/*! 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 9 Find expressions for the followi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 9

Find expressions for the following algebraic fractions, simplifying your answers as far as possible. (a) \(\frac{5}{x-1} \times \frac{x-1}{x+2}\) (b) \(\frac{x^{2}}{x+10} \div \frac{x}{x+1}\) (c) \(\frac{4}{x+1}+\frac{1}{x+1}\) (d) \(\frac{2}{x+1}-\frac{1}{x+2}\)

Short Answer

Expert verified
a) \(\frac{5}{x+2}\) \ b) \(\frac{x^{2} + x}{x + 10}\) \ c) \(\frac{5}{x+1}\) \ d) \(\frac{x+3}{(x+1)(x+2)}\)

Step by step solution

01

- Simplify the first expression (a)

Consider the expression \(\frac{5}{x-1} \times \frac{x-1}{x+2}\). First, cancel the common factor \(x-1\) in the numerator and denominator. The simplified result is: \(\frac{5}{x+2}\).
02

- Simplify the second expression (b)

Consider the expression \(\frac{x^{2}}{x+10} \div \frac{x}{x+1}\). Division of fractions is equivalent to multiplying by the reciprocal. Thus, the expression becomes: \(\frac{x^{2}}{x+10} \times \frac{x+1}{x}\). Next, cancel the common factor \x\ in the numerator and denominator, resulting in: \(\frac{x(x+1)}{x+10}\). The simplified result is: \( \frac{x^{2} + x}{x + 10}\).
03

- Simplify the third expression (c)

Consider the expression \(\frac{4}{x+1}+\frac{1}{x+1}\). Both fractions have the same denominator. Combine the numerators, resulting in: \(\frac{4+1}{x+1}\). The simplified result is: \(\frac{5}{x+1}\).
04

- Simplify the fourth expression (d)

Consider the expression \(\frac{2}{x+1}-\frac{1}{x+2}\).The fractions have different denominators. In this case, use a common denominator \(x+1)(x+2)\). Rewrite the fractions: \( \frac{2(x+2)}{(x+1)(x+2)} - \frac{1(x+1)}{(x+1)(x+2)}\). Combine the numerators over the common denominator: \( \frac{2x+4-x-1}{(x+1)(x+2)} \). Simplify the numerator: \( \frac{x+3}{(x+1)(x+2)} \).

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 91影视!

Key Concepts

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

Fraction Simplification
Simplifying fractions means making them as simple as possible. The goal is to make both the numerator and the denominator smaller while keeping the fraction equivalent to the original one. For algebraic fractions, simplification often involves canceling out common factors in both parts of the fraction.

Let's look at an example: \(\frac{5}{x-1} \times \frac{x-1}{x+2}\).
First, identify common factors in both the numerator and denominator. In this case, \(x - 1\) appears in both parts, so we can cancel it out.

This leaves us with \(\frac{5}{x+2}\). Simple, right?
Common Denominator
When adding or subtracting fractions, having a common denominator is essential. A common denominator is like a common ground for the fractions to stand on, making it easier to combine them.

Take the example: \(\frac{2}{x+1} - \frac{1}{x+2}\).
Here, the denominators are different. To subtract these fractions, we need to find a common denominator, which is often the product of the given denominators: \(x+1)(x+2)\).

Rewrite each fraction with this common denominator: \(\frac{2(x+2)}{(x+1)(x+2)} - \frac{1(x+1)}{(x+1)(x+2)}\).
Now, it's much easier to combine them into one fraction: \(\frac{2x+4-x-1}{(x+1)(x+2)}\), which simplifies to \(\frac{x+3}{(x+1)(x+2)}\).
Numerator and Denominator
Understanding the roles of numerators and denominators is crucial for working with fractions. The numerator is the top part of a fraction, indicating how many parts we have. The denominator is the bottom part, showing how many parts make a whole.

For instance, in the fraction \( \frac{4}{x+1} + \frac{1}{x+1} \), both fractions have the same denominator (\(x+1\). This denominator tells us that each part of the fraction is divided by \(x+1\).

To add these fractions, simply add the numerators (4 and 1), keeping the common denominator: \( \frac{4+1}{x+1} \). The result is \( \frac{5}{x+1} \).

In another example like \( \frac{x^{2}}{x+10} \), the numerator (\(x^{2}\)) tells us how many parts are being considered, and the denominator (\(x+10\)) tells us that those parts are out of \(x+10\) equal parts.

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

Find expressions for the following fractions: (a) \(\frac{x^{2}+6 x}{x-2} \times \frac{x-2}{x}\) (b) \(\frac{1}{x} \div \frac{1}{x+1}\) (c) \(\frac{2}{x y}+\frac{3}{x y}\) (d) \(\frac{x}{2}+\frac{x+1}{3}\) (e) \(\frac{3}{x}+\frac{4}{x+1}\) (f) \(\frac{3}{x}+\frac{5}{x^{2}}\) (g) \(x-\frac{2}{x+1}\) (h) \(\frac{5}{x(x+1)}-\frac{2}{x}+\frac{3}{x+1}\)

Attempt to solve the following systems of equations (a) \(3 x-6 y=-2\) (b) \(-5 x+y=4\) $$ -4 x+8 y=-1 $$ $$ 10 x-2 y=-8 $$ Comment on the nature of the solution in each case.

(Excel) The supply and demand functions of a good are given by $$ \begin{aligned} &P=-Q_{\mathrm{D}}+240 \\ &P=60+2 Q_{\mathrm{s}} \end{aligned} $$ where \(P, Q_{D}\) and \(Q_{5}\) denote price, quantity demanded and quantity supplied, respectively. Sketch graphs of both functions on the same diagram, on the range \(0 \leq Q \leq 80\) and hence find the equilibrium price. The government now imposes a fixed tax, \(\$ 60\), on each good. Draw the new supply equation on the same diagram and hence find the new equilibrium price. What fraction of the \(\$ 60\) tax is paid by the consumer? Consider replacing the demand function by the more general equation $$ P=-k Q_{\mathrm{D}}+240 $$ By repeating the calculations above, find the fraction of the tax paid by the consumer for the case when \(k\) is (a) 2 (b) 3 (c) 4 State the connection between this fraction and the value of \(k\). Use this connection to predict how much tax is paid by the consumer when \(k=6\).

Starting with the true statement $$ 6>3 $$ decide which of the following are valid operations when performed on both sides: (a) add 6 (b) multiply by 2 (c) subtract 3 (d) add -3 (e) divide by 3 (f) multiply by \(-4\) (g) multiply by \(-1\) (h) divide by \(-3\) (i) add - 10

The demand function of a good is $$ Q=100-P+2 Y+1 / 2 A $$ where \(Q, P, Y\) and A denote quantity demanded, price, income and advertising expenditure respectively. (a) Calculate the demand when \(P=10, Y=40\) and \(A=6\). Assuming that price and income are fixed, calculate the additional advertising expenditure needed to raise demand to 179 units. (b) Is this good inferior or superior?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation
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.