/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Simplify the expression. $$\fr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Simplify the expression. $$\frac{\frac{x+2}{x}-\frac{a+2}{a}}{x-a}$$

Short Answer

Expert verified
\( \frac{-2}{xa} \)

Step by step solution

01

Identify the Complex Fraction

The expression given is a complex fraction: \( \frac{\frac{x+2}{x}-\frac{a+2}{a}}{x-a} \). Our goal is to simplify it by reducing it to simpler terms. The numerator of the complex fraction involves subtracting two separate fractions.
02

Find a Common Denominator

To simplify the expression \( \frac{x+2}{x} - \frac{a+2}{a} \), we must first find a common denominator for the fractions in the numerator. The common denominator of \( x \) and \( a \) is \( xa \).
03

Rewrite Each Fraction

Each fraction must be rewritten with the common denominator: \[ \frac{x+2}{x} = \frac{(x+2)a}{xa} \] \[ \frac{a+2}{a} = \frac{(a+2)x}{xa} \] So, the expression becomes: \[ \frac{(x+2)a - (a+2)x}{xa} \]
04

Subtract the Numerators

With the common denominator, subtract the numerators of the fractions:\[ (x+2)a - (a+2)x = xa + 2a - ax - 2x \] Simplify this to: \[ xa - ax + 2a - 2x = 2a - 2x \] Now the expression under the common denominator is: \[ \frac{2a - 2x}{xa} \]
05

Simplify the Complex Fraction

Insert the simplified numerator back into the original complex fraction: \[ \frac{\frac{2a - 2x}{xa}}{x-a} \] This is equal to multiplying by the reciprocal: \[ \frac{2a - 2x}{xa} \times \frac{1}{x-a} \] Thus, it simplifies to: \[ \frac{2(a-x)}{xa(x-a)} \]
06

Final Simplification

Notice that \( 2(a-x) = -2(x-a) \). Substitute and simplify: \[ \frac{-2(x-a)}{xa(x-a)} \] Cancel out \( (x-a) \) from the numerator and the denominator, giving:\[ \frac{-2}{xa} \]
07

Verify the Result

It's essential to check if cancelling is valid. Since \( x eq a \), division by \( x-a \) is valid, confirming the simplification is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simplifying Expressions
When we talk about simplifying expressions, especially complex fractions, the goal is to make them easier to handle and understand. A complex fraction is one where the numerator, the denominator, or both, are also fractions. To simplify such expressions, we work through a series of steps to turn them into simpler forms.

For example, given a complex fraction like \( \frac{\frac{x+2}{x}-\frac{a+2}{a}}{x-a} \), we aim to reduce it to a more straightforward form. This involves a process that includes finding common denominators, rewriting fractions with these denominators, and performing basic fraction operations like subtraction.

The ultimate goal of simplifying complex expressions is to make them easier to work with, understand, and solve. This process enhances clarity and facilitates further algebraic manipulation or evaluation. It’s a crucial skill in solving algebraic equations and problems effectively.
Common Denominator
Finding a common denominator is a key step in simplifying fractions, particularly when dealing with a complex fraction. The common denominator serves as a single base that allows you to add or subtract fractions easily.

In our example \( \frac{x+2}{x} - \frac{a+2}{a} \), the denominators are \( x \) and \( a \). The simplest common denominator for these terms is \( xa \), as it is the product of the two denominators.

By rewriting each fraction over this common denominator, we align them
  • \( \frac{x+2}{x} = \frac{(x+2)a}{xa} \)
  • \( \frac{a+2}{a} = \frac{(a+2)x}{xa} \)
This sets up the fractions for proper subtraction or addition. Using common denominators simplifies the process of combining fractions and is a vital technique in algebra. It ensures that we manipulate expressions correctly and consistently.
Fraction Operations
Performing operations with fractions involves basic arithmetic skills such as addition, subtraction, multiplication, and division. Once you have fractions with a common denominator, it becomes straightforward to perform these operations.

