Problem 5 - Algebra and Trigonometry

Question

Consider the expression \(\frac{1}{x}-\frac{2}{x+1}-\frac{x}{(x+1)^{2}}\) (a) How many terms does this expression have? (b) Find the least common denominator of all the terms. (c) Perform the addition and simplify.

Step-by-Step Solution

Answer

(a) 3 terms. (b) LCD is \(x(x+1)^2\). (c) Simplified expression: \(\frac{x^2 - x + 1}{x(x+1)^2}\).

Step 1: Identify Terms in the Expression

The expression is \( \frac{1}{x} - \frac{2}{x+1} - \frac{x}{(x+1)^{2}} \). Identify each separate fraction as a term: 1) \( \frac{1}{x} \), 2) \( -\frac{2}{x+1} \), 3) \( -\frac{x}{(x+1)^{2}} \). Therefore, there are three terms in this expression.

Step 2: Find the Least Common Denominator (LCD)

The denominators in the expression are \( x \), \( x+1 \), and \( (x+1)^{2} \). The LCD must include each factor at its highest power. Therefore, the LCD is \( x (x+1)^{2} \).

Step 3: Rewrite Each Term with the LCD

Rewrite each fraction with the new denominator \( x (x+1)^{2} \). - \( \frac{1}{x} \) becomes \( \frac{(x+1)^{2}}{x(x+1)^{2}} \).- \( -\frac{2}{x+1} \) becomes \( \frac{-2x}{x(x+1)^{2}} \).- \( -\frac{x}{(x+1)^{2}} \) becomes \( \frac{-x}{x(x+1)^{2}} \).

Step 4: Combine the Fractions

Combine all the fractions under the common denominator:\[ \frac{(x+1)^{2} - 2x - x}{x(x+1)^{2}} \].

Step 5: Simplify the Expression

Expand \( (x+1)^{2} \) to get \( x^2 + 2x + 1 \). Substitute back: \[ \frac{x^2 + 2x + 1 - 2x - x}{x(x+1)^2} \]. Simplify the numerator: \[ x^2 + 2x + 1 - 2x - x = x^2 - x + 1 \].Final simplified expression: \[ \frac{x^2 - x + 1}{x(x+1)^2} \].

Key Concepts

Understanding Fractions

Fractions represent a part of a whole and are defined by two main components: the numerator, which is the top number, and the denominator, which is the bottom number. In the expression given,

• \( \frac{1}{x} \), where "1" is the numerator and "x" is the denominator.
• \(-\frac{2}{x+1} \), where "-2" is the numerator and "x+1" is the denominator.
• \(-\frac{x}{(x+1)^2} \), where "-x" is the numerator and "(x+1)^2" is the denominator.

Fractions can express different values based on the relationship between their numerators and denominators. In algebra, fractions are used to represent division and can be combined or simplified when they share a common denominator. These fractional expressions often need a bit more work when their denominators differ. Instead, they can be added or subtracted by first identifying a common denominator.

Finding the Least Common Denominator (LCD)

Finding the least common denominator (LCD) is crucial when combining fractions with different denominators. The LCD is the smallest multiple that the denominators of each fraction can divide into evenly without leaving a remainder. For the expression \( \frac{1}{x} - \frac{2}{x+1} - \frac{x}{(x+1)^2} \), the denominators are "x", "x+1", and "(x+1)^2".To determine the LCD:

• List each unique factor.
• Choose each factor at its highest power present in any term.
• The factors "x" and "(x+1)^2" are identified, with "(x+1)^2" being the highest power of "x+1".

Thus, the LCD in this case is \( x(x+1)^2 \). This unified denominator is then used to rewrite each fraction in terms of a common denominator, allowing you to combine them more easily.

Simplification of Expressions

Simplification involves the process of condensing expressions into their simplest form. After establishing a common denominator, you can rewrite each term to have this denominator. Using the LCD \( x(x+1)^2 \) from the previous step:- Rewrite \( \frac{1}{x} \) as \( \frac{(x+1)^2}{x(x+1)^2} \).- Rewrite \( -\frac{2}{x+1} \) as \( \frac{-2x}{x(x+1)^2} \).- Rewrite \( -\frac{x}{(x+1)^2} \) as \( \frac{-x}{x(x+1)^2} \).Combine these fractions into a single expression: \[\frac{(x+1)^2 - 2x - x}{x(x+1)^2}.\]Next, expand the binomial \((x+1)^2\) as \( x^2 + 2x + 1 \) and substitute back into the expression. Simplifying the numerator \( x^2 + 2x + 1 - 2x - x \),we get \( x^2 - x + 1 \).Thus, the simplified form of the initial expression is \( \frac{x^2 - x + 1}{x(x+1)^2} \). This process highlights the importance of being able to manage complex fractions by combining and simplifying until you get a more straightforward expression.