/*! 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 81 For which values of \(a\) is \(x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 81

For which values of \(a\) is \(x=1\) a solution of the following inequality? $$\frac{2 a+x}{x-2 a}<1$$

Short Answer

Expert verified
For \(x=1\) to be a solution, \(a < 0\).

Step by step solution

01

Substitute x=1 into the inequality

To determine if \(x=1\) is a solution, substitute \(x=1\) into the inequality. The original inequality is \[ \frac{2a + x}{x-2a} < 1 \]. Substituting \(x=1\) gives us \[ \frac{2a + 1}{1 - 2a} < 1 \].
02

Simplifying the inequality

Start by simplifying the inequality obtained from the substitution: \[ \frac{2a + 1}{1 - 2a} < 1 \]. Multiply both sides by \((1-2a)\) to eliminate the fraction (being mindful of the possible sign change): \[ 2a + 1 < 1 - 2a \]. This is valid as long as \(1 - 2a eq 0\), which means \(a eq \frac{1}{2}\).
03

Solve the inequality

Rearrange the inequality \[ 2a + 1 < 1 - 2a \] by adding \(2a\) to both sides and subtracting 1 from both sides: \[ 4a < 0 \]. Dividing both sides by 4 gives: \[ a < 0 \].
04

Checking exception

Recall that for the inequality you multiplied by \((1-2a)\), it was established that \(a eq \frac{1}{2}\) to ensure it was not zero and thus valid. Therefore, the solution remains valid in this range without needing to reassess the case when \(a = \frac{1}{2}\), confirming \(a < 0\) 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 91影视!

Key Concepts

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

Algebraic Manipulation
Algebraic manipulation is a fundamental skill in solving mathematical problems and involves rearranging and simplifying expressions to uncover solutions and insights. In this exercise, algebraic manipulation allowed us to handle the original inequality by dealing with fractions and rearranging formulae.

Key steps involved:
  • Simplifying expressions: When an expression contains fractions, it is often useful to eliminate these by multiplying through by a common denominator.
  • Maintaining equation balance: Anytime you add, subtract, multiply, or divide both sides of an equation or inequality, you must perform every action in a balanced way.
  • Substitution: By substituting a specific value into an expression, you can transform the original expression into a more workable form.
Understanding algebraic manipulation enables solutions that might not be obvious at first glance. With practice, it becomes an intuitive part of solving various algebra problems.
Substitution Method
The substitution method is a powerful tool in solving equations and inequalities. It involves replacing a variable with a given value to simplify the problem.

In this particular exercise:
  • We substituted the given value of 1 for the variable \( x \) in the inequality.
  • This substitution helped transform the inequality into a form that was easier to manipulate and solve, showcasing how the inequality behaves when \( x = 1 \).
  • The resulting expression was then simplified to determine for which values of \( a \) the condition holds true.
Using substitution is especially useful in verifying potential solutions and simplifying the process of solving. In more complex problems, it can break down a challenge into manageable steps by focusing on specific variable values one at a time.
Solving Inequalities
Solving inequalities requires a nuanced approach compared to solving equations, as it deals with ranges or sets of values rather than a single solution. The key characteristics involve:
  • Direction of Inequality: Remember that multiplying or dividing by a negative number reverses the inequality sign. This is critical when simplifying inequalities.
  • Multiple Solutions: Unlike equations, inequalities often have a range of possible solutions, which provides a broader understanding of variable behavior under the given conditions.
  • Graphical Interpretation: Solutions for inequalities can often be interpreted on a number line, giving a visual representation of solution sets.
In the exercise provided, solving the inequality required attention to these elements. As we manipulated the inequality, we confirmed the validity of our steps by understanding how our operations affected the inequality's range and its conditions. Ultimately, this guided us to determine that \( a < 0 \) while ensuring no invalid operations occurred, such as division by zero.
Rational Expressions
Rational expressions involve fractions where the numerator and denominator are polynomials. Working with these requires careful attention to their algebraic properties. In this problem, the key tasks were:
  • Identifying the rational component: The expression \( \frac{2a + x}{x-2a} \) is rational because both the numerator and the denominator are linear expressions.
  • Managing Denominators: Be cautious of division by zero. In our exercise, it was crucial to ensure that \( 1 - 2a eq 0 \), otherwise the expression would be undefined.
  • Simplifying Rational Expressions: Often involves clearing the fractions by multiplying through by the common denominator, enabling further simplification and solution extraction.
Understanding rational expressions aids in breaking down complex algebraic relationships. The practice of simplifying these expressions ensures you are working within defined mathematical rules and conventions, crucial for valid and efficient problem solving.

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

Solve for the indicated letter. $$-16 t^{2}+v_{0} t=0 ; \text { for } t$$

Find all real solutions of each equation. For Exercises \(31-36,\) give two forms for each answer: an exact answer (involving a radical) and a calculator approximation rounded to two decimal places. $$4 x^{-4}-33 x^{-2}-27=0$$

Find the value(s) of k such that the equation has exactly one real root. $$k x^{2}+k x+1=0$$

(a) Use a graph to estimate the solution set for each inequality. Zoom in far enough so that you can estimate the relevant endpoints to the nearest thousandth. (b) Exercises \(61-70\) can be solved algebraically using the techniques presented in this section. Carry out the algebra to obtain exact expressions for the endpoints that you estimated in part (a). Then use a calculator to check that your results are consistent with the previous estimates. $$x^{4}-2 x^{2}-1>0$$

Let \(a, b,\) and \(c\) be nonnegative numbers. Follow steps (a) through (e) to show that $$\sqrt[3]{a b c} \leq \frac{a+b+c}{3}$$ with equality holding if and only if \(a=b=c\). This result is known as the arithmetic-geometric mean inequality for three numbers. (Applications are developed in the projects at the ends of Sections 4.6 and \(4.7 .\) ) (a) By multiplying out the right-hand side, show that the following equation holds for all real numbers \(A, B\) and \(C\) $$\begin{aligned} 3 A B C=& A^{3}+B^{3}+C^{3}-\frac{1}{2}(A+B+C) \\ & \times\left[(A-B)^{2}+(B-C)^{2}+(C-A)^{2}\right] \quad \quad (1) \end{aligned}$$ (b) Now assume for the remainder of this exercise that \(A\) \(B,\) and \(C\) are nonnegative numbers. Use equation (1) to explain why $$3 A B C \leq A^{3}+B^{3}+C^{3}\quad \quad (2)$$ (c) Make the following substitutions in inequality ( 2 ): \(A^{3}=a, B^{3}=b,\) and \(C^{3}=c .\) Show that the result can be written $$\sqrt[3]{a b c} \leq \frac{a+b+c}{3} \quad \quad (3)$$ (d) Assuming that \(a=b=c,\) show that inequality (3) becomes an equality. (e) Assuming \(\sqrt[3]{a b c}=\frac{a+b+c}{3},\) show that \(a=b=c\) Hint: In terms of \(A, B,\) and \(C,\) the assumption becomes \(A B C=\frac{A^{3}+B^{3}+C^{3}}{3} .\) Use this to substitute for \(A B C\) on the left-hand side of equation (1). Then use the resulting equation to deduce that \(A=B=C,\) and consequently \(a=b=c\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

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.