{x-1}\) and \(\frac{1}{2 x+1}\). To combine them, we change both fractions to have their LCD, which we've identified as \(x-1)(2x+1)\). You achieve this by multiplying the numerator and denominator of each fraction by the factor that it's missing from the LCD. It's like balancing the scales to make sure each side is even.
The clever trick here is ensuring you perform the same operation on the top (numerator) and bottom (denominator) of each fraction. This use of equivalent fractions is a fundamental skill in algebra that simplifies complex expressions and helps solve equations with greater ease.
Algebraic Fractions
Finally, combining the concept of Least Common Denominators and Equivalent Fractions brings us to algebraic fractions. These are simply fractions that include algebraic expressions within their numerators or denominators (or both). They follow the same rules as ordinary fractions but require some additional algebraic manipulation to simplify.
With algebraic fractions, we encounter variables and expressions that may seem daunting at first. To manage them, we apply all the rules of arithmetic fractions—finding common denominators, creating equivalent fractions, adding, subtracting, multiplying, and dividing—as needed.
However, there’s an extra layer of complexity, as we can have polynomial expressions within the fractions. Whenever you're working with algebraic fractions, it's important to simplify the numerator and denominator as much as possible. This often involves expanding and factoring polynomials, just as we combined and simplified \(\frac{x(2x+1) - (x-1)}{(x-1)(2x+1)}\) to get \(\frac{2x^2+1}{(x-1)(2x+1)}\) in our step-by-step solution.
Developing confidence with algebraic fractions opens up a world of solving complex equations and understanding deeper mathematical concepts. Each algebraic fraction is a mini-puzzle to be worked out—and with practice, solving these can become as intuitive as working with simple numerical fractions.
With algebraic fractions, we encounter variables and expressions that may seem daunting at first. To manage them, we apply all the rules of arithmetic fractions—finding common denominators, creating equivalent fractions, adding, subtracting, multiplying, and dividing—as needed.
However, there’s an extra layer of complexity, as we can have polynomial expressions within the fractions. Whenever you're working with algebraic fractions, it's important to simplify the numerator and denominator as much as possible. This often involves expanding and factoring polynomials, just as we combined and simplified \(\frac{x(2x+1) - (x-1)}{(x-1)(2x+1)}\) to get \(\frac{2x^2+1}{(x-1)(2x+1)}\) in our step-by-step solution.
Developing confidence with algebraic fractions opens up a world of solving complex equations and understanding deeper mathematical concepts. Each algebraic fraction is a mini-puzzle to be worked out—and with practice, solving these can become as intuitive as working with simple numerical fractions.
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