/*! 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 51 Solve the inequality. Then graph... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 51

Solve the inequality. Then graph the solution set. $$\frac{3}{x-1}+\frac{2 x}{x+1}>-1$$

Short Answer

Expert verified
The solution to the inequality is \(x \in (-鈭�, -4) \cup (-4, -1)\), excluding \(x = 1\) and \(x = -1\).

Step by step solution

01

Combine the fractions

To solve this inequality, first rewrite it as \[ \frac{3(x+1) + 2x(x-1)}{(x-1)(x+1)} > -1 \]Which simplifies to \[ \frac{5x+3}{x^2 - 1} > - 1 \]
02

Get rid of fraction

Multiply through by \(x^2 - 1\) to get rid of the fraction. Be careful here because \(x^2 - 1\) could be negative, depending on the value of \(x\). We'll have to consider it in the next step. For now, just multiply through: \[ 5x+3 > - (x^2 - 1) \] which simplifies to\[ x^2 + 5x + 4 < 0 \]
03

Factorise the quadratic

Factoring the left-hand side yields:\[ (x+4)(x+1) < 0 \]
04

Find the intervals

To find the intervals, we take the values of \(x\) that make the inequality equal to zero, namely \(x = -4\) and \(x = -1\) Thus, our intervals are \((-鈭�, -4)\), \((-4, -1)\), and \((-1, 鈭�)\)
05

Test the interval

Take a test point in each interval and substitute it into the inequality. If the inequality is satisfied, then the interval is in our solution set. \nFrom our test, we find that \((-4, -1)\) and \((-鈭�, -4)\) satisfy the inequality.
06

Consider excluded values

Also, remember from Step 2 that we can't have \(x = 1\) or \(x = -1\) because those values would make the denominator zero. So, we add \(x = 1\) and \(x = -1\) to our excluded values. This doesn't change the solution set.
07

Graph the solution set

Mark -4 and -1 on the number line and draw an interval from -鈭� to -4, excluding -4 and from -4 to -1, excluding -1. The result is a graphical representation of our solution set.

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!

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

Write the polynomial (a) as the product of factors that are irreducible over the rationals, (b) as the product of linear and quadratic factors that are irreducible over the reals, and (c) in completely factored form. $$f(x)=x^{4}+6 x^{2}-27$$

Use synthetic division to verify the upper and lower bounds of the real zeros of \(f\) \(f(x)=x^{3}-4 x^{2}+1\) (a) Upper: \(x=4\) (b) Lower: \(x=-1\)

Describe the error. $$\sqrt{-6} \sqrt{-6}=\sqrt{(-6)(-6)}=\sqrt{36}=6$$

Decide whether the statement is true or false. Justify your answer. If \(x=-i\) is a zero of the function \(f(x)=x^{3}+i x^{2}+i x-1\) then \(x=i\) must also be a zero of \(f\)

Use the given zero to find all the zeros of the function. Function \(f(x)=x^{3}-x^{2}+4 x-4\) Zero \(2 i\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Statistics

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.