In the exercise, after finding the common denominator \( xa \), we subtract the numerators:
  • Subtract \((x+2)a\) from \((a+2)x\) resulting in \( xa + 2a - ax - 2x \).
  • Simplify to \(2a - 2x\).
These operations transform the initial complex expression into something more manageable.

The process also involves multiplying the simplified fraction by the reciprocal of the denominator when dealing with division, as seen when transforming \( \frac{2a - 2x}{xa} \times \frac{1}{x-a} \) into \( \frac{2(a-x)}{xa(x-a)} \). Understanding these basic fraction operations is fundamental for solving complex algebraic problems.
Algebraic Manipulation
Algebraic manipulation involves altering expressions using techniques that adhere to algebraic rules, transforming them into equivalent structures that are simpler or more useful. In simplifying complex fractions, these manipulations play a crucial role.

For the expression \( \frac{-2(x-a)}{xa(x-a)} \), the manipulation involves:
  • Recognizing that \(2(a-x) = -2(x-a)\).
  • Canceling out common terms, \((x-a)\), in both the numerator and the denominator.
It’s important to ensure the term you’re canceling does not equal zero, which in our case is confirmed as valid since \( x eq a \).

Canceling elements that repeat in the numerator and denominator is a powerful technique, as it significantly simplifies expressions. These manipulations require understanding and applying fundamental algebraic operations and properties to achieve the desired simplicity and clarity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Simplify the expression, and rationalize the denominator when appropriate. $$\sqrt[5]{-64}$$

The table gives the numbers of minutes of daylight occurring at various latitudes in the northern hemisphere at the summer and winter solstices. $$\begin{array}{|c|c|c|}\hline \text { Latitude } & \text { Summer } & \text { Winter } \\\\\hline 0^{\circ} & 720 & 720 \\\\\hline 10^{\circ} & 755 & 685 \\\\\hline 20^{\circ} & 792 & 648 \\\\\hline 30^{\circ} & 836 & 604 \\\\\hline 40^{\circ} & 892 & 548 \\\\\hline 50^{\circ} & 978 & 462 \\\\\hline 60^{\circ} & 1107 & 333 \\\\\hline\end{array}$$ (a) Which of the following equations more accurately predicts the length of day at the summer solstice at latitude \(L ?\) (1) \(D_{1}=6.096 L+685.7\) (2) \(D_{2}=0.00178 L^{3}-0.072 L^{2}+4.37 L+719\) (b) Approximate the length of daylight at \(35^{\circ}\) at the summer solstice.

On a clear day, the distance \(d\) (in miles) that can be seen from the top of a tall building of height \(h\) (in feet) can be approximated by \(d=1.2 \sqrt{h} .\) Approximate the distance that can be seen from the top of the Chicago Sears Tower, which is 1454 feet tall.

The formula occurs in the indicated application. Solve for the specified variable. \(S=\pi r \sqrt{r^{2}+h^{2}}\) for \(h \quad\) (surface area of a cone)

Calorie reguirements The basal energy requirement for an individual indicates the minimum number of calories necessary to maintain essential life-sustaining processes such as circulation, regulation of body temperature, and respiration. Given a person's sex, weight \(w\) (in kilograms), height \(h\) (in centimeters), and age \(y\) (in years), we can estimate the basal energy requirement in calories using the following formulas, where \(C_{f}\) and \(C_{m}\) are the calories necessary for females and males, respectively: $$\begin{array}{l}C_{f}=66.5+13.8 w+5 h-6.8 y \\\C_{m}=655+9.6 w+1.9 h-4.7 y\end{array}$$ (a) Determine the basal energy requirements first for a 25 -year-old female weighing 59 kilograms who is 163 centimeters tall and then for a 55 -year-old male weighing 75 kilograms who is 178 centimeters tall. Discuss why, in both formulas, the coefficient for \(y\) is negative but the other coefficients are positive.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